Free math worksheets

- Work backward from the rectangle area formula. - What number multiplied by \(12\) equals \(96\)?

1. Divide the area by the length: \(96\div12=8\,\text{m}\).

The room is \(8\,\text{m}\) wide.

- What formula gives the area of a parallelogram? - What does it mean for two figures to have the same area? - How can you find a missing factor when you know the product?

1. Find the area of the parallelogram: \(A=bh=12\times5=60\,\text{cm}^2\). 2. The rectangle also has area \(60\,\text{cm}^2\). 3. Let \(w\) be the missing side length. Then \(10w=60\), so \(w=60\div10=6\,\text{cm}\).

The other side of the rectangle is \(6\,\text{cm}\) long.

- Use the area formula for a parallelogram. - Use the area formula for a rectangle. - Equal-area figures have the same numerical area even when their shapes differ.

1. The parallelogram's area is \(80\times35=2800\,\text{m}^2\). 2. The rectangle has the same area, so its missing side length is \(2800\div70=40\,\text{m}\).

a) The field has area \(2800\,\text{m}^2\). b) The rectangle's other side is \(40\,\text{m}\) long.

- Which measurements are needed for area, and which are needed for perimeter? - Be careful to distinguish a side length from its corresponding height. - Compare the two pairs of results.

1. For parallelogram A, \(A_A=8\times5=40\,\text{cm}^2\), and \(P_A=2(8+6)=28\,\text{cm}\). 2. For parallelogram B, \(A_B=10\times4=40\,\text{cm}^2\), and \(P_B=2(10+5)=30\,\text{cm}\). 3. The parallelograms have equal areas but different perimeters.

a) Parallelogram A: \(A=40\,\text{cm}^2\), \(P=28\,\text{cm}\) Parallelogram B: \(A=40\,\text{cm}^2\), \(P=30\,\text{cm}\) b) The areas are equal, but the perimeters are different.

- Which formula gives the area of a trapezoid? - Rearrange the formula so the height is alone. - What is the average of the two parallel side lengths?

1. Use \(A=\frac{1}{2}(b_1+b_2)h\). 2. Substitute: \(48=\frac{1}{2}(10+6)h=8h\). 3. Divide by \(8\): \(h=48\div8=6\,\text{cm}\).

The height is \(6\,\text{cm}\).

- Express all lengths in the same unit before calculating. - Use the area formula for a trapezoid. - Identify the parallel sides and the perpendicular distance between them.

1. Convert the height: \(80\,\text{cm}=0.80\,\text{m}\). 2. Use the trapezoid area formula: \(A=\frac{2.40+1.60}{2}\times0.80\). 3. The average of the parallel side lengths is \(2.00\,\text{m}\), so \(A=2.00\times0.80=1.60\,\text{m}^2\).

The side panel has area \(1.60\,\text{m}^2\).

- Compare the area formulas for rectangles and parallelograms. - What happens to a product when one factor is multiplied by a number? - What happens when both factors are changed?

1. Both figures have area \(bh\), so their areas are equal. 2. With base \(3b\), the new parallelogram has area \((3b)h=3bh\). Its area is \(3\) times the rectangle's area. 3. With base \(2b\) and height \(4h\), the area is \((2b)(4h)=8bh\). The area increases by a factor of \(8\).

a) The areas are equal. b) \(3\) times as great c) A factor of \(8\)

- Which measurements in the area formula are known? - Rearrange the formula to isolate the height. - Check that the units are consistent.

1. Use \(A=\frac{1}{2}(b_1+b_2)h\). 2. Substitute: \(0.50=\frac{1}{2}(1.40+0.60)h\). 3. The average of the parallel side lengths is \(\frac{2.00}{2}=1.00\,\text{m}\). 4. Therefore, \(0.50=1.00h\), so \(h=0.50\,\text{m}\).

The height is \(0.50\,\text{m}\), or \(50\,\text{cm}\).

- Track how changing a square's side changes both factors in its area. - Compare the area structures of a rectangle and parallelogram with the same base and perpendicular height. - Calculate the two areas in part c) only after identifying the relevant measurements.

1. Statement a) is false. If the original area is \(s^2\), the new area is \((2s)^2=4s^2\), so the area becomes four times as great. 2. Statement b) is true. A rectangle and a parallelogram both have area \(bh\) when they have the same base \(b\) and perpendicular height \(h\). 3. Statement c) is true. The trapezoid's area is \(\frac{1}{2}(5+3)\times4=16\,\text{cm}^2\), and the parallelogram's area is \(4\times4=16\,\text{cm}^2\).

a) False; the area becomes four times as great. b) True; both areas equal \(bh\). c) True; both areas are \(16\,\text{cm}^2\).

- Can the quadrilateral change shape without changing its side lengths? - Which area formulas apply to figures with four equal sides? - What happens to the height when the figure is pushed sideways?

1. A quadrilateral with four congruent sides is a rhombus. A square is a special type of rhombus. 2. A square with side length \(5\,\text{cm}\) has area \(25\,\text{cm}^2\). 3. A non-square rhombus has area \(A=bh\), and its height can vary while all four side lengths stay \(5\,\text{cm}\). 4. Therefore, the area is not uniquely determined without another measurement such as a height or an angle.

a) A rhombus; a square is a special case. b) No. Rhombi with the same side length can have different heights and therefore different areas.

- Write the area formula for each figure. - What difference do you see between the formulas? - How many times does the triangle's area fit into the parallelogram's area?

1. The parallelogram's area is \(A_P=8\times4.5=36\,\text{cm}^2\). 2. The triangle's area is \(A_T=\frac{1}{2}\times8\times4.5=18\,\text{cm}^2\). 3. The triangle's area is one-half the parallelogram's area, so the parallelogram's area is twice the triangle's area.

The parallelogram has area \(36\,\text{cm}^2\), and the triangle has area \(18\,\text{cm}^2\). The triangle's area is one-half the parallelogram's area.

- Use the area formula for a rectangle. - Find both areas before comparing them. - Line up decimal places when finding the difference.

1. Find the area of Rug A: \(5.0\times3.6=18.0\), so its area is \(18.0\,\text{ft}^2\). 2. Find the area of Rug B: \(6.4\times2.8=17.92\), so its area is \(17.92\,\text{ft}^2\). 3. Since \(18.0>17.92\), Rug A has the greater area. 4. The difference is \(18.00-17.92=0.08\,\text{ft}^2\).

Rug A has the greater area. The difference is \(0.08\,\text{ft}^2\).

- How do you find the area of a parallelogram? - Which quantity stays the same in part b)? - Rearrange the formula to find the missing height.

1. The first area is \(A=8\times4.5=36\,\text{cm}^2\). 2. For the second parallelogram, \(36=12h\), so \(h=36\div12=3\,\text{cm}\).

a) \(36\,\text{cm}^2\) b) \(3\,\text{cm}\)

- Express all measurements in the same unit. - Use the trapezoid area formula. - What does \(\frac{b_1+b_2}{2}\) represent?

1. For part a), \(A=\frac{1}{2}(4.5+7.5)\times4=24\,\text{cm}^2\). 2. For part b), convert \(60\,\text{cm}=0.6\,\text{m}\). Then \(A=\frac{1}{2}(1.2+0.8)\times0.6=0.6\,\text{m}^2\).

a) \(24\,\text{cm}^2\) b) \(0.6\,\text{m}^2\)

- Which trapezoid measurements appear in the area formula? - Distinguish the slanted side lengths from the perpendicular height. - Use the trapezoid area formula.

1. The area formula uses the two parallel side lengths and the perpendicular height. The slanted side lengths are not needed. 2. The area is \(A=\frac{1}{2}(9+5)\times3.5=7\times3.5=24.5\,\text{cm}^2\).

a) The parallel side lengths \(9\,\text{cm}\) and \(5\,\text{cm}\), and the height \(3.5\,\text{cm}\) b) \(24.5\,\text{cm}^2\)

- Start with the trapezoid area formula. - Rearrange the formula to isolate the sum of the parallel side lengths. - What must be true of \(b_1\) and \(b_2\) for the trapezoid not to be a parallelogram?

1. Use the trapezoid area formula: \(24=\frac{1}{2}(b_1+b_2)\times4\). 2. Simplify: \(24=2(b_1+b_2)\), so \(b_1+b_2=12\,\text{cm}\). 3. For the trapezoid not to be a parallelogram, choose unequal positive parallel side lengths whose sum is \(12\,\text{cm}\). Two examples are \(7\,\text{cm}\) and \(5\,\text{cm}\), or \(8\,\text{cm}\) and \(4\,\text{cm}\).

a) \(b_1+b_2=12\,\text{cm}\) b) Sample answers: \(b_1=7\,\text{cm}\), \(b_2=5\,\text{cm}\); or \(b_1=8\,\text{cm}\), \(b_2=4\,\text{cm}\).

- For the rectangle, use length times width. - For the triangle, use one-half times the base times the height. - Split the step-shaped figure into smaller rectangles. - Compare the three numerical areas after you calculate them.

1. Figure A is a rectangle with width \(3\) units and height \(2\) units, so its area is \(3 \times 2 = 6\) square units. 2. Figure B is a right triangle with base \(4\) units and height \(3\) units. Its area is \(\frac{1}{2} \times 4 \times 3 = 6\) square units. 3. Figure C can be split into three vertical rectangles with areas \(3\), \(2\), and \(1\) square units. Its total area is \(3 + 2 + 1 = 6\) square units. 4. All three figures have the same area: \(6\) square units.

a) Figure A: \(6\) square units Figure B: \(6\) square units Figure C: \(6\) square units b) All three figures have the same area even though they have different shapes.

- Find the area of each figure separately. - Use one-half times the base times the height for the triangle. - Split the L-shaped figure into rectangles. - Compare the three results.

1. Figure A is a square with side length \(2\) units, so its area is \(2 \times 2 = 4\) square units. 2. Figure B is a right triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units. 3. Figure C can be split into two nonoverlapping rectangles with areas \(2\) square units and \(2\) square units, so its total area is \(2 + 2 = 4\) square units. 4. Therefore, all three figures have the same area.

All three figures have the same area: \(4\) square units each.

- Use length times width for the rectangle. - For the triangle, imagine a rectangle with the same base and height. - What fraction of that rectangle is the triangle?

1. Figure A is a \(4 \times 2\) rectangle, so its area is \(4 \times 2 = 8\) square units. 2. Figure B is a triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units.

Figure A has an area of \(8\) square units, and Figure B has an area of \(4\) square units.

- Split each figure into simpler shapes. - For a slanted part, look for a right triangle. - Use one-half times the base times the height for the triangle, then compare the two total areas.

1. Split Figure A into a \(3 \times 1\) rectangle and a \(1 \times 2\) rectangle above its left end. Its area is \(3 + 2 = 5\,\text{cm}^2\). 2. Split Figure B into a \(2 \times 2\) square and a right triangle with base \(1\,\text{cm}\) and height \(2\,\text{cm}\). Its area is \(4 + \frac{1}{2} \times 1 \times 2 = 5\,\text{cm}^2\). 3. The areas are equal, so statement C is true.

C) The two figures have the same area: \(5\,\text{cm}^2\) each.

- Split the figure into a rectangle and a triangle. - Find the area of each part separately. - Add the two areas.

1. Split the figure into a lower rectangle and an upper triangle. 2. The rectangle is \(4\) units wide and \(2\) units high, so its area is \(4 \times 2 = 8\) square units. 3. The triangle has base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units. 4. The total area is \(8 + 4 = 12\) square units.

The area of the figure is \(12\) square units.

- For Figure A, imagine moving a triangular piece from one side to the other to form a rectangle. - Split Figure B into two rectangles. - Compare the two total areas.

1. Move the triangular piece on the left side of parallelogram A to the right side. This forms a \(4 \times 2\) rectangle, so Figure A has area \(4 \times 2 = 8\) square units. 2. Split Figure B into a \(4 \times 1\) rectangle and a \(2 \times 2\) square. Its area is \(4 + 4 = 8\) square units. 3. The two figures have equal areas.

Figures A and B each have an area of \(8\) square units, so their areas are equal.

- Split the trapezoid into a rectangle and triangles. - Split the step-shaped figure into rectangles. - Compare the two total areas.

1. Split trapezoid A into a \(2 \times 2\) rectangle and two right triangles, each with base \(1\) unit and height \(2\) units. Its area is \(4 + 1 + 1 = 6\) square units. 2. Split Figure B into a \(4 \times 1\) rectangle and a \(1 \times 2\) rectangle. Its area is \(4 + 2 = 6\) square units. 3. The figures have the same area even though their shapes are different.

a) Figure A has an area of \(6\) square units. b) Figure B has an area of \(6\) square units. c) The two figures have the same area.

- Use length times width for the rectangle. - Imagine completing the triangle to make a rectangle. - What fraction of that rectangle is the triangle?

1. Rectangle a) has side lengths \(4\) units and \(1\) unit, so its area is \(4 \times 1 = 4\) square units. 2. Triangle b) has base \(2\) units and height \(4\) units, so its area is \(\frac{1}{2} \times 2 \times 4 = 4\) square units. 3. The two figures have equal areas.

The two figures have the same area: \(4\) square units each.

- Find a base and its corresponding height for each figure. - Which area formulas apply to parallelograms and triangles? - Remember the factor of \(\frac{1}{2}\) in the triangle area formula.

1. Figure a) is a parallelogram with base \(3\) units and height \(2\) units, so its area is \(3\times2=6\) square units. 2. Figure b) is a triangle with base \(4\) units and height \(3\) units, so its area is \(\frac{1}{2}\times4\times3=6\) square units. 3. Both figures have an area of \(6\) square units, so their areas are equal.

Yes. Both figures have an area of \(6\) square units.

