Free math worksheets

- Read each side length from the diagram. - Perimeter is the total distance around the outside edge. - Add all four side lengths once.

1. The four side lengths are \(5\,\text{cm}\), \(4\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). 2. Add all four side lengths: \(5 + 4 + 4 + 5 = 18\). 3. The perimeter is \(18\,\text{cm}\).

The perimeter of the quadrilateral is \(18\,\text{cm}\).

- Perimeter is the total length around the outside of a shape. - Count how many equal side lengths must be included. - Use multiplication to combine the \(8\) equal lengths.

1. The perimeter includes all \(8\) equal sides. 2. Multiply the number of sides by the length of each side: \(8 \times 3 = 24\).

The perimeter is \(24\,\text{cm}\).

- Trace the entire outside boundary of each figure. - Count each horizontal or vertical step between neighboring pegs as \(1\) unit. - Add the boundary lengths, then compare the totals.

1. The perimeter of Figure A is \(4 + 1 + 3 + 2 + 1 + 3 = 14\) units. 2. The perimeter of Figure B is \(1 + 2 + 1 + 1 + 3 + 1 + 1 + 2 = 12\) units. 3. The perimeter of Figure C is \(3 + 1 + 1 + 1 + 1 + 1 + 1 + 3 = 12\) units. 4. Figure A has the greatest perimeter.

Figure A: \(14\) units Figure B: \(12\) units Figure C: \(12\) units Figure A has the greatest perimeter, \(14\) units.

- Perimeter is the distance around the outside of a figure. - Count the unit-length segments between pegs along each boundary. - Add all the boundary lengths before comparing.

1. Figure a) is a rectangle with side lengths \(3\) units and \(2\) units. Its perimeter is \(3 + 2 + 3 + 2 = 10\) units. 2. Tracing Figure b) gives side lengths \(2\), \(1\), \(1\), \(1\), \(1\), and \(2\) units. Its perimeter is \(2 + 1 + 1 + 1 + 1 + 2 = 8\) units. 3. Since \(10 > 8\), Figure a) has the greater perimeter.

Figure a) has a perimeter of \(10\) units. Figure b) has a perimeter of \(8\) units. Figure a) has the greater perimeter.

- Add the number of unit squares in each column. - For perimeter, count every unit-length segment around the outside boundary.

1. The four columns have heights \(4\), \(3\), \(2\), and \(1\). The area is \(4 + 3 + 2 + 1 = 10\) square units. 2. The bottom and left sides are each \(4\) units long. The stepped boundary has four horizontal unit segments and four vertical unit segments. The perimeter is \(4 + 4 + 4 + 4 = 16\) units.

a) \(10\) square units b) \(16\) units

- Read the side length from the figure. - A square has four sides of equal length. - Find the total distance around all four sides.

1. The figure shows a square with side length \(70\,\text{cm}\). 2. A square has four equal sides, so its perimeter is \(4 \times 70 = 280\,\text{cm}\).

\(280\,\text{cm}\) of lace trim is needed.

- Count the rows and columns in the tile array. - Multiply rows by columns to find the area. - Only the outside tile edges count toward the perimeter.

1. The diagram has \(5\) rows and \(8\) columns, so the area is \(5 \times 8 = 40\) tiles. 2. The rectangle has two sides of length \(8\) and two sides of length \(5\). Its perimeter is \(8 + 5 + 8 + 5 = 26\) tile-side lengths.

Area: \(40\) tiles Perimeter: \(26\) tile-side lengths

- Trace every side that forms the rectangle's boundary. - Check whether each side length occurs more than once.

1. A rectangle has two sides of each length, so Sam counted only half of the boundary. 2. Add all four sides: \(9 + 4 + 9 + 4 = 26\).

Sam counted only one length and one width. The correct perimeter is \(26\,\text{cm}\).

- The same side length occurs four times around a square. - Compare each completed perimeter with the side length in the same row.