- Find the area of each figure separately. - For each triangle, use one-half times the base times the height. - Compare all four results.

1. Figure a) is a \(2 \times 2\) square, so its area is \(4\) square units. 2. Figure b) is a \(4 \times 1\) rectangle, so its area is \(4\) square units. 3. Figure c) is a triangle with base \(3\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 3 \times 2 = 3\) square units. 4. Figure d) is a triangle with base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 4 \times 2 = 4\) square units. 5. Figures a), b), and d) have the same area.

Figures a), b), and d) have the same area: \(4\) square units each.

- Recall the area formula for a square. - For the slanted parallelogram, identify a base and the perpendicular height. - Compare the number of grid squares covered by each figure.

1. Figure 1 is a square with side length \(2\,\text{cm}\), so its area is \(2\times2=4\,\text{cm}^2\). 2. Figure 2 is a parallelogram with base \(2\,\text{cm}\) and height \(2\,\text{cm}\), so its area is \(2\times2=4\,\text{cm}^2\). 3. The figures have equal areas.

Both figures have an area of \(4\,\text{cm}^2\).

- Split the figure into a rectangle and triangles. - Count any whole square units directly. - Combine equal triangular parts when helpful.

1. The figure is a trapezoid with bases \(6\) and \(2\) and height \(2\). 2. Split it into a \(2 \times 2\) rectangle and two right triangles, each with base \(2\) and height \(2\). 3. The area is \(4+\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 2 \times 2=4+2+2=8\) square units.

\(8\) square units

- Which sides of the figure are parallel? - What is the perpendicular distance between the parallel sides? - Use the trapezoid area formula.

1. The parallel sides have lengths \(5\) units and \(3\) units, and the height is \(4\) units. 2. The area is \(A=\frac{1}{2}(5+3)\times4=16\) square units.

The area is \(16\) square units.

- Split the figure into a rectangle and a triangle. - Find the area of each part. - Add the two areas.

1. Split the figure into a \(2 \times 2\) rectangle and a right triangle. 2. The rectangle has area \(4\) square units. 3. The triangle has base \(2\) units and height \(2\) units, so its area is \(\frac{1}{2} \times 2 \times 2 = 2\) square units. 4. The total area is \(4 + 2 = 6\) square units.

The blue figure covers \(6\) square units.

- Find each area separately. - For Figure b), imagine moving a triangular piece from one side to the other to form a rectangle. - Compare the two results.

1. Figure a) is a \(4 \times 2\) rectangle, so its area is \(4 \times 2 = 8\) square units. 2. Rearranging a triangular piece of parallelogram b) forms a \(3 \times 3\) rectangle, so its area is \(3 \times 3 = 9\) square units. 3. Since \(9 > 8\), Figure b) has the greater area.

Figure b) has the greater area: \(9\) square units compared with \(8\) square units for Figure a).

- Split the figure into a rectangle and two triangles. - The two triangular pieces can be combined to form a rectangle. - Add the areas of the parts.

1. Split the figure into a central \(2 \times 2\) rectangle and two right triangles. 2. The rectangle has area \(4\) square units. 3. Each triangle has base \(1\) unit and height \(2\) units, so each has area \(1\) square unit. Together, the triangles have area \(2\) square units. 4. The total area is \(4 + 2 = 6\) square units.

The figure has an area of \(6\) square units.

- Identify the triangular piece created by the slanted side. - Imagine moving that piece to fill the gap on the other side. - Read the dimensions of the resulting rectangle from the grid.

1. Cut off the triangular end piece on one side of the parallelogram and move it to the gap on the opposite side. 2. The rearranged figure is a rectangle with width \(4\) units and height \(3\) units. 3. Rearranging pieces does not change area, so the parallelogram has area \(4\times3=12\) square units.

The end triangle can be moved to the opposite side to make a \(4\times3\) rectangle, so the parallelogram's area is \(12\) square units.

- Focus on the triangular overhang and matching gap in the parallelogram. - What rectangle appears after that piece is moved? - Find the area only after establishing that the rearrangement preserves it.

1. In parallelogram b), cut off the triangular piece on the left of the top edge and move it to the matching gap on the right. 2. The rearranged figure is the same \(4\)-unit by \(2\)-unit rectangle shown in a). 3. Rearranging pieces preserves area, so the two figures have equal area. 4. Their common area is \(4\times2=8\) square units.

Moving the left triangular piece of b) to the right makes rectangle a), so the figures have the same area: \(8\) square units each.

- Split each figure into rectangles and triangles. - Use one-half times the base times the height for triangular parts. - Compare the two total areas.

1. Split trapezoid A into a \(2 \times 2\) rectangle and two right triangles, each with base \(1\) unit and height \(2\) units. Its area is \(4 + 1 + 1 = 6\) square units. 2. Split Figure B into a \(4 \times 1\) rectangle and a triangle with base \(4\) units and height \(1\) unit. Its area is \(4 + \frac{1}{2} \times 4 \times 1 = 6\) square units. 3. The figures have equal areas.

Figures A and B each have an area of \(6\) square units.

- Choose a side that lies along a grid line as the base. - Find the perpendicular distance to the opposite side. - You can also imagine moving a triangular piece to turn the parallelogram into a rectangle.

1. The horizontal base is \(5\) units long. 2. The perpendicular distance between the horizontal sides is \(3\) units, so the height is \(3\) units. 3. The area is \(A=5\times3=15\) square units.

The area of the parallelogram is \(15\) square units.

- Which given lengths appear in the parallelogram area formula? - Distinguish the slanted side from the perpendicular height. - Use the base and corresponding height.

1. The area of a parallelogram uses the base and its corresponding perpendicular height. The slanted side length is not needed. 2. The area is \(A=25\times10=250\,\text{cm}^2\).

The area of the parallelogram is \(250\,\text{cm}^2\).

- Find the dollar amount of the increase first. - Percent increase compares the increase with the original price, not with the final price. - Convert the resulting decimal rate to a percent.

1. Find the increase: \(\$881.28 - \$816.00 = \$65.28\). 2. Compare the increase with the original pretax price: \(65.28 \div 816 = 0.08\). 3. Convert to a percent: \(0.08 = 8\%\).

The tax increased the price by \(8\%\).

- How are a parallelogram's base, corresponding height, and area related? - Which measurement should the area be divided by to recover the base? - Check that your base times the given height reproduces the stated area.

1. The base and corresponding height multiply to the area, so \(48=6b\). 2. Divide the area by the height: \(b=48\div6=8\,\text{cm}\).

The base is \(8\,\text{cm}\) long.

- Find the average of the two parallel side lengths. - How does that average combine with the height to give trapezoid area? - Work backward from the area once the average side length is known.

1. The average of the parallel side lengths is \((10+8)\div2=9\,\text{cm}\). 2. Since area equals average parallel-side length times height, the height is \(45\div9=5\,\text{cm}\).

The height is \(5\,\text{cm}\).

- Choose a side whose length is easy to read from the grid. - Find the perpendicular distance between that side and the opposite side. - Use the parallelogram area formula.

1. The horizontal side \(AB\) is \(5\) units long. 2. The perpendicular distance between \(AB\) and \(CD\) is \(3\) units. 3. The area is \(A=5\times3=15\) square units.

The area of the parallelogram is \(15\) square units.

- Recall the trapezoid area formula. - Identify the two parallel sides and the height from the given measurements.

1. Use the trapezoid area formula: \(A=\frac{1}{2}(b_1+b_2)h\). 2. Substitute the measurements: \(A=\frac{1}{2}(8+4)\times5=30\,\text{cm}^2\).

The area is \(30\,\text{cm}^2\).

- What does the factor \(\frac{b_1+b_2}{2}\) represent geometrically? - Compare the student's expression with the structure of the trapezoid area formula. - Check whether the student's result is reasonable compared with rectangles of widths \(8\,\text{cm}\) and \(14\,\text{cm}\) and the same height.

1. The student used the sum of the parallel sides as though it were the trapezoid's average width. 2. The average of the parallel side lengths is \(\frac{8+14}{2}=11\,\text{cm}\). 3. The correct area is \(11\times6=66\,\text{cm}^2\).

The student forgot to divide the sum of the parallel sides by \(2\). The correct area is \(66\,\text{cm}^2\).

- Express all lengths in the same unit. - Find the floor area and the area of one tile. - Divide to find the number of tiles, then account for the package size.

1. The floor area is \(18\times 12=216\,\text{ft}^2\). 2. Each tile has side length \(18\,\text{in.}=1.5\,\text{ft}\), so its area is \(1.5\times 1.5=2.25\,\text{ft}^2\). 3. The number of tiles needed is \(216\div 2.25=96\). 4. Since \(96\div 12=8\), eight packages are needed.

The minimum number is \(8\) packages.

- Find the area of one square panel. - Convert centimeters to meters before finding the area. - A more compact rectangle has a smaller perimeter than a long, narrow rectangle with the same area. - List the factor pairs of \(36\).

1. Convert the panel side length: \(40\,\text{cm}=0.4\,\text{m}\). 2. One panel has area \(0.4\times0.4=0.16\,\text{m}^2\), so the total area is \(36\times0.16=5.76\,\text{m}^2\). 3. The possible rectangular arrays are \(1\times36\), \(2\times18\), \(3\times12\), \(4\times9\), and \(6\times6\). The most compact array, \(6\times6\), has the smallest perimeter. 4. Each side of the display is \(6\times0.4=2.4\,\text{m}\).

a) The total area is \(5.76\,\text{m}^2\). b) A \(6\times6\) arrangement gives the smallest perimeter. The side lengths are \(2.4\,\text{m}\) and \(2.4\,\text{m}\).

- Determine how many stones fit across the \(1.20\,\text{m}\) width. - Use the number of stones in each row to find how many rows the original walkway has. - For part b), find how many new rows fit in \(1.50\,\text{m}\).

1. Each stone is \(0.30\,\text{m}\) wide, so the number of stones across the walkway is \(1.20\div0.30=4\). 2. The \(80\) stones make \(80\div4=20\) rows along the walkway. Its length is \(20\times0.30=6\,\text{m}\). 3. Extending the walkway by \(1.50\,\text{m}\) requires \(1.50\div0.30=5\) more rows. 4. Each row uses \(4\) stones, so the extension requires \(5\times4=20\) additional stones.

a) The walkway is \(6\,\text{m}\) long. b) The extension requires \(20\) additional stones.

- Use the relationship among a rectangle's area, length, and width. - Divide the area by the known side length. - Consider what happens to a product when one factor stays fixed and the product doubles.

1. Divide the area by the length: \(7\frac{1}{2}\div5=\frac{15}{2}\times\frac{1}{5}=\frac{3}{2}=1.5\). The width is \(1.5\,\text{m}\). 2. Since \(A=l\times w\), doubling the area while keeping \(l\) constant doubles \(w\). The new width would be \(3\,\text{m}\).

a) The width is \(1.5\,\text{m}\), or \(1\frac{1}{2}\,\text{m}\). b) The width would double to \(3\,\text{m}\) because area is proportional to width when length is fixed.

- Divide the area by the known side length to find the missing side. - Use the perimeter formula for a rectangle. - The fence must go around all four sides.

1. Divide the area by the width: \(22.1\div6.5=3.4\). The length is \(3.4\,\text{m}\). 2. Find the perimeter: \(P=2\times(6.5+3.4)=2\times9.9=19.8\,\text{m}\).

a) The length is \(3.4\,\text{m}\). b) The garden bed requires \(19.8\,\text{m}\) of fencing.

- First find the side length of the square. - Then find the area shared by the square and rectangle. - Use that area and the rectangle's known side to find the missing side.

1. Find the side length of the square: \(18.4\div4=4.6\,\text{cm}\). 2. Find the square's area: \(4.6\times4.6=21.16\,\text{cm}^2\). 3. Divide that area by the known side of the rectangle: \(21.16\div4=5.29\,\text{cm}\).

The rectangle's other side is \(5.29\,\text{cm}\) long.

- Look for pairs of numbers whose product is \(30\). - In each pair, make sure at least one number is a decimal or fraction. - Compare the two perimeter calculations.

1. One possible pair is \(a=4\,\text{cm}\) and \(b=7.5\,\text{cm}\), since \(4\times7.5=30\). Another is \(a=2.5\,\text{cm}\) and \(b=12\,\text{cm}\), since \(2.5\times12=30\). 2. For the first pair, \(P=2\times(4+7.5)=23\,\text{cm}\). 3. For the second pair, \(P=2\times(2.5+12)=29\,\text{cm}\). 4. The rectangles have the same area but different perimeters.

a) One possible response is \(4\,\text{cm}\) by \(7.5\,\text{cm}\) and \(2.5\,\text{cm}\) by \(12\,\text{cm}\). b) The perimeters are \(23\,\text{cm}\) and \(29\,\text{cm}\). Rectangles with the same area can have different perimeters.

- First determine the sum of one length and one width. - Use the rectangle's perimeter to find the missing width. - Once you know both side lengths, find the area.

1. Half the perimeter is \(40\div2=20\,\text{m}\), so the length and width together total \(20\,\text{m}\). 2. The width is \(20-12=8\,\text{m}\). 3. The area is \(12\times8=96\,\text{m}^2\).

The chicken run is \(8\,\text{m}\) wide and has area \(96\,\text{m}^2\).

- Find the missing side length of each garden bed first. - For a square, all four side lengths are equal. - Then calculate and compare the two areas.

1. For garden bed A, half the perimeter is \(12\,\text{m}\), so its width is \(12-8=4\,\text{m}\). Its area is \(8\times4=32\,\text{m}^2\). 2. Since garden bed B is a square, each side is \(24\div4=6\,\text{m}\). Its area is \(6\times6=36\,\text{m}^2\). 3. Since \(36>32\), garden bed B has the greater area.

Garden bed B has the greater area: \(36\,\text{m}^2\), compared with \(32\,\text{m}^2\) for garden bed A.

- Compare the area formulas for a parallelogram and a triangle. - With the same base, what happens to a triangle's area when its height doubles? - How far must the third vertex be from line \(PQ\)?

1. The base length is \(6-1=5\) units, and the height is \(6-2=4\) units. The parallelogram has area \(A=5\times4=20\) square units. 2. For triangle \(PQT\), \(A=\frac{1}{2}bh\). Set its area equal to \(20\): \(\frac{1}{2}\times5\times h=20\). 3. Solving gives \(2.5h=20\), so \(h=8\) units. 4. Since base \(PQ\) lies on \(y=2\), point \(T\) must lie on a line parallel to \(PQ\) at a perpendicular distance of \(8\) units. Therefore, \(T\) may be any point on \(y=10\) or \(y=-6\).

a) \(A=20\) square units b) Point \(T\) can be any point on \(y=10\) or \(y=-6\). The triangle needs a height of \(8\) units because the factor \(\frac{1}{2}\) in the triangle area formula requires twice the parallelogram's height for the same base and area.

- In the trapezoid area formula, which expression represents the average width? - Since the heights are equal, which part of the trapezoid formula must match the rectangle's width? - What do you get when you add the two parallel side lengths and divide by \(2\)?

1. The trapezoid's area is \(A=\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(9+3)\times5=6\times5=30\,\text{cm}^2\). 2. If \(w\) is the rectangle's width, then \(30=w\times5\), so \(w=30\div5=6\,\text{cm}\). 3. Because the figures have the same height, the rectangle's width must equal the average of the trapezoid's parallel side lengths: \(w=\frac{9+3}{2}=6\,\text{cm}\).

a) The rectangle is \(6\,\text{cm}\) wide. b) Average the lengths of the parallel sides: \(w=\frac{b_1+b_2}{2}\).

- Remember that the logo has two congruent parts. - Determine how many times the area covered by one can fits into the total area. - When a division result is not a whole number of cans, you must round up.

1. One parallelogram has area \(1.20\times0.50=0.60\,\text{m}^2\). 2. The total area is \(2\times0.60=1.20\,\text{m}^2\). 3. The number of cans needed is \(1.20\div0.4=3\).

a) The logo has a total area of \(1.20\,\text{m}^2\). b) At least \(3\) cans of paint are needed.

- Divide the parallelogram's area by its base to find the height. - Find the area of the square. - Compare the two numerical areas.

1. For garden bed A, divide the area by the base: \(120\div15=8\,\text{m}\). Its height is \(8\,\text{m}\). 2. Garden bed B has area \(11\times11=121\,\text{m}^2\). 3. Since \(121>120\), garden bed B has the greater area.

a) The height of garden bed A is \(8\,\text{m}\). b) Garden bed B has the greater area: \(121\,\text{m}^2\), compared with \(120\,\text{m}^2\) for garden bed A.

- Divide the area by the base to find the height. - Keep the parallelogram's height separate from its slanted side length. - Find and compare both perimeters.

1. The height is \(1800\div45=40\,\text{m}\). 2. The parallelogram's perimeter is \(2\times(45+42)=174\,\text{m}\). 3. The rectangle measures \(45\,\text{m}\) by \(40\,\text{m}\), so its perimeter is \(2\times(45+40)=170\,\text{m}\). 4. Since \(170<174\), the rectangular practice field has the smaller perimeter.

a) The height is \(40\,\text{m}\). b) The parallelogram's perimeter is \(174\,\text{m}\). c) The rectangular practice field has the smaller perimeter, \(170\,\text{m}\).

- A rhombus is a type of parallelogram. Which parallelogram area formula can you use? - What two formulas can be used to find the area of a rhombus? - Once you know the area, how can you rearrange the diagonal formula to find a missing length?

1. Use the parallelogram area formula: \(A=bh=10\times9.6=96\,\text{cm}^2\). 2. A rhombus also has area \(A=\frac{1}{2}d_1d_2\). 3. Substitute the known values: \(96=\frac{1}{2}\times12\times d_2=6d_2\). 4. Divide by \(6\): \(d_2=96\div6=16\,\text{cm}\).

a) \(A=96\,\text{cm}^2\) b) \(d_2=16\,\text{cm}\)

- How are a base, its corresponding height, and area related? - Divide the area by each possible base length. - Compare each side length with its corresponding height.

1. The height corresponding to the \(8\,\text{cm}\) side is \(48\div8=6\,\text{cm}\). 2. The height corresponding to the \(6\,\text{cm}\) side is \(48\div6=8\,\text{cm}\). 3. With a fixed area, a longer base has a shorter corresponding height because the product of the base and height must remain constant.

a) The height corresponding to the \(8\,\text{cm}\) side is \(6\,\text{cm}\), and the height corresponding to the \(6\,\text{cm}\) side is \(8\,\text{cm}\). b) For a fixed area, the longer side has the shorter corresponding height.

- Picture how the shape changes when it is pushed sideways. - In the rectangle, which side is also the height to the \(12\,\text{cm}\) base? - Do any of the wooden sides become longer or shorter? - Which measurement in the area formula changes?

1. The rectangle's area is \(12\times5=60\,\text{cm}^2\). 2. For the parallelogram, use the \(12\,\text{cm}\) side as the base and \(4\,\text{cm}\) as the height: \(A=12\times4=48\,\text{cm}^2\). 3. The perimeter stays the same because none of the four side lengths changes. 4. Area depends on the base and perpendicular height, not only on the side lengths. The height decreases from \(5\,\text{cm}\) to \(4\,\text{cm}\), so the area decreases.

a) \(60\,\text{cm}^2\) b) \(48\,\text{cm}^2\) c) The perimeter stays the same because all four side lengths are unchanged. d) The area decreases because the perpendicular height decreases while the base stays the same.

- Use the area and the given height to find the corresponding base. - What is the perimeter formula for a parallelogram? - Once one side is known, use the perimeter to find the other side. - Use the information one step at a time.

1. Let \(x\) be the side corresponding to the \(3\,\text{cm}\) height. Since area equals base times height, \(24=3x\), so \(x=8\,\text{cm}\). 2. Let \(y\) be the other side length. Use the perimeter formula: \(22=2(x+y)=2(8+y)\). 3. Divide by \(2\): \(11=8+y\), so \(y=3\,\text{cm}\). 4. The side lengths are \(8\,\text{cm}\) and \(3\,\text{cm}\). These measurements describe a rectangle, which is also a parallelogram.

The side lengths are \(8\,\text{cm}\) and \(3\,\text{cm}\).

- A parallelogram has two base-height pairs, but both give the same area. - Which measurements are needed to find the area? - Once you know the area, how can you find a missing height? - Recall the perimeter formula for a parallelogram.

1. The area is \(A=bh=12\times6=72\,\text{cm}^2\). 2. The same area also equals the product of the \(8\,\text{cm}\) side and its corresponding height. Therefore, the missing height is \(72\div8=9\,\text{cm}\). 3. The perimeter is \(P=2(12+8)=40\,\text{cm}\).

a) \(A=72\,\text{cm}^2\) b) The height corresponding to the \(8\,\text{cm}\) side is \(9\,\text{cm}\). c) \(P=40\,\text{cm}\)

- First express all lengths in the same unit. - Use the perimeter and the known side to find the other side. - Which of the two sides is shorter? - Use the area and the shorter side to find its corresponding height.

1. Convert the perimeter to centimeters: \(1.4\,\text{m}=140\,\text{cm}\). 2. Let \(x\) be the other side length. From \(P=2(45+x)\), \(140=2(45+x)\). Thus, \(70=45+x\), so \(x=25\,\text{cm}\). 3. The shorter side is \(25\,\text{cm}\). 4. Let \(h\) be the height corresponding to the shorter side. Since \(A=bh\), \(900=25h\), so \(h=900\div25=36\,\text{cm}\). 5. The result is geometrically possible because \(36\,\text{cm}\le45\,\text{cm}\).

The height corresponding to the shorter side is \(36\,\text{cm}\).

- Write the area formula for each figure. - Substitute the known base and area into each formula. - How does the factor \(\frac{1}{2}\) affect the required height?

1. For the parallelogram, \(24=8h_P\), so \(h_P=24\div8=3\,\text{cm}\). 2. For the triangle, \(24=\frac{1}{2}\times8\times h_T=4h_T\), so \(h_T=24\div4=6\,\text{cm}\). 3. The triangle formula includes a factor of \(\frac{1}{2}\). With the same base and area, the triangle's height must therefore be twice the parallelogram's height.

The parallelogram's height is \(3\,\text{cm}\), and the triangle's height is \(6\,\text{cm}\). The triangle needs twice the height because its area formula includes the factor \(\frac{1}{2}\).

- Draw a segment parallel to one slanted side of the trapezoid. - Subtract the shorter base from the longer base to find the triangle's base. - The two component figures have the same height as the trapezoid.

1. The trapezoid's area is \(\frac{1}{2}(10+6)\times4=32\,\text{cm}^2\). 2. The parallelogram has area \(6\times4=24\,\text{cm}^2\). 3. The triangle's base is \(10-6=4\,\text{cm}\), and its height is \(4\,\text{cm}\). Its area is \(\frac{1}{2}\times4\times4=8\,\text{cm}^2\). 4. The component areas add to \(24+8=32\,\text{cm}^2\), which equals the trapezoid's area.

The parallelogram has area \(24\,\text{cm}^2\), and the triangle has area \(8\,\text{cm}^2\). Their sum is \(32\,\text{cm}^2\), the area of the trapezoid.

- First find the area of the triangular flower bed. - Compare the area formulas for a triangle and a parallelogram. - With equal bases and equal areas, how must the heights compare?

1. The triangular flower bed has area \(A=\frac{1}{2}\times12\times5=30\,\text{m}^2\). 2. For the parallelogram, \(30=12h\). 3. Therefore, \(h=30\div12=2.5\,\text{m}\).

The parallelogram-shaped flower bed has a height of \(2.5\,\text{m}\).

- Express all lengths in the same unit before calculating. - Which formula gives the area of a trapezoid? - The final answer must be in square meters.

1. Convert all lengths to meters: \(85\,\text{cm}=0.85\,\text{m}\) and \(60\,\text{cm}=0.60\,\text{m}\). 2. Use the trapezoid area formula: \(A=\frac{1}{2}(b_1+b_2)h\). 3. Substitute: \(A=\frac{1}{2}(0.85+1.15)\times0.60\). 4. Since \(0.85+1.15=2.00\), \(A=1.00\times0.60=0.60\,\text{m}^2\).

\(0.60\,\text{m}^2\)

- Substitute the known values into the area formula and solve for the height. - What happens to a product when one factor is doubled?

1. The average of the parallel side lengths is \(\frac{7+5}{2}=6\,\text{m}\). 2. Use \(A=\frac{1}{2}(b_1+b_2)h\): \(24=6h\), so \(h=4\,\text{m}\). 3. In the area formula, the height is a factor. If the parallel sides stay fixed and the height is multiplied by \(2\), the area is also multiplied by \(2\).

a) \(h=4\,\text{m}\) b) Yes. Doubling the height doubles the area when the parallel side lengths stay the same.

- Decide which measurements appear in the trapezoid area formula and which are extra information. - Distinguish the perpendicular height from the slanted side lengths. - For part c), calculate the new area and compare it with the original area.

1. For part a), the area formula uses the two parallel widths and the perpendicular height. The slanted side lengths are not needed. 2. The original area is \(A=\frac{1}{2}(3+9.5)\times4=25\,\text{m}^2\). 3. The new bottom width is \(10.5\,\text{m}\), so the new area is \(A_{\text{new}}=\frac{1}{2}(3+10.5)\times4=27\,\text{m}^2\). 4. The increase is \(27-25=2\,\text{m}^2\).

a) The top width, bottom width, and perpendicular height b) \(25\,\text{m}^2\) c) The area increases by \(2\,\text{m}^2\).

- In the area formula, what happens when the numerator is doubled? - Do you need the individual parallel side lengths in part b)? - How are the sum of the parallel side lengths and the area related when the height is fixed?

1. Substitute the known sum into \(A=\frac{1}{2}(b_1+b_2)h\): \(36=\frac{1}{2}(12)h=6h\). 2. Therefore, \(h=6\,\text{cm}\). 3. With the height fixed, area is directly proportional to the sum of the parallel side lengths. Doubling that sum doubles the area. 4. The second area is \(2\times36=72\,\text{cm}^2\).

a) \(h=6\,\text{cm}\) b) \(72\,\text{cm}^2\). The area doubles because the sum of the parallel side lengths doubles while the height stays fixed.

- Start with the general trapezoid area formula. - Identify which factors change and which remain fixed. - Substitute the scale factors into the formula and compare with the original area.

1. The area formula is \(A=\frac{1}{2}(b_1+b_2)h\). 2. In part a), replace \(b_1+b_2\) with \(2(b_1+b_2)\): \(A_{\text{new}}=\frac{1}{2}[2(b_1+b_2)]h=2A\). The area doubles. 3. In part b), replace \(b_1+b_2\) with \(\frac{1}{3}(b_1+b_2)\) and \(h\) with \(3h\): \(A_{\text{new}}=\frac{1}{2}\left[\frac{1}{3}(b_1+b_2)\right](3h)=A\). The area stays the same.

a) The area doubles. b) The area stays the same.

- Which part of the trapezoid area formula contains the parallel side lengths? - What happens to their sum when the same amount is added to one and subtracted from the other? - Test the claim using convenient numbers.