1. A square has four sides of the same length. 2. The perimeters are \(4 \times 2 = 8\) inches, \(4 \times 5 = 20\) inches, and \(4 \times 9 = 36\) inches. 3. In every row, the perimeter is four times the side length.

<table><tr><th>Side length</th><th>Perimeter</th></tr><tr><td>\(2\,\text{in.}\)</td><td>\(8\,\text{in.}\)</td></tr><tr><td>\(5\,\text{in.}\)</td><td>\(20\,\text{in.}\)</td></tr><tr><td>\(9\,\text{in.}\)</td><td>\(36\,\text{in.}\)</td></tr></table> A square's perimeter is four times its side length.

- Count how many equal sides each regular shape has. - Compare the total distances around the two shapes after finding them.

1. The pentagon's perimeter is \(5 \times 4 = 20\) centimeters. 2. The square's perimeter is \(4 \times 4 = 16\) centimeters. 3. The difference is \(20 - 16 = 4\) centimeters.

The pentagon's perimeter is \(20\,\text{cm}\), and the square's perimeter is \(16\,\text{cm}\). The pentagon's perimeter is \(4\,\text{cm}\) greater.

- Think of the perimeter as a total divided into equal side lengths. - Find how many \(2\)-foot lengths fit in \(24\) feet.

1. The perimeter is made of equal \(2\)-foot side lengths. 2. The number of sides is \(24 \div 2 = 12\).

The polygon has \(12\) sides.

- Count all sides of each figure. - Compare the two boundary totals after calculating them.

1. The square's perimeter is \(4 \times 6 = 24\) units. 2. The rectangle's perimeter is \(8 + 4 + 8 + 4 = 24\) units. 3. The two different shapes have the same perimeter.

Both perimeters are \(24\) units. The square and rectangle have the same perimeter.

- Decide whether the new segment lies on the outside boundary. - Count only lengths traveled in one trip around the edge.

1. Perimeter includes only the outside boundary, not an interior diagonal. 2. The six outside sides each measure \(6\,\text{cm}\). 3. The perimeter is \(6 \times 6 = 36\,\text{cm}\).

The diagonal does not change the perimeter. The perimeter is \(36\,\text{cm}\).

- Separate the sides already traveled from the side not yet traveled. - A complete perimeter includes every side exactly once.

1. Moving along \(8\) sides covers \(8 \times 3 = 24\,\text{cm}\). 2. One side remains, so \(3\,\text{cm}\) more completes the perimeter. 3. The full perimeter is \(24 + 3 = 27\,\text{cm}\).

The token has moved \(24\,\text{cm}\) and needs \(3\,\text{cm}\) more to complete the perimeter.

- Count each side of the quadrilateral exactly once. - Add the four side lengths to find the total distance around the shape. - Keep meters as the unit in the answer.

1. Perimeter is the total length of all four sides. 2. Add: \(5 + 6 + 7 + 8 = 26\).

The perimeter is \(26\,\text{m}\).

- Rotation changes orientation, not the lengths of the boundary sides. - Add the same four side lengths before or after the turn.

1. Turning the rectangle does not change any side length. 2. The perimeter is \(11 + 3 + 11 + 3 = 28\) units.

No. The perimeter remains \(28\) units.

- Compare the total length of the two new segments with the old side length. - A greater number of labeled boundary pieces does not necessarily mean a longer boundary.

1. The original perimeter is \(10 \times 3 = 30\) units. 2. The split side still has total length \(1 + 2 = 3\) units. 3. Dividing one side into two labeled segments changes the segment count but not the boundary length.

The perimeter has not changed. It is still \(30\) units.

- Count the unit squares inside each figure to find its area. - To find perimeter, trace the entire outside boundary and count each unit-length segment. - Be sure to count the inward and outward segments of Figure C.