1. The trapezoid area formula is \(A=\frac{1}{2}(b_1+b_2)h\), so the area depends on the sum \(b_1+b_2\). 2. After the change, the sum is \((b_1+2)+(b_2-2)=b_1+b_2\). The sum of the parallel side lengths does not change. 3. If the height also stays the same, the area stays the same. 4. For example, let \(b_1=6\,\text{cm}\), \(b_2=4\,\text{cm}\), and \(h=5\,\text{cm}\). The original area is \(\frac{1}{2}(6+4)\times5=25\,\text{cm}^2\). After the change, the area is \(\frac{1}{2}(8+2)\times5=25\,\text{cm}^2\). 5. Lucas is not correct.

Lucas is not correct. When the height stays the same, the area stays the same because the sum of the parallel side lengths is unchanged.

- Think of trapezoid area as average parallel-side length times height. - Use the area and height to find that average length first. - How can the average of two lengths help you find their sum?

1. The average length of the two parallel sides is \(24.5\div7=3.5\,\text{cm}\). 2. Therefore, the sum of the parallel side lengths is \(2\times3.5=7\,\text{cm}\). 3. Subtract the known side: \(7-4.2=2.8\,\text{cm}\).

The other parallel side is \(2.8\,\text{cm}\) long.

- First express all lengths in the same unit. - Which formulas give the area and perimeter of a trapezoid? - Substitute the given measurements into the formulas.

1. Convert the height: \(40\,\text{mm}=4\,\text{cm}\). 2. The area is \(A=\frac{1}{2}(14+8)\times4=11\times4=44\,\text{cm}^2\). 3. The perimeter is \(P=14+5+8+5=32\,\text{cm}\).

\(A=44\,\text{cm}^2\) \(P=32\,\text{cm}\)

- Convert the height to meters before working with the area in square meters. - Think of trapezoid area as average parallel-edge length times height. - After finding the average of the two parallel edges, how can you recover their total length?

1. Convert the height: \(50\,\text{cm}=0.50\,\text{m}\). 2. The average length of the parallel edges is \(0.75\div0.50=1.50\,\text{m}\). 3. Their total length is \(2\times1.50=3.00\,\text{m}\). 4. The bottom edge is \(3.00-1.20=1.80\,\text{m}\).

The bottom edge must be \(1.80\,\text{m}\) long.

- First find the area that both flower beds must have. - Use that area in the trapezoid formula for the second flower bed. - Notice what happens to the division by \(2\) when the height is \(2\).

1. The first flower bed has area \(\frac{5+3}{2}\times2.5=4\times2.5=10\,\text{m}^2\). 2. Let \(x\) be the unknown parallel side of the second flower bed. Since its area is also \(10\,\text{m}^2\), \(10=\frac{x+4}{2}\times2\). 3. The factors of \(2\) cancel, so \(10=x+4\). 4. Therefore, \(x=6\,\text{m}\).

The other parallel side of the second flower bed is \(6\,\text{m}\) long.

- How does the sum of the parallel side lengths appear in the trapezoid area formula? - Calculate both areas before deciding. - Compare the new area with twice the original area.

1. The original area is \(A=\frac{1}{2}(3+5)\times4=16\,\text{cm}^2\). 2. After doubling the \(3\,\text{cm}\) side to \(6\,\text{cm}\), the area is \(A_1=\frac{1}{2}(6+5)\times4=22\,\text{cm}^2\). 3. Twice the original area would be \(32\,\text{cm}^2\), not \(22\,\text{cm}^2\). Leon is not correct because the formula uses the sum of both parallel side lengths.

1) \(16\,\text{cm}^2\) 2) \(22\,\text{cm}^2\) 3) Leon is not correct. The new area is not twice the original area.

- First find the area of the parallelogram. - For a trapezoid, area equals height times the average of the parallel side lengths. - Once you know that average, how can you find the total of the two parallel side lengths?

1. The parallelogram's area is \(12\times6=72\,\text{cm}^2\). 2. For the trapezoid to have area \(72\,\text{cm}^2\) with height \(6\,\text{cm}\), the average of its parallel side lengths must be \(72\div6=12\,\text{cm}\). 3. The two parallel sides must therefore total \(2\times12=24\,\text{cm}\). 4. The other parallel side is \(24-15=9\,\text{cm}\).

The other parallel side is \(9\,\text{cm}\) long.

- Which parts of the trapezoid area formula must stay constant? - What must \(\frac{b_1+b_2}{2}\) equal when the height is \(4\) and the area is \(24\)? - Try different positive numbers with the same sum.

1. The area is \(A=\frac{1}{2}(9+3)\times4=24\,\text{cm}^2\). 2. Since the height is fixed, the sum of the parallel side lengths must remain \(12\,\text{cm}\). Possible pairs are \(8\,\text{cm}\) and \(4\,\text{cm}\), or \(7\,\text{cm}\) and \(5\,\text{cm}\). 3. In \(A=\frac{1}{2}(b_1+b_2)h\), a fixed area and height require \(b_1+b_2\) to remain constant.

a) \(24\,\text{cm}^2\) b) Possible pairs are \(8\,\text{cm}\) and \(4\,\text{cm}\), or \(7\,\text{cm}\) and \(5\,\text{cm}\). c) The sum must remain \(12\,\text{cm}\).

- How does doubling \(b_1+b_2\) affect the area? - What additional effect comes from doubling the height? - In \(A=\frac{1}{2}(b_1+b_2)h\), what happens when two factors are doubled?

1. The original area is \(A=\frac{1}{2}(b_1+b_2)h\). 2. After the changes, the sum of the parallel side lengths is \(2b_1+2b_2=2(b_1+b_2)\), and the height is \(2h\). 3. The new area is \(A_{\text{new}}=\frac{1}{2}[2(b_1+b_2)](2h)=4\left[\frac{1}{2}(b_1+b_2)h\right]=4A\).

The new area is \(4\) times the original area.

- How can you express the longer side in terms of the shorter side? - Use the trapezoid area formula. - Substitute so the equation has only one unknown. - Solve the resulting equation.

1. Let the shorter parallel side be \(x\). Then the longer parallel side is \(3x\). 2. Use the area formula: \(36=\frac{1}{2}(x+3x)\times6\). 3. Simplify: \(36=\frac{1}{2}(4x)\times6=12x\). 4. Thus, \(x=36\div12=3\,\text{cm}\), and the longer side is \(3x=9\,\text{cm}\).

The parallel side lengths are \(3\,\text{cm}\) and \(9\,\text{cm}\).

- Identify which parts of the area formula are known. - In part a), isolate the height. - In part b), you only need the sum, not the individual side lengths. - What do you get when you divide the area by the height?

1. For the first trapezoid, \(40=\frac{1}{2}(7+13)h=10h\), so \(h=4\,\text{cm}\). 2. For the second trapezoid, \(40=\frac{1}{2}(b_1+b_2)\times5\). 3. Divide by \(5\): \(8=\frac{b_1+b_2}{2}\), so \(b_1+b_2=16\,\text{cm}\).

a) \(h=4\,\text{cm}\) b) \(b_1+b_2=16\,\text{cm}\)

- Work backward through the area formula to isolate the unknown side. - What happens to a product when one factor is doubled? - Use the side length found in part a) for part c). - What is the average of the two parallel side lengths?

1. Let the unknown parallel side be \(x\). Then \(25=\frac{1}{2}(4+x)\times5\). 2. Divide by \(5\) and multiply by \(2\): \(10=4+x\), so \(x=6\,\text{cm}\). 3. Doubling the height doubles the area because the height is a factor in the formula. The new area is \(50\,\text{cm}^2\). 4. For an area of \(60\,\text{cm}^2\), \(60=\frac{1}{2}(4+6)h=5h\), so \(h=12\,\text{cm}\).

a) \(6\,\text{cm}\) b) The area doubles to \(50\,\text{cm}^2\). c) \(h=12\,\text{cm}\)

- Sketch the rectangle and the two congruent triangles. - Subtract the two \(2\,\text{cm}\) segments from the longer base. - Use either the trapezoid formula or the areas of the three component figures.

1. The two triangles use \(2\,\text{cm}\) at each end of the longer base, so the shorter base is \(10-2\times2=6\,\text{cm}\). 2. Use the trapezoid area formula: \(A=\frac{1}{2}(10+6)\times4=32\,\text{cm}^2\).

a) The shorter base is \(6\,\text{cm}\). b) The area of the trapezoid is \(32\,\text{cm}^2\).

- Write the two area formulas side by side. - What factor appears in the triangle formula but not the rectangle formula? - How do the scale factors \(2\) and \(3\) combine? - You may test with simple values for \(b\) and \(h\).

1. The rectangle has area \(A_R=bh\), and the triangle has area \(A_T=\frac{1}{2}bh\). Therefore, the rectangle's area is twice the triangle's area. 2. The new triangle has area \(A_{\text{new}}=\frac{1}{2}(2b)(3h)=6\left(\frac{1}{2}bh\right)=6A_T\).

a) The rectangle's area is twice the triangle's area, a ratio of \(2\) to \(1\). b) The new triangle's area is \(6\) times the original area.

- Use the area formula to analyze changes in base and height. - Perimeter measures boundary length, while area measures the region inside. - Consider how a diagonal divides a parallelogram.

1. Statement a) is true. The new area is \(\frac{1}{2}(2b)\left(\frac{1}{2}h\right)=\frac{1}{2}bh\), which equals the original area. 2. Statement b) is false. A \(10\,\text{cm}\times2\,\text{cm}\) rectangle has area \(20\,\text{cm}^2\) and perimeter \(24\,\text{cm}\). A \(5\,\text{cm}\times4\,\text{cm}\) rectangle has the same area but perimeter \(18\,\text{cm}\). Rectangles are parallelograms. 3. Statement c) is true. The diagonal creates two triangles with equal base lengths and equal corresponding heights, so their areas are equal.

a) True b) False c) True

- First find the parallelogram's area. - How does the triangle area formula differ from the parallelogram area formula? - Set the two areas equal and solve for the missing height.

1. The parallelogram's area is \(A=12\times5=60\,\text{cm}^2\). 2. For the triangle, \(60=\frac{1}{2}\times15\times h=7.5h\). 3. Therefore, \(h=60\div7.5=8\,\text{cm}\).

The triangle's height is \(8\,\text{cm}\).

- Express all lengths in the same unit before finding the trapezoid's area. - The square must have the same area as the trapezoid. - Which whole-number side length multiplied by itself gives the required square area?

1. Convert \(1.3\,\text{m}\) to \(130\,\text{cm}\). 2. The trapezoid's area is \(A=\frac{1}{2}(130+50)\times40=3600\,\text{cm}^2\). 3. A square with side length \(60\,\text{cm}\) has area \(60\times60=3600\,\text{cm}^2\). 4. Therefore, the equal-area square has side length \(60\,\text{cm}\).

The square's side length is \(60\,\text{cm}\).

- Convert all lengths to the same unit before calculating. - Use the correct formula for each figure. - How do the triangle and parallelogram area formulas differ?

1. Convert the triangle's base: \(1.2\,\text{m}=120\,\text{cm}\). Its area is \(\frac{1}{2}\times120\times40=2400\,\text{cm}^2\). 2. Convert the parallelogram's base: \(0.5\,\text{m}=50\,\text{cm}\). Its area is \(50\times50=2500\,\text{cm}^2\). 3. The parallelogram has the greater area. The difference is \(2500-2400=100\,\text{cm}^2\).

The parallelogram has the greater area by \(100\,\text{cm}^2\).

- The same area results from either base-height pair. - What happens to a product when one factor is doubled?

1. The area is \(A=bh=10\times4.5=45\,\text{cm}^2\). 2. Let \(h\) be the height corresponding to the \(6\,\text{cm}\) side. Then \(45=6h\), so \(h=45\div6=7.5\,\text{cm}\). 3. If the base is doubled while its corresponding height stays the same, the product of base and height doubles. Therefore, the area doubles.

a) \(45\,\text{cm}^2\) b) The height corresponding to the \(6\,\text{cm}\) side is \(7.5\,\text{cm}\). c) The area doubles.

- Use the square's perimeter to find its side length. - Half of a parallelogram's perimeter is the sum of two adjacent side lengths. - For a fixed base, how does the perpendicular height compare with the adjacent side length?

1. For a), the square's side length is \(32\div4=8\,\text{m}\), so its area is \(8\times8=64\,\text{m}^2\). 2. For b), half of the parallelogram's perimeter is \(32\div2=16\,\text{m}\). The two adjacent side lengths total \(16\,\text{m}\), so the other side is \(16-10=6\,\text{m}\). 3. For c), using the \(10\,\text{m}\) side as the base, the corresponding height cannot exceed the adjacent \(6\,\text{m}\) side. Its area is therefore at most \(10\times6=60\,\text{m}^2\), which is less than \(64\,\text{m}^2\).

a) Side length \(8\,\text{m}\); area \(64\,\text{m}^2\) b) \(6\,\text{m}\) c) No. Its greatest possible area is \(60\,\text{m}^2\).

- Draw a diagonal and identify the base of each triangle. - Both triangles have the same perpendicular height as the trapezoid. - Compare their total with the trapezoid area formula.

1. Both triangles have height \(4\,\text{cm}\). The triangle whose base is \(7\,\text{cm}\) has area \(\frac{1}{2}\times7\times4=14\,\text{cm}^2\). 2. The triangle whose base is \(3\,\text{cm}\) has area \(\frac{1}{2}\times3\times4=6\,\text{cm}^2\). 3. Their total area is \(14+6=20\,\text{cm}^2\). 4. The trapezoid formula gives \(\frac{1}{2}(7+3)\times4=20\,\text{cm}^2\), so the two methods agree.

The triangle areas are \(14\,\text{cm}^2\) and \(6\,\text{cm}^2\). Their sum and the trapezoid's area are both \(20\,\text{cm}^2\).

- Divide the area by the known width to find each length. - Use the perimeter formula for a rectangle. - Perimeter represents the total length around the garden bed. - Compare the two perimeter values.

1. For the first garden bed, the length is \(14.4\div3=4.8\,\text{m}\). Its perimeter is \(2\times(3+4.8)=15.6\,\text{m}\). 2. For the second garden bed, the length is \(14.4\div1.2=12\,\text{m}\). Its perimeter is \(2\times(1.2+12)=26.4\,\text{m}\). 3. Since \(26.4>15.6\), the second garden bed requires more fencing.

a) The length is \(4.8\,\text{m}\), and the perimeter is \(15.6\,\text{m}\). b) The length is \(12\,\text{m}\), and the perimeter is \(26.4\,\text{m}\). c) The second garden bed requires more fencing.

- Find the whole-number factor pairs of \(12\). - Use the perimeter formula for each pair. - Compare how close the two side lengths are in each rectangle. - Look for a relationship between the shape and its perimeter.

1. The whole-number factor pairs of \(12\) are \(1\) and \(12\), \(2\) and \(6\), and \(3\) and \(4\). 2. Their perimeters are \(2\times(1+12)=26\,\text{m}\), \(2\times(2+6)=16\,\text{m}\), and \(2\times(3+4)=14\,\text{m}\). 3. The perimeter decreases as the side lengths become closer together. The \(3\,\text{m}\times4\,\text{m}\) rug has the smallest perimeter.

a) The side-length pairs are \(1\,\text{m}\) and \(12\,\text{m}\), \(2\,\text{m}\) and \(6\,\text{m}\), and \(3\,\text{m}\) and \(4\,\text{m}\). b) The perimeters are \(26\,\text{m}\), \(16\,\text{m}\), and \(14\,\text{m}\), respectively. c) The more square-shaped rectangle has the smaller perimeter.

- Which parts of the trapezoid area formula are equal for both figures? - What happens when the sums of the parallel sides are equal? - Can you compare the areas before calculating them?

1. For trapezoid 1, the sum of the parallel side lengths is \(4+8=12\,\text{cm}\). 2. For trapezoid 2, the sum is \(5+7=12\,\text{cm}\). 3. The trapezoids have equal heights and equal sums of parallel side lengths, so \(A=\frac{1}{2}(b_1+b_2)h\) gives equal areas. 4. Each area is \(\frac{1}{2}\times12\times5=30\,\text{cm}^2\).

The trapezoids have equal areas, \(30\,\text{cm}^2\), because their heights and the sums of their parallel side lengths are equal.

- Distinguish between the parallelogram and triangle area formulas. - A parallelogram's perimeter is found like a rectangle's perimeter. - How must the triangle's height account for the factor \(\frac{1}{2}\)?

1. Let \(h\) be the height corresponding to the \(8\,\text{cm}\) side. From \(40=8h\), \(h=5\,\text{cm}\). 2. The perimeter is \(P=2(8+6)=28\,\text{cm}\). 3. For the triangle, \(40=\frac{1}{2}\times8\times h=4h\), so \(h=10\,\text{cm}\).

a) \(5\,\text{cm}\) b) \(P=28\,\text{cm}\) c) \(h=10\,\text{cm}\)

- The cut ends halfway along a \(30\,\text{cm}\) side. - The triangle's height is the rectangle's shorter side. - Subtract the triangle's area from the rectangle's area to find the trapezoid's area.

1. The triangle's base is half of \(30\,\text{cm}\), or \(15\,\text{cm}\), and its height is \(12\,\text{cm}\). Its area is \(\frac{1}{2}\times15\times12=90\,\text{cm}^2\). 2. The full rectangle has area \(30\times12=360\,\text{cm}^2\). 3. The trapezoid is the remaining piece, so its area is \(360-90=270\,\text{cm}^2\).

The triangle has area \(90\,\text{cm}^2\), and the trapezoid has area \(270\,\text{cm}^2\).

- Express all side lengths in meters before comparing the areas. - Find the area of each rectangle. - Subtract the smaller area from the larger area.

1. Garden bed A has area \(4.5\times2=9\,\text{m}^2\). 2. Convert the dimensions of garden bed B: \(600\,\text{cm}=6\,\text{m}\) and \(160\,\text{cm}=1.6\,\text{m}\). 3. Garden bed B has area \(6\times1.6=9.6\,\text{m}^2\). 4. The difference is \(9.6-9=0.6\,\text{m}^2\), so garden bed B is larger.

Garden bed B has the greater area. The difference is \(0.6\,\text{m}^2\).

- First find the area shared by both trapezoids. - Compare the average widths of the two trapezoids. - How must height change when average width increases but area stays fixed?

1. Trapezoid 1 has area \(A=\frac{1}{2}(10+14)\times6=72\,\text{cm}^2\). 2. For trapezoid 2, \(72=\frac{1}{2}(8+22)h_2=15h_2\), so \(h_2=72\div15=4.8\,\text{cm}\). 3. The parallel side sums are \(24\,\text{cm}\) and \(30\,\text{cm}\). Since trapezoid 2 has a greater average width, its height must be smaller to keep the same area.

a) \(4.8\,\text{cm}\) b) The sums are \(24\,\text{cm}\) and \(30\,\text{cm}\). The greater sum for trapezoid 2 requires a smaller height for the same area.

- Choose the correct area formula for each figure. - Use the area from part a) in the other equations. - Rearrange each formula to isolate the requested measurement.

1. The parallelogram's area is \(10\times6=60\,\text{cm}^2\). 2. For the triangle, \(60=\frac{1}{2}\times12\times h=6h\), so \(h=10\,\text{cm}\). 3. For the trapezoid, \(60=\frac{1}{2}(b_1+b_2)\times5\). Thus, \(120=5(b_1+b_2)\), so \(b_1+b_2=24\,\text{cm}\).

a) \(60\,\text{cm}^2\) b) \(10\,\text{cm}\) c) \(b_1+b_2=24\,\text{cm}\)

- Average the parallel side lengths and multiply by the height. - What happens when one factor in a product doubles? - Picture the upper side shrinking until its endpoints meet.

1. The original area is \(A=\frac{1}{2}(9+5)\times4=28\,\text{cm}^2\). 2. Doubling the height doubles the area to \(56\,\text{cm}^2\) because height is a factor in the formula. 3. When one parallel side shrinks to a point, the figure becomes a triangle with base \(9\,\text{cm}\) and height \(4\,\text{cm}\). Its area is \(\frac{1}{2}\times9\times4=18\,\text{cm}^2\).

a) \(28\,\text{cm}^2\) b) The area doubles to \(56\,\text{cm}^2\). c) A triangle results, with area \(18\,\text{cm}^2\).

- Which pairs of numbers have a product of \(36\)? - What happens when an area is divided into two equal parts? - Compare the rectangle and parallelogram area formulas.

1. Since \(A=bh\), any positive pair with product \(36\) works. Examples are \(6\,\text{cm}\) and \(6\,\text{cm}\), or \(9\,\text{cm}\) and \(4\,\text{cm}\). 2. A diagonal divides a parallelogram into two congruent triangles, so each triangle has area \(36\div2=18\,\text{cm}^2\). 3. Both a rectangle and a parallelogram have area \(bh\). If their areas and bases are equal, their heights must also be equal.

a) Possible pairs include \(6\,\text{cm}\) and \(6\,\text{cm}\), or \(9\,\text{cm}\) and \(4\,\text{cm}\). b) \(18\,\text{cm}^2\) each c) The heights are equal.

- Find the area shared by both lots first. - Use that area to find the rectangle's width. - Find both perimeters before comparing the fencing needed. - Keep area and perimeter measurements separate.

1. The square lot has area \(21\times21=441\,\text{m}^2\). 2. The rectangular lot has the same area, so its width is \(441\div24.5=18\,\text{m}\). 3. The square's perimeter is \(4\times21=84\,\text{m}\). 4. The rectangle's perimeter is \(2\times(24.5+18)=85\,\text{m}\). 5. The rectangular lot requires \(85-84=1\,\text{m}\) more fencing.

a) The rectangular lot is \(18\,\text{m}\) wide. b) The rectangular lot requires more fencing, by \(1\,\text{m}\).

- Which side can you find from the area and the given height? - What does half of a parallelogram's perimeter represent? - Once one side is known, how much of that semiperimeter remains?

1. The side corresponding to the \(6\,\text{cm}\) height is \(48\div6=8\,\text{cm}\). 2. Half of the perimeter is \(40\div2=20\,\text{cm}\), so the two adjacent side lengths total \(20\,\text{cm}\). 3. The other side is \(20-8=12\,\text{cm}\).

The side lengths are \(8\,\text{cm}\) and \(12\,\text{cm}\).

- Express all lengths in meters before calculating. - Use the area formula for a trapezoid. - Multiply the total area by the cost per square meter.

1. Convert the height: \(1200\,\text{cm}=12\,\text{m}\). 2. The trapezoid's area is \(\frac{24+16}{2}\times12=20\times12=240\,\text{m}^2\). 3. The total cost is \(240\times\$0.80=\$192.00\).

a) The play area has an area of \(240\,\text{m}^2\). b) The grass seed costs \(\$192.00\).

- Calculate each area separately. - What is the value of \(\frac{b_1+b_2}{2}\) for the trapezoid? - Compare that value with the parallelogram's base.

1. The parallelogram's area is \(A_P=6\times4=24\,\text{cm}^2\). 2. The trapezoid's area is \(A_T=\frac{1}{2}(8+4)\times4=6\times4=24\,\text{cm}^2\). 3. The average of the trapezoid's parallel side lengths is \(6\,\text{cm}\), equal to the parallelogram's base. Since the heights are also equal, the areas are equal.

a) Both areas are \(24\,\text{cm}^2\). b) The trapezoid's average width equals the parallelogram's base, and the heights are equal.

- The trapezoid's midsegment length is the average of its parallel side lengths. - Area equals midsegment times height. - Express all lengths in the same unit before calculating. - Rearrange the formula when one factor is missing.

1. For part a), \(m=\frac{12+18}{2}=15\,\text{cm}\). Then \(A=mh=15\times5=75\,\text{cm}^2\). 2. For part b), convert \(40\,\text{cm}=0.4\,\text{m}\). Then \(A=mh=0.7\times0.4=0.28\,\text{m}^2\). 3. For part c), \(h=A\div m=100\div20=5\,\text{cm}\).

a) \(m=15\,\text{cm}\), \(A=75\,\text{cm}^2\) b) \(A=0.28\,\text{m}^2\) c) \(h=5\,\text{cm}\)

- Work one part at a time and decide which relationship connects the known measurements to the unknown measurement. - The midsegment is the average of the two parallel side lengths, so twice the midsegment gives their total length. - Area equals midsegment times height.

1. For a), \(m=\frac{4.5+3.5}{2}=4\,\text{m}\). Then \(A=mh=4\times2=8\,\text{m}^2\). 2. For b), the two parallel sides total \(2\times10=20\,\text{cm}\), so \(b_2=20-8=12\,\text{cm}\). Also, \(h=50\div10=5\,\text{cm}\). 3. For c), the two parallel sides total \(2\times15=30\,\text{mm}\), so \(b_1=30-12=18\,\text{mm}\). Then \(A=mh=15\times10=150\,\text{mm}^2\).

a) \(m=4\,\text{m}\), \(A=8\,\text{m}^2\) b) \(b_2=12\,\text{cm}\), \(h=5\,\text{cm}\) c) \(b_1=18\,\text{mm}\), \(A=150\,\text{mm}^2\)

- First find the triangle's area. - For the trapezoid, divide the area by the height to find the average of the parallel side lengths. - How can the average and one known side determine the other side?

1. The triangle's area is \(A=\frac{1}{2}\times10\times8=40\,\text{cm}^2\). 2. For the trapezoid to have area \(40\,\text{cm}^2\) and height \(5\,\text{cm}\), the average of its parallel side lengths must be \(40\div5=8\,\text{cm}\). 3. The two parallel sides therefore total \(2\times8=16\,\text{cm}\). 4. The other parallel side is \(16-6=10\,\text{cm}\).

The other parallel side is \(10\,\text{cm}\) long.

- Express the area and height using compatible units. - Represent the shorter side with a variable and the longer side as twice that variable. - Substitute both expressions into the trapezoid area formula.

1. Convert the area: \(0.9\,\text{m}^2=9000\,\text{cm}^2\). 2. Let \(x\) be the top side length. Then the bottom side length is \(2x\). 3. Substitute into the trapezoid area formula: \(9000=\frac{x+2x}{2}\times60\). 4. Simplify: \(9000=90x\), so \(x=9000\div90=100\,\text{cm}\). 5. The bottom side is \(2\times100=200\,\text{cm}\).

The top parallel side is \(100\,\text{cm}\) long, and the bottom parallel side is \(200\,\text{cm}\) long.

- Use the area formula for a trapezoid. - Find the total area of all five pieces. - Determine half the area of the original sheet. - Compare the minimum unused area with half the sheet.

1. One trapezoid has area \(\frac{20+30}{2}\times15=25\times15=375\,\text{cm}^2\). 2. All \(5\) pieces require \(5\times375=1875\,\text{cm}^2\). 3. The wood sheet has area \(50\times80=4000\,\text{cm}^2\), so at least \(4000-1875=2125\,\text{cm}^2\) will remain unused. 4. Half the sheet is \(4000\div2=2000\,\text{cm}^2\). Since \(2125>2000\), more than half the sheet remains even in the best possible arrangement. Lina's claim is false.

a) One piece has area \(375\,\text{cm}^2\). b) The \(5\) pieces require \(1875\,\text{cm}^2\) of wood. c) Lina's claim is false. At least \(2125\,\text{cm}^2\) remains, which is more than half of the \(4000\,\text{cm}^2\) sheet.

- Find the area that the entire banner must cover. - Find the area of one piece of each type. - Add the areas of all four pieces and compare the total with the banner area.