1. Figure A is a \(4 \times 2\) rectangle, so its area is \(8\) square units. 2. Counting or decomposing Figure B gives an area of \(8\) square units. 3. Counting or decomposing Figure C gives an area of \(8\) square units. Therefore, all three figures have the same area. 4. The perimeter of Figure A is \(4 + 2 + 4 + 2 = 12\) units. 5. The perimeter of Figure B is \(5 + 1 + 2 + 1 + 3 + 2 = 14\) units. 6. The perimeter of Figure C is \(4 + 3 + 1 + 2 + 2 + 2 + 1 + 3 = 18\) units. 7. Figure C has the greatest perimeter.

a) Each figure has an area of \(8\) square units. b) Figure A: \(12\) units Figure B: \(14\) units Figure C: \(18\) units Figure C has the greatest perimeter.

- Count the unit squares inside each figure to find area. - Trace the entire boundary and count each unit-length segment to find perimeter. - Compare the areas separately from the perimeters.

1. Rectangle 1 has \(4\) unit squares in each of \(3\) rows, so its area is \(4 + 4 + 4 = 12\) square units. 2. Figure 2 has \(4\) unit squares in the bottom row, \(1\) in the middle row, and \(4\) in the top row, so its area is \(4 + 1 + 4 = 9\) square units. 3. The perimeter of Rectangle 1 is \(4 + 3 + 4 + 3 = 14\) units. 4. Adding the boundary lengths of Figure 2 gives \(4 + 1 + 3 + 1 + 3 + 1 + 4 + 3 = 20\) units. 5. Rectangle 1 has the greater area, while Figure 2 has the greater perimeter.

a) Rectangle 1 has an area of \(12\) square units. Figure 2 has an area of \(9\) square units. Rectangle 1 has the greater area. b) Figure 2 has the greater perimeter: \(20\) units compared with \(14\) units for Rectangle 1.

- Count the rows and columns in each array. - Perimeter uses only the outside edge lengths. - Compare the two perimeter totals after finding each one.

1. Rectangle A has \(3\) rows and \(8\) columns, so its perimeter is \(3 + 8 + 3 + 8 = 22\) tile-side lengths. 2. Rectangle B has \(4\) rows and \(6\) columns, so its perimeter is \(4 + 6 + 4 + 6 = 20\) tile-side lengths. 3. Rectangle B has the smaller perimeter. Both arrays contain \(24\) tiles, so this also shows that equal areas can have different perimeters.

Rectangle B has the smaller perimeter: \(20\) tile-side lengths. Rectangle A has a perimeter of \(22\) tile-side lengths.

- The perimeter is the sum of all the side lengths. - Find how much of the perimeter is left after accounting for the two known sides.

1. Add the known sides: \(8 + 10 = 18\). 2. Subtract from the perimeter: \(29 - 18 = 11\).

The third side is \(11\,\text{cm}\) long.

- Only lengths on the outside boundary count toward perimeter. - Decide what happens to the two sides where the triangles touch.

1. The joined \(6\)-centimeter sides are inside the new figure, so they are not part of its perimeter. 2. The outside boundary has four sides of \(5\,\text{cm}\). 3. Multiply: \(4 \times 5 = 20\).

The perimeter of the new figure is \(20\,\text{cm}\).

- Identify which two side lengths are used in each route. - Check whether the two routes together include every side of the rectangle exactly once. - Compare the combined route length with the complete boundary length.

1. Each route uses one \(9\)-meter side and one \(5\)-meter side. 2. Each student walks \(9 + 5 = 14\) meters. 3. The two routes use all four boundary sides exactly once, so together they measure \(14 + 14 = 28\) meters. 4. This equals the perimeter: \(9 + 5 + 9 + 5 = 28\) meters.

Each student walks \(14\,\text{m}\). Together, the two routes make the full \(28\)-meter perimeter.

- First determine the perimeter shared by both shapes. - Then find the part of the triangle’s perimeter not accounted for by its known sides.

1. Find the square’s perimeter: \(4 \times 6 = 24\) centimeters. 2. The known triangle sides total \(7 + 8 = 15\) centimeters. 3. Subtract: \(24 - 15 = 9\) centimeters.

The third side of the triangle is \(9\,\text{cm}\).

- Find the perimeter shared by both rectangles. - After accounting for both known length sides, divide the remaining boundary equally between the widths.

1. Rectangle A's perimeter is \(8 + 4 + 8 + 4 = 24\) units. 2. Rectangle B's two length sides total \(7 + 7 = 14\) units. 3. Its two width sides total \(24 - 14 = 10\) units. 4. Each width is \(10 \div 2 = 5\) units.

Rectangle B's width is \(5\) units.

- A square needs four equal side lengths. - Check whether the total length can be separated into four equal whole-number parts.

1. A square would divide the \(22\)-unit perimeter into \(4\) equal whole-number sides. 2. Four sides of \(5\) units use \(20\) units, leaving \(2\) units. 3. Therefore, \(22\) cannot be shared equally into \(4\) whole-number side lengths.

No. A \(22\)-unit loop cannot form a square with whole-number side lengths.

- Find the first triangle’s perimeter from its three equal sides. - Think about what happens to the total when each of the three side lengths is doubled. - Compare the second perimeter with twice the first perimeter.

1. The first perimeter is \(3 \times 8 = 24\,\text{in.}\). 2. Every side of the second triangle is twice as long, so its perimeter is twice as long. 3. Double the first perimeter: \(24 + 24 = 48\,\text{in.}\).

The perimeters are \(24\,\text{in.}\) and \(48\,\text{in.}\). Doubling every side doubles the perimeter.

- Repeated side lengths still belong to separate sides of the polygon. - Count how many times each length occurs before adding.

1. Perimeter includes every side, even when several sides have the same length. 2. There is one \(4\)-centimeter side, two \(5\)-centimeter sides, and three \(7\)-centimeter sides. 3. The perimeter is \(4 + (2 \times 5) + (3 \times 7) = 4 + 10 + 21 = 35\) centimeters.

The student counted each different length only once instead of counting every side. The perimeter is \(35\,\text{cm}\).

- Trace only the outside boundary of each joined figure. - Use the labeled dimensions to account for every outside segment. - Compare the two complete boundary totals.

1. Figure a) is a \(6\)-by-\(2\) rectangle, so its perimeter is \(6 + 2 + 6 + 2 = 16\) units. 2. For figure b), add the six outside side lengths: \(4 + 2 + 2 + 2 + 2 + 4 = 16\) units. 3. The two figures have equal perimeters.

Both figures have a perimeter of \(16\) units.

- Track how each side change affects the total boundary length. - Combine the increase and decrease before applying the net change.

1. Lengthening one side adds \(3\) centimeters to the perimeter. 2. Shortening another side removes \(1\) centimeter. 3. The net change is \(3 - 1 = 2\) centimeters. 4. The new perimeter is \(25 + 2 = 27\) centimeters.

The new perimeter is \(27\,\text{cm}\).

- The same side-length change occurs four times. - Use the given square only to verify the general reasoning.

1. A square has \(4\) sides, and each side increases by \(2\) units. 2. The perimeter increase is \(4 \times 2 = 8\) units. 3. For the check, the original perimeter is \(4 \times 5 = 20\) units and the new perimeter is \(4 \times 7 = 28\) units; \(28 - 20 = 8\).

The perimeter increases by \(8\) units.

- Think about how one total is shared among equal sides. - Check the proposed side length by rebuilding the perimeter.

1. The perimeter is shared equally among \(5\) sides; the number of sides is not subtracted from the perimeter. 2. Divide: \(35 \div 5 = 7\,\text{cm}\). 3. A check is \(5 \times 7 = 35\,\text{cm}\).

Morgan should divide by \(5\), not subtract \(5\). Each side is \(7\,\text{cm}\) long.