1. The banner has area \(120\times40=4800\,\text{cm}^2\). 2. One type A piece has area \(\frac{40+20}{2}\times40=1200\,\text{cm}^2\). 3. One type B piece has area \(\frac{30+10}{2}\times40=800\,\text{cm}^2\). 4. The four pieces have total area \(2\times1200+2\times800=4000\,\text{cm}^2\). 5. Since \(4000<4800\), the pieces do not provide enough fabric. They are short by \(4800-4000=800\,\text{cm}^2\).

No. The banner has area \(4800\,\text{cm}^2\), but the four pieces have total area \(4000\,\text{cm}^2\). They are short by \(800\,\text{cm}^2\).

- First find the missing height of the parallelogram. - How are the heights of the two figures related? - Substitute the trapezoid's height and area into its area formula.

1. The parallelogram's height is \(40\div8=5\,\text{cm}\). 2. The trapezoid's height is \(2\times5=10\,\text{cm}\). 3. Let \(b_1\) and \(b_2\) be the trapezoid's parallel side lengths. Then \(40=\frac{1}{2}(b_1+b_2)\times10=5(b_1+b_2)\). 4. Therefore, \(b_1+b_2=40\div5=8\,\text{cm}\).

The sum of the parallel side lengths is \(8\,\text{cm}\).

- Write an expression for the original area first. - List the new parallel side lengths and the new height. - Calculate the new area and compare the two expressions.

1. The original area is \(A_1=\frac{1}{2}(12+8)h=10h\,\text{cm}^2\). 2. The new measurements are \(6\,\text{cm}\), \(4\,\text{cm}\), and \(2h\,\text{cm}\). 3. The new area is \(A_2=\frac{1}{2}(6+4)(2h)=5(2h)=10h\,\text{cm}^2\). 4. Since \(A_1=A_2\), the area does not change.

The area stays the same: both the original and new areas are \(10h\,\text{cm}^2\).

- Find how many strips fit across the wall's width. - Find how many wall-height strips can be cut from one roll. - The purchase must be a whole number of rolls.

1. Find the number of vertical strips needed across the wall: \(12\div2=6\) strips. 2. Each strip must be \(8\,\text{ft}\) long. One roll provides \(32\div8=4\) strips. 3. One roll gives 4 of the 6 needed strips, leaving 2 strips still to cover. 4. A second roll provides those remaining strips, so \(2\) rolls are needed.

\(2\) rolls are needed.

- Think of covering the floor with long strips placed side by side. - Find how many strips are needed across the room. - Find the total strip length and compare it with one roll's length.

1. Find the number of strips needed across the room: \(12\div6=2\) strips. 2. Each strip must be \(15\,\text{ft}\) long, so the total length needed is \(2\times15\,\text{ft}=30\,\text{ft}\). 3. One roll is \(36\,\text{ft}\) long. Since \(30\,\text{ft}<36\,\text{ft}\), one roll is enough.

\(1\) roll is needed.

- Identify the parallel sides and the height of the trapezoid. - You may also split the figure into a rectangle and a right triangle to find the area. - What are the dimensions of the rectangle and triangle in that decomposition? - To find the perimeter, add all four outside side lengths.

1. The parallel sides are \(8\,\text{m}\) and \(5\,\text{m}\), and the height is \(4\,\text{m}\). The area is \(A=\frac{1}{2}(8+5)\times4=26\,\text{m}^2\). 2. The four side lengths are \(8\,\text{m}\), \(5\,\text{m}\), \(5\,\text{m}\), and \(4\,\text{m}\). The perimeter is \(8+5+5+4=22\,\text{m}\).

a) \(26\,\text{m}^2\) b) \(22\,\text{m}\) of fencing

- Split each figure into rectangles and triangles, or enclose it in a rectangle and subtract. - Use \(\frac{1}{2}bh\) for each triangle. - A parallelogram can be rearranged into a rectangle with the same base and height.

1. Figure D can be split into a \(5\times 3\) rectangle and a triangle with base \(5\) and height \(2\). Its area is \(5\times 3+\frac{1}{2}\times 5\times 2=15+5=20\) square units. 2. Figure E fits inside a \(4\times 4\) square with a triangular cutout of base \(4\) and height \(2\). Its area is \(4\times 4-\frac{1}{2}\times 4\times 2=16-4=12\) square units. 3. Figure F is a parallelogram with base \(3\) and height \(3\), so its area is \(3\times 3=9\) square units. It can also be rearranged into a \(3\times 3\) rectangle.

D: \(20\) square units E: \(12\) square units F: \(9\) square units

- Count the unit squares in Figure A. - Split Figure B into a rectangle and a triangle. - Use one-half times the base times the height for the triangle. - Find a pair of whole numbers whose product is the area from part b).

1. Counting the unit squares in Figure A gives an area of \(5\) square units. 2. Split Figure B into a \(4 \times 2\) rectangle and a triangle with base \(4\) units and height \(2\) units. Its area is \(8 + \frac{1}{2} \times 4 \times 2 = 12\) square units. 3. A rectangle with area \(12\) square units can have side lengths \(3\) units and \(4\) units. Other possible whole-number pairs are \(2\) and \(6\), or \(1\) and \(12\).

a) \(5\) square units b) \(12\) square units c) One possible pair is \(3\) units by \(4\) units.

- Imagine a vertical line where the rectangular shaft meets the triangular point. - Find the rectangle's area. - Use \(\frac{1}{2}bh\) for the triangle, then add.

1. The rectangular part is \(4\) units wide and \(2\) units high, so its area is \(4\times 2=8\) square units. 2. The triangular point has base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2}\times 4\times 2=4\) square units. 3. The total area is \(8+4=12\) square units.

\(12\) square units

- Decompose each figure into rectangles, squares, and triangles. - A diagonal divides a unit square into two congruent triangles. - Combine partial areas before comparing the figures. - Subtract the smaller area from the larger area.

1. Figure A can be split into a \(4\times2\) rectangle and a triangle with base \(4\) and height \(2\). Its area is \(4\times2+\frac{1}{2}\times4\times2=8+4=12\) square units. 2. Figure B can be split into a \(3\times3\) square, a \(1\times1\) square, and two half-unit triangles. Its area is \(9+1+1=11\) square units. 3. Figure A is larger by \(12-11=1\) square unit.

Figure A has area \(12\) square units. Figure B has area \(11\) square units. Figure A is larger by \(1\) square unit.

- Break each figure into rectangles or triangles. - Compare each figure with a rectangle that surrounds it. - Use one-half of base times height for triangles. - Compare your three results.

1. Figure A is a \(3 \times 2\) rectangle, so its area is \(3 \times 2=6\) square units. 2. Figure B is a right triangle with base \(4\) and height \(3\), so its area is \(\frac{1}{2} \times 4 \times 3=6\) square units. 3. Figure C is a trapezoid. Split it into a \(2 \times 2\) rectangle and two triangles with base \(1\) and height \(2\). Its area is \(4+1+1=6\) square units. 4. All three figures have the same area even though their shapes are different.

a) A: \(6\) square units; B: \(6\) square units; C: \(6\) square units b) All three figures have the same area.

- Split each figure into familiar shapes. - For Figure A, look for one rectangle and one triangle. - For Figure B, look for one lower rectangle and two upper rectangles. - Add the areas of the parts and compare the totals.

1. Figure A can be split into a \(4\times 2\) rectangle and a triangle with base \(4\) and height \(2\). Their areas are \(4\times 2=8\) and \(\frac{1}{2}\times 4\times 2=4\) square units. Therefore, Figure A has area \(8+4=12\) square units. 2. Figure B can be split into a \(4\times 2\) rectangle and two \(1\times 2\) rectangles. Their total area is \(4\times 2+2\times(1\times 2)=8+4=12\) square units. 3. Both figures have an area of \(12\) square units.

Each figure has an area of \(12\) square units.

- Enclose the pentagon in a larger, simple shape. - Identify the regions inside that shape but outside the pentagon. - Find the areas of those triangles. - Subtract their total area from the enclosing area.

1. Enclose the pentagon in a \(4\,\text{cm} \times 4\,\text{cm}\) square. The square has area \(4 \times 4=16\,\text{cm}^2\). 2. The four corner regions outside the pentagon are right triangles with areas \(\frac{1}{2} \times 1 \times 1=0.5\), \(\frac{1}{2} \times 1 \times 3=1.5\), \(\frac{1}{2} \times 3 \times 1=1.5\), and \(\frac{1}{2} \times 1 \times 3=1.5\) square centimeters. 3. Their total area is \(0.5+1.5+1.5+1.5=5\,\text{cm}^2\). 4. The pentagon area is \(16-5=11\,\text{cm}^2\).

\(11\,\text{cm}^2\)

- Imagine a vertical or horizontal segment that divides each figure into familiar shapes. - Use \(\frac{1}{2}bh\) for a right triangle. - For Figure B, consider a surrounding rectangle and subtract the missing corner. - Add the two figure areas for part b.

1. Figure A can be split into a \(2\times 3\) rectangle and a right triangle with base \(2\) and height \(3\). Its area is \(2\times 3+\frac{1}{2}\times 2\times 3=6+3=9\) square units. 2. Figure B fits inside a \(4\times 3\) rectangle. The missing upper-left triangle has base \(1\) and height \(2\), so its area is \(\frac{1}{2}\times 1\times 2=1\) square unit. Figure B has area \(4\times 3-1=11\) square units. 3. Together, the figures have area \(9+11=20\) square units.

a) Figure A: \(9\) square units Figure B: \(11\) square units b) \(20\) square units

- Split each figure into rectangles and triangles, or rearrange a piece to make a rectangle. - Use one-half times the base times the height for each triangle. - Compare the three areas after calculating them.

1. Split trapezoid A into a \(2 \times 2\) rectangle and two right triangles, each with base \(1\) unit and height \(2\) units. Its area is \(2 \times 2 + 2\left(\frac{1}{2} \times 1 \times 2\right) = 6\) square units. 2. Triangle B has base \(4\) units and height \(3\) units. Its area is \(\frac{1}{2} \times 4 \times 3 = 6\) square units. 3. Rearranging a triangular piece of parallelogram C forms a \(3 \times 2\) rectangle. Its area is \(3 \times 2 = 6\) square units. 4. All three figures have the same area.

a) Figure A: \(6\) square units Figure B: \(6\) square units Figure C: \(6\) square units b) All three figures have the same area.

- Divide each figure into smaller familiar shapes. - Use \(\frac{1}{2}bh\) for each right triangle. - Add the areas of all parts in each figure. - Compare the two totals.

1. Figure a) contains a \(3\times 2\) rectangle and two right triangles, each with base \(1\) and height \(2\). Its area is \(3\times 2+2\times\left(\frac{1}{2}\times 1\times 2\right)=6+2=8\) square units. 2. Figure b) can be split into two right triangles, each with base \(2\) and height \(4\). Its area is \(2\times\left(\frac{1}{2}\times 2\times 4\right)=8\) square units. 3. The two figures have equal areas.

1. Figure a): \(8\) square units 2. Figure b): \(8\) square units 3. The figures have the same area.

- Split each figure into rectangles and triangles. - Two half-squares make one whole square. - Use one-half times the base times the height for each triangular part. - Subtract the smaller total area from the greater total area.

1. Split the house into a \(4 \times 2\) rectangle and a triangle with base \(4\) units and height \(2\) units. Its area is \(4 \times 2 + \frac{1}{2} \times 4 \times 2 = 8 + 4 = 12\) square units. 2. The boat hull has area \(4\) square units. The sail is a right triangle with base \(2\) units and height \(3\) units, so its area is \(\frac{1}{2} \times 2 \times 3 = 3\) square units. The boat has total area \(4 + 3 = 7\) square units. 3. Since \(12 - 7 = 5\), the house has the greater area by \(5\) square units.

a) Figure A has an area of \(12\) square units, and Figure B has an area of \(7\) square units. b) Figure A has the greater area by \(5\) square units.

- Decompose each figure into rectangles and triangles. - A triangle with the same base and height as a rectangle has half the rectangle's area. - After finding both areas, compare them and find the difference.

1. Figure a) can be split into a \(4\times 2\) rectangle and a triangle with base \(4\) and height \(2\). Its area is \(4\times 2+\frac{1}{2}\times 4\times 2=8+4=12\) square units. 2. Figure b) can be split into a \(5\times 3\) rectangle, a \(2\times 2\) rectangle, and a right triangle with base \(1\) and height \(2\). Its area is \(5\times 3+2\times 2+\frac{1}{2}\times 1\times 2=15+4+1=20\) square units. 3. Figure b) has the greater area. The difference is \(20-12=8\) square units.

1) Figure a): \(12\) square units; figure b): \(20\) square units 2) Figure b) is greater by \(8\) square units.

- Find the trapezoid's area first. - Use the appropriate area formula for each other figure. - Compare all four results.

1. Trapezoid a) has bases \(4\) and \(2\) and height \(2\), so its area is \(\frac{1}{2}(4+2)\times2=6\) square units. 2. Triangle b) has base \(4\) and height \(3\), so its area is \(\frac{1}{2}\times4\times3=6\) square units. 3. Rectangle c) has area \(3\times2=6\) square units. 4. Triangle d) has area \(\frac{1}{2}\times3\times3=4.5\) square units. 5. Therefore, figures b) and c) have the same area as figure a).

Figures b) and c) have the same area as figure a): \(6\) square units.

- For the parallelogram, move a triangular end piece to the other side. - For the triangle, imagine using two congruent copies to form a rectangle. - Read each base and perpendicular height from the grid.

1. Figure 1 is a parallelogram with base \(4\) units and height \(2\) units. Moving the triangular piece from one side to the other forms a \(4\times2\) rectangle, so the area is \(4\times2=8\) square units. 2. Figure 2 is a triangle with base \(4\) units and height \(4\) units. Two copies form a \(4\times4\) rectangle, so one triangle has area \(\frac{1}{2}\times4\times4=8\) square units.

1) \(8\) square units 2) \(8\) square units

- Look for rectangles and triangles. - Consider the \(4\times 3\) rectangle surrounding the figure. - Each missing triangle is half of a \(2\times 1\) rectangle.

1. Enclose the figure in a \(4\times 3\) rectangle with area \(12\) square units. 2. Two right triangles are missing from the top. Each has base \(2\) and height \(1\), so each area is \(\frac{1}{2}\times 2\times 1=1\) square unit. 3. The figure's area is \(12-1-1=10\) square units. 4. Equivalently, the figure consists of a \(4\times 2\) rectangle and two triangles of area \(1\) square unit each, giving \(8+1+1=10\) square units.

\(10\) square units

- Find the area of each figure separately. - Split slanted figures into rectangles and triangles, or rearrange a triangular piece to form a rectangle. - Compare the three results.

1. Figure a) is a \(3 \times 2\) rectangle, so its area is \(3 \times 2 = 6\) square units. 2. Split Figure b) into a \(2 \times 2\) rectangle and a right triangle with base \(2\) units and height \(2\) units. Its area is \(4 + \frac{1}{2} \times 2 \times 2 = 6\) square units. 3. Rearranging a triangular piece of Figure c) forms a \(2 \times 2\) rectangle, so its area is \(4\) square units. 4. Figures a) and b) have the same area.

Figures a) and b) have the same area: \(6\) square units each.

- Find each area separately. - Account for the triangular parts in Figure 2. - Compare the three numerical areas.

1. Figure 1 is a square with side length \(2\) units, so its area is \(2 \times 2 = 4\) square units. 2. Figure 2 can be split into a \(3 \times 1\) rectangle and a triangle with base \(3\) units and height \(1\) unit. Its area is \(3 + \frac{1}{2} \times 3 \times 1 = 4.5\) square units. 3. Figure 3 can be split into rectangles with areas \(3\) square units and \(2\) square units. Its total area is \(5\) square units. 4. Since \(4 < 4.5 < 5\), the order is Figure 1, Figure 2, Figure 3.

Figure 1: \(4\) square units Figure 2: \(4.5\) square units Figure 3: \(5\) square units Order from least to greatest: Figure 1, Figure 2, Figure 3.

- Check each design separately. - Break the figures into rectangles, parallelograms, or triangles. - Use the grid to identify base and height.

1. Design a) is a triangle with base \(4\) and height \(2\), so its area is \(\frac{1}{2} \times 4 \times 2=4\) square units. 2. Design b) is a parallelogram with base \(3\) and height \(2\), so its area is \(3 \times 2=6\) square units. 3. Design c) can be split into a \(2 \times 2\) rectangle and a right triangle with base \(1\) and height \(2\). Its area is \(4+\frac{1}{2} \times 1 \times 2=5\) square units. 4. Design c) meets the requirement.

Design c), because its area is \(5\) square units.

- Split the figure into one rectangle and one triangle. - Add the two areas. - Subtract the current area from the target area.

1. The lower rectangle measures \(5\times 2\), so its area is \(10\) square units. The upper triangle has base \(5\) and height \(1\), so its area is \(\frac{1}{2}\times 5\times 1=2.5\) square units. 2. The current area is \(10+2.5=12.5\) square units. 3. The additional area needed is \(15-12.5=2.5\) square units.

1. \(12.5\) square units 2. \(2.5\) square units must be added.

- Imagine vertical segments from the endpoints of the upper base. - Identify the resulting rectangle and right triangles. - Add their areas.

1. Decompose the trapezoid into a \(2\,\text{cm}\times 4\,\text{cm}\) rectangle, a left right triangle with base \(1\,\text{cm}\) and height \(4\,\text{cm}\), and a right right triangle with base \(2\,\text{cm}\) and height \(4\,\text{cm}\). 2. The areas are \(2\times 4=8\,\text{cm}^2\), \(\frac{1}{2}\times 1\times 4=2\,\text{cm}^2\), and \(\frac{1}{2}\times 2\times 4=4\,\text{cm}^2\). 3. The total area is \(8+2+4=14\,\text{cm}^2\).

\(14\,\text{cm}^2\)

- Imagine a vertical segment that splits the figure into a rectangle and a right triangle. - Find the dimensions of both parts. - Add their areas.

1. Decompose the figure into a \(4\,\text{cm}\times 3\,\text{cm}\) rectangle and a right triangle with base \(1\,\text{cm}\) and height \(3\,\text{cm}\). 2. The rectangle's area is \(4\times 3=12\,\text{cm}^2\). 3. The triangle's area is \(\frac{1}{2}\times 1\times 3=1.5\,\text{cm}^2\). 4. The total area is \(12+1.5=13.5\,\text{cm}^2\).

\(13.5\,\text{cm}^2\)

- Identify the base and perpendicular height. - Imagine a vertical partition from the upper-left vertex to the base. - Compare the triangular region on the left with the gap on the right.

1. The parallelogram has base \(4\,\text{cm}\) and height \(3\,\text{cm}\), so its area is \(4\times 3=12\,\text{cm}^2\). 2. A vertical partition from the upper-left vertex to the base identifies a triangular region on the left. That region has the same shape and size as the gap beside the right slanted side. 3. Placing the corresponding triangular region in that gap forms a rectangle measuring \(4\,\text{cm}\times 3\,\text{cm}\), so the area remains \(12\,\text{cm}^2\).

a) \(12\,\text{cm}^2\) b) The left triangular region fills the right-side gap, forming a \(4\,\text{cm} \times 3\,\text{cm}\) rectangle.

- Think of a \(4\times 4\) square with a triangular piece removed. - Find the missing triangle's base and height. - Subtract the triangle's area from the square's area.

1. Enclose the figure in a \(4\,\text{cm}\times 4\,\text{cm}\) square with area \(16\,\text{cm}^2\). 2. The missing upper triangle has base \(4\,\text{cm}\) and height \(2\,\text{cm}\), so its area is \(\frac{1}{2}\times 4\times 2=4\,\text{cm}^2\). 3. The figure's area is \(16-4=12\,\text{cm}^2\).

\(12\,\text{cm}^2\)

- Imagine a horizontal segment that separates a rectangle and a right triangle. - Determine the rectangle's dimensions. - Determine the triangle's base and height, then add the two areas.

1. A horizontal segment separates a \(4\times 1\) rectangle from a right triangle with base \(4\) and height \(2\). 2. The rectangle's area is \(4\times 1=4\) square units. 3. The triangle's area is \(\frac{1}{2}\times 4\times 2=4\) square units. 4. The total area is \(4+4=8\) square units.

\(8\) square units

- Split the figure into two triangles with a common base. - Count spaces between pegs to find the shared base and each triangle's height. - Add the areas of the two triangles.

1. Split the kite along its horizontal diagonal. This creates two triangles with a common base of \(2\,\text{cm}\). 2. The lower triangle has height \(2\,\text{cm}\), so its area is \(\frac{1}{2}\times2\times2=2\,\text{cm}^2\). 3. The upper triangle has height \(3\,\text{cm}\), so its area is \(\frac{1}{2}\times2\times3=3\,\text{cm}^2\). 4. The kite's area is \(2+3=5\,\text{cm}^2\).

The area of the kite is \(5\,\text{cm}^2\).

- Separate the rectangular shaft from the triangular point. - Identify the triangle's vertical base and horizontal height. - Add the rectangle and triangle areas.

1. The rectangular shaft measures \(4\times 2\), so its area is \(8\) square units. 2. The triangular point has base \(6\) and perpendicular height \(4\), so its area is \(\frac{1}{2}\times 6\times 4=12\) square units. 3. The total area is \(8+12=20\) square units.

\(20\) square units

- Imagine vertical segments from the endpoints of the shorter base. - Find the area of the central rectangle and the two right triangles. - Add the three areas.

1. Decompose the trapezoid into a \(3\times 3\) rectangle and two right triangles, each with base \(1\) and height \(3\). 2. The rectangle's area is \(3\times 3=9\) square units. Each triangle's area is \(\frac{1}{2}\times 1\times 3=1.5\) square units. 3. The total area is \(9+1.5+1.5=12\) square units.

\(12\) square units

- Imagine a perpendicular partition from the upper-left vertex to the base. - Compare the triangular region on the left with the gap on the right. - Read the base from the diagram and use it with the given height.

1. A perpendicular partition from the upper-left vertex to the base identifies a triangular region on the left. That region has the same shape and size as the gap beside the right slanted side. 2. The corresponding rectangle has length \(10\,\text{cm}\) and width \(6\,\text{cm}\). This rearrangement does not change the area. 3. Therefore, the parallelogram's area is \(10\times6=60\,\text{cm}^2\).

The left triangular region corresponds to the right-side gap, forming a \(10\,\text{cm}\times6\,\text{cm}\) rectangle. The area is \(60\,\text{cm}^2\).

- Find the patio area first. - Multiply the area by the cost per square meter. - An \(8\%\) sales tax means the final cost is \(108\%\) of the pretax cost.

1. The patio area is \(7.5\times4.4=33\,\text{m}^2\). 2. The cost before tax is \(33\times\$28.00=\$924.00\). 3. Including an \(8\%\) sales tax, the total is \(\$924.00\times1.08=\$997.92\).

The patio has area \(33\,\text{m}^2\), and the total material cost is \(\$997.92\).

- The final price includes the whole original price plus an additional \(8\%\) of that original price. - Think about what percent of the original price the final amount represents. - Work backward from the final amount to the original amount.

1. After an \(8\%\) increase, the final price is \(108\%\) of the original price. 2. Let the original price be \(x\). Then \(1.08x = 54\). 3. Divide by \(1.08\): \(x = 54 \div 1.08 = 50\). 4. The price before tax was \(\$50.00\).

\(\$50.00\)

- Split the figure into a rectangle and a triangle. - The triangle's base is the difference between the \(14\,\text{m}\) and \(9\,\text{m}\) parallel sides.

1. Split the trapezoid into a \(9\,\text{m}\times6\,\text{m}\) rectangle and a right triangle. 2. The triangle's base is \(14-9=5\,\text{m}\), and its height is \(6\,\text{m}\). 3. The rectangle's area is \(9\times6=54\,\text{m}^2\). The triangle's area is \(\frac{1}{2}\times5\times6=15\,\text{m}^2\). 4. The total area is \(54+15=69\,\text{m}^2\).

The garden bed's area is \(69\,\text{m}^2\).

- The figure is still a trapezoid even though its parallel sides are vertical. - A trapezoid's height is the perpendicular distance between its parallel sides. - Mentally rotate the figure if that makes the base-and-height relationship easier to see.

1. The figure is a trapezoid with vertical parallel sides \(KL\) and \(MN\). 2. Their lengths are \(6\) units and \(2\) units. The perpendicular distance between them is \(4\) units. 3. The area is \(A=\frac{1}{2}(6+2)\times4=16\) square units.

The area of the figure is \(16\) square units.

- Which whole-number side length gives a square area of \(100\,\text{cm}^2\)? - Connecting the midpoints leaves four congruent right triangles in the corners. - Subtract the total corner-triangle area from the original square.

1. Since \(10\times10=100\), the original square has side length \(10\,\text{cm}\). 2. Each midpoint is \(5\,\text{cm}\) from the nearest vertices, so the four corner regions are congruent right triangles with legs \(5\,\text{cm}\) and \(5\,\text{cm}\). 3. Each corner triangle has area \(\frac{1}{2}\times5\times5=12.5\,\text{cm}^2\), so all four have area \(4\times12.5=50\,\text{cm}^2\). 4. The inner quadrilateral has area \(100-50=50\,\text{cm}^2\).

\(50\,\text{cm}^2\)

- How does \(BM\) compare with \(BC\)? - Compare the triangle and parallelogram area formulas using the same height.

1. Since \(M\) is the midpoint of \(BC\), \(BM=\frac{1}{2}BC\). 2. Using \(BM\) as the triangle's base, triangle \(ABM\) has the same corresponding height as the parallelogram with base \(BC\). 3. Thus, \(A_{\triangle ABM}=\frac{1}{2}\times\frac{1}{2}BC\times h=\frac{1}{4}A_{ABCD}\). 4. Therefore, \(A_{ABCD}=4\times12=48\,\text{cm}^2\).

The area of the parallelogram is \(48\,\text{cm}^2\).

- Does the triangle's height change as \(K\) moves along \(BC\)? - Compare the triangle and parallelogram area formulas.

1. Use \(AD\) as the triangle's base. Because \(K\) lies on the opposite side \(BC\), the triangle and parallelogram have the same base length and corresponding height. 2. A triangle has half the area of a parallelogram with the same base and height. 3. Therefore, \(A_{\triangle AKD}=\frac{1}{2}\times60=30\,\text{cm}^2\).

The area of triangle \(AKD\) is \(30\,\text{cm}^2\).

- Use the fact that each diagonal is cut into two equal segments at \(O\). - Look for pairs of triangles with equal bases lying on the same diagonal and a shared height. - After establishing equal areas, scale one small triangle's area to the whole parallelogram.

1. Triangles \(AOB\) and \(BOC\) have equal bases \(AO=OC\) on diagonal \(AC\) and share the same perpendicular height from \(B\), so they have equal area. 2. Triangles \(AOD\) and \(DOC\) also have equal bases \(AO=OC\) and share the same height from \(D\), so they have equal area. 3. Similarly, using \(BO=OD\) on diagonal \(BD\), the triangles on opposite sides of \(BD\) match in area. Therefore all four small triangles have equal area. 4. The parallelogram area is \(4\times12=48\,\text{cm}^2\).

a) The diagonal-bisection relationships give equal bases with common heights, so all four small triangles have equal area. b) \(48\,\text{cm}^2\)

- Read the side length of the outer square and the marked offset from the diagram. - Determine the two leg lengths of a corner triangle. - Subtract the total area of the four corner triangles from the area of the square.

1. The area of the entire garden bed is \(10\,\text{m}\times10\,\text{m}=100\,\text{m}^2\). 2. Each corner triangle is a right triangle with leg lengths \(3\,\text{m}\) and \(10\,\text{m}-3\,\text{m}=7\,\text{m}\). 3. The area of one corner triangle is \(\frac{1}{2}\times3\times7=10.5\,\text{m}^2\). 4. The four corner triangles have total area \(4\times10.5=42\,\text{m}^2\). 5. The grass section has area \(100-42=58\,\text{m}^2\).

The grass section has area \(58\,\text{m}^2\).

- Find the number of grid cells in the whole rectangle. - Use the total area to find the area of one cell. - Split the orange region into two triangles. - Convert the triangle areas from grid cells to square centimeters.

1. The rectangle has \(8 \times 5=40\) grid cells, so each cell has area \(80 \div 40=2\,\text{cm}^2\). 2. The left orange triangle has a base of \(5\) grid units and a height of \(2\) grid units, so its area is \(\frac{1}{2} \times 5 \times 2=5\) grid cells. 3. The right orange triangle has a base of \(5\) grid units and a height of \(6\) grid units, so its area is \(\frac{1}{2} \times 5 \times 6=15\) grid cells. 4. The orange region covers \(5+15=20\) grid cells, so its area is \(20 \times 2=40\,\text{cm}^2\).

\(40\,\text{cm}^2\)

- Find the area represented by one grid square. - Find each shaded triangle's base and height in grid units. - Convert the total number of grid squares to square centimeters.

1. The rectangle contains \(12\times 6=72\) grid squares, so each grid square represents \(144\div 72=2\,\text{cm}^2\). 2. The lower shaded triangle has base \(12\) grid units and height \(3\) grid units, so its area is \(\frac{1}{2}\times 12\times 3=18\) grid squares. 3. The upper shaded triangle has base \(6\) grid units and height \(3\) grid units, so its area is \(\frac{1}{2}\times 6\times 3=9\) grid squares. 4. The shaded regions cover \(18+9=27\) grid squares, representing \(27\times 2=54\,\text{cm}^2\).

\(54\,\text{cm}^2\)

- Count the grid cells in the entire rectangle. - Split the yellow region into two triangles and the middle quadrilateral. - Compare the yellow area in grid cells with the total number of grid cells. - Use that fraction with the given area.

1. The rectangle contains \(10\times6=60\) grid cells. 2. Each outer yellow triangle covers \(\frac{1}{2}\times6\times3=9\) grid cells. 3. The middle rhombus has diagonals \(4\) and \(6\), so its area is \(\frac{1}{2}\times4\times6=12\) grid cells. 4. The yellow regions cover \(9+9+12=30\) of the \(60\) grid cells, so they make up one-half of the garden bed. 5. The entire area is \(2\times48=96\,\text{m}^2\).

\(96\,\text{m}^2\)

- The shaded area is not the area of the outer rectangle alone. - Treat the unshaded interior rectangle as an area that must be removed. - After finding the required coverage, consider what a non-whole-number container count means in context.

1. The outer rectangle has area \(12\times9=108\,\text{m}^2\). 2. The planting bed has area \(4\times3=12\,\text{m}^2\). 3. The area to seal is \(108-12=96\,\text{m}^2\). 4. The number of containers is \(96\div5=19.2\). Since only whole containers can be purchased, \(20\) containers are needed.

Area to seal: \(96\,\text{m}^2\) Containers needed: \(20\)

- Convert all lengths to meters before finding the area of one paver. - Divide the driveway's total area by the area of one paver. - Determine how the area of a rectangle changes when both side lengths double.

1. One section has area \(2.40\times1.50=3.60\,\text{m}^2\), so the total area is \(2\times3.60=7.20\,\text{m}^2\). 2. Convert the paver dimensions: \(20\,\text{cm}=0.20\,\text{m}\) and \(10\,\text{cm}=0.10\,\text{m}\). One paver has area \(0.20\times0.10=0.020\,\text{m}^2\). 3. The number of pavers is \(7.20\div0.020=360\). 4. Doubling both dimensions makes each paver's area \(2\times2=4\) times as large. Therefore, only one-fourth as many pavers are needed: \(360\div4=90\).

a) The driveway has a total area of \(7.20\,\text{m}^2\). b) The driveway requires \(360\) pavers. c) Only one-fourth as many pavers would be needed, so \(90\) pavers would be required.

- What product must the two diagonal lengths have in part a)? - Make sure your second pair is not just the first pair written in reverse order. - In part b), replace each diagonal in the formula by twice its original length and compare the new expression with the original area.

1. Since \(A=\frac{1}{2}d_1d_2\), the diagonal lengths must satisfy \(d_1d_2=2A=40\). 2. Two genuinely different pairs are \(d_1=4\,\text{cm}\), \(d_2=10\,\text{cm}\) and \(d_1=5\,\text{cm}\), \(d_2=8\,\text{cm}\). 3. If both diagonals are doubled, the new area is \(A_{\text{new}}=\frac{1}{2}(2d_1)(2d_2)=4\left(\frac{1}{2}d_1d_2\right)=4A\). 4. Therefore, the student's claim is correct.

a) Possible pairs include \(4\,\text{cm}\) and \(10\,\text{cm}\), and \(5\,\text{cm}\) and \(8\,\text{cm}\). b) The student is correct. Doubling each diagonal multiplies their product, and therefore the area, by \(2\times2=4\).

- Use the area and height to find the sum of the parallel side lengths first. - Once you know the sum, think about two numbers with that sum and a difference of \(4\). - Start from two equal numbers and adjust them in opposite directions without changing their sum.

1. Since \(A=\frac{1}{2}(b_1+b_2)h\), the sum of the two parallel side lengths is \((2\times60)\div6=20\,\text{cm}\). 2. If the two sides were equal, each would be \(20\div2=10\,\text{cm}\). 3. To keep the sum \(20\,\text{cm}\) while making one side \(4\,\text{cm}\) longer, subtract \(2\,\text{cm}\) from one and add \(2\,\text{cm}\) to the other. 4. The parallel sides are \(8\,\text{cm}\) and \(12\,\text{cm}\).

The longer parallel side is \(12\,\text{cm}\), and the shorter parallel side is \(8\,\text{cm}\).

- Add the parallel side lengths, divide by \(2\), and multiply by the height. - What happens to a product when one factor doubles? - When every length is halved, calculate the area step by step.

1. The original area is \(A=\frac{1}{2}(4+6)\times5=5\times5=25\,\text{cm}^2\). 2. If the height doubles to \(10\,\text{cm}\), the new area is \(A_1=\frac{1}{2}(4+6)\times10=50\,\text{cm}^2\). The area doubles. 3. If every length is halved, the new measurements are \(2\,\text{cm}\), \(3\,\text{cm}\), and \(2.5\,\text{cm}\). 4. The new area is \(A_2=\frac{1}{2}(2+3)\times2.5=6.25\,\text{cm}^2\). 5. Since \(6.25=\frac{1}{4}\times25\), the area is one-fourth of the original area.

a) \(25\,\text{cm}^2\) b) The area doubles to \(50\,\text{cm}^2\). c) The area becomes one-fourth of the original, or \(6.25\,\text{cm}^2\).

- First find the parallelogram's area. - Compare the parallelogram and trapezoid area formulas. - What must \(\frac{b_1+b_2}{2}\) equal? - Is there only one pair with the required sum?

1. The parallelogram's area is \(A=10\times4=40\,\text{cm}^2\). 2. For the trapezoid, \(A=\frac{1}{2}(b_1+b_2)\times4\). Set this equal to \(40\): \(\frac{1}{2}(b_1+b_2)\times4=40\). 3. This simplifies to \(b_1+b_2=20\,\text{cm}\). 4. Two possible pairs are \(12\,\text{cm}\) and \(8\,\text{cm}\), or \(15\,\text{cm}\) and \(5\,\text{cm}\).

The condition is \(b_1+b_2=20\,\text{cm}\). Possible pairs include \(12\,\text{cm}\) and \(8\,\text{cm}\), or \(15\,\text{cm}\) and \(5\,\text{cm}\).

- If a product must stay constant and one factor increases, what must happen to the other? - Use variables such as \(b\) and \(h\) to test a general claim. - What is \((2b)(2h)\) compared with \(bh\)?

1. The triangle's area is \(A=\frac{1}{2}\times10\times6=30\,\text{cm}^2\). 2. For the new triangle, \(30=\frac{1}{2}\times20\times h=10h\), so \(h=3\,\text{cm}\). The height must be halved. 3. A parallelogram with base \(b\) and height \(h\) has area \(A=bh\). Doubling both gives \(A_{\text{new}}=(2b)(2h)=4bh=4A\). The claim is correct.

a) \(30\,\text{cm}^2\) b) The height must be halved to \(3\,\text{cm}\). c) The claim is correct; the area is multiplied by \(4\).

- Start with the parallelogram area formula. - Think about how changing one factor affects a product. - To keep a product constant, an increase in one factor must be offset by a decrease in the other. - Test your reasoning with simple values for \(b\) and \(h\).

1. The area is \(A=bh\). 2. In part a), \(A_{\text{new}}=(2b)h=2A\), so the area doubles. 3. In part b), \((3b)h_{\text{new}}=bh\), so \(h_{\text{new}}=\frac{1}{3}h\). 4. In part c), require \(b_{\text{new}}\left(\frac{1}{4}h\right)=2bh\). Therefore, \(b_{\text{new}}=8b\).

a) The area doubles. b) The height must be divided by \(3\). c) The base must be multiplied by \(8\).

- Identify the central rectangle and the four right triangles. - For the second method, use the \(7\times 3\) enclosing rectangle. - Use \(\frac{1}{2}bh\) for each triangle.

1. Decomposition: The central rectangle is \(3\,\text{cm}\times 3\,\text{cm}\), with area \(9\,\text{cm}^2\). The two upper triangles each have area \(\frac{1}{2}\times 2\times 2=2\,\text{cm}^2\). The two lower triangles each have area \(\frac{1}{2}\times 2\times 1=1\,\text{cm}^2\). The total is \(9+2\times 2+2\times 1=15\,\text{cm}^2\). 2. Completion: The enclosing rectangle is \(7\,\text{cm}\times 3\,\text{cm}\), with area \(21\,\text{cm}^2\). The four corner triangles total \(2\times 2+2\times 1=6\,\text{cm}^2\). The figure's area is \(21-6=15\,\text{cm}^2\).

a) \(15\,\text{cm}^2\) b) \(15\,\text{cm}^2\); both methods agree.

- Separate the arrow's shaft from its point. - Use the rectangle and triangle area formulas. - For Figure B, compare one vertical partition with one horizontal partition. - The total area should not change when the decomposition changes.

1. Figure A has a rectangular shaft measuring \(2\times 3\), with area \(6\) square units. Its triangular point has base \(6\) and height \(3\), with area \(\frac{1}{2}\times 6\times 3=9\) square units. The total area is \(6+9=15\) square units. 2. One decomposition of Figure B uses a \(2\times 5\) rectangle and a \(2\times 2\) rectangle. The area is \(2\times 5+2\times 2=10+4=14\) square units. 3. Another decomposition uses a \(4\times 2\) rectangle and a \(2\times 3\) rectangle. The area is \(4\times 2+2\times 3=8+6=14\) square units. Both decompositions agree.

a) Rectangle: \(6\) square units; triangle: \(9\) square units; total: \(15\) square units b) Both decompositions give \(14\) square units.

- Mentally split the figure at the inward corner with a horizontal segment. - Split each resulting part into a rectangle and a right triangle. - Add all the partial areas.

1. Split the figure with a horizontal segment from \((0, 2)\) to \((1, 2)\). 2. The lower part is a \(2\times2\) rectangle plus a right triangle with base \(1\) and height \(2\), so its area is \(4+1=5\) square units. 3. The upper part is a \(1\times2\) rectangle plus a right triangle with base \(1\) and height \(2\), so its area is \(2+1=3\) square units. 4. The total area is \(5+3=8\) square units.

\(8\) square units

- Imagine drawing horizontal and vertical lines through the center of the outer square. - How do the sides of the inner square divide the smaller squares? - What fraction of the outer square is covered by the inner square?

1. Draw a horizontal and a vertical line through the center of the outer square. This divides the outer square into four congruent smaller squares. 2. Each side of the inner square is a diagonal of one smaller square and divides that smaller square into two equal triangles. 3. The inner square is made of four such half-squares, so its area equals the area of two of the four smaller squares. 4. Therefore, the inner square has half the area of the outer square. 5. The outer area is \(2 \times 32 = 64\,\text{cm}^2\).

The area of the outer square is \(64\,\text{cm}^2\).

- Compare the area of the original rectangle with the area that remains. - The missing corner is itself a rectangle, so its area can determine its unknown side length. - After finding the missing height, identify every boundary segment of the L-shape before adding the perimeter.

1. The original rectangle has area \(15\times10=150\,\text{in.}^2\). 2. The removed corner has area \(150-126=24\,\text{in.}^2\). 3. Since the removed rectangle is \(6\,\text{in.}\) wide, its height is \(24\div6=4\,\text{in.}\). 4. The remaining right vertical segment is \(10-4=6\,\text{in.}\), and the remaining top horizontal segment is \(15-6=9\,\text{in.}\). 5. The L-shape perimeter is \(15+6+6+4+9+10=50\,\text{in.}\).

Missing height: \(4\,\text{in.}\) Perimeter: \(50\,\text{in.}\)