- Shared sides become interior and are not part of the new boundary. - Determine the dimensions of the combined rectangle before finding its perimeter.

1. The joined lengths combine to make \(5 + 4 = 9\) units. 2. The larger rectangle is \(9\) units by \(3\) units. 3. Its perimeter is \(9 + 3 + 9 + 3 = 24\) units.

The larger rectangle has a perimeter of \(24\) units.

- A correct pair must satisfy both the perimeter total and the equal-length condition. - Find the unused perimeter and divide it into two equal parts.

1. The four known sides total \(6 + 8 + 10 + 12 = 36\,\text{cm}\). 2. The two missing sides must total \(50 - 36 = 14\,\text{cm}\). 3. The proposed lengths \(6\,\text{cm}\) and \(8\,\text{cm}\) have the correct total, but they do not satisfy the condition that the two sides are equal. 4. Splitting \(14\,\text{cm}\) equally gives \(14 \div 2 = 7\,\text{cm}\) for each side.

The student's lengths have the correct total but are not equal. The missing sides are \(7\,\text{cm}\) and \(7\,\text{cm}\).

- Find each complete perimeter using its own number of sides and side length. - Compare the products rather than comparing only the number of sides.

1. The \(10\)-sided display has perimeter \(10 \times 4 = 40\,\text{in.}\). 2. The octagonal display has perimeter \(8 \times 5 = 40\,\text{in.}\). 3. Their perimeters are equal, so having more sides does not by itself guarantee a greater perimeter.

Both perimeters are \(40\,\text{in.}\). No, more sides do not necessarily mean a greater perimeter.

- Find the original boundary length before changing any sides. - Replace the total length of the two old sides with the shortcut's length.

1. The original perimeter is \(4 + 5 + 6 + 7 + 8 + 9 = 39\) units. 2. The two replaced sides total \(6 + 7 = 13\) units. 3. Remove those sides and add the shortcut: \(39 - 13 + 10 = 36\) units.

The new perimeter is \(36\) units.

- Compare the sum of the known sides with the original perimeter. - A change to one side changes the total boundary by the same amount.

1. The known sides total \(7 + 9 + 12 = 28\) units. 2. The fourth side is \(36 - 28 = 8\) units. 3. Lengthening one side by \(3\) units increases the perimeter by \(3\) units, so the new perimeter is \(36 + 3 = 39\) units.

The fourth side is \(8\) units long. The new perimeter is \(39\) units.

- First find the length of the complete rectangular boundary. - Sections left open are not included in the border length.

1. The complete perimeter is \(8 + 5 + 8 + 5 = 26\) units. 2. The two openings total \(2 + 1 = 3\) units. 3. The border length is \(26 - 3 = 23\) units.

The display uses \(23\) units of border.

- Start with the complete perimeter of the regular polygon. - Replace the old side length with the total length of the new boundary segments.

1. The original perimeter is \(5 \times 6 = 30\) units. 2. Removing the original side leaves \(30 - 6 = 24\) units. 3. The replacement segments add \(2 + 5 = 7\) units. 4. The new perimeter is \(24 + 7 = 31\) units.

The new perimeter is \(31\) units.

- Decide whether the triangles share a segment with measurable length. - Find the perimeter of one triangle. - Add the two perimeters when no boundary edge is hidden.

1. Each triangle has perimeter \(3 \times 5 = 15\) units. 2. A single touching point has no side length, so no boundary segment becomes internal. 3. The total outside boundary length is \(15 + 15 = 30\) units.

The total outside boundary length is \(30\) units. The touching point has no length, so it does not remove any part of either boundary.

- Perimeter depends on the sum of the side lengths. - Compare the collection of lengths, not the order in which they appear.

1. Each octagon has four \(6\)-unit sides, contributing \(4 \times 6 = 24\) units. 2. Each also has four \(4\)-unit sides, contributing \(4 \times 4 = 16\) units. 3. The perimeter is \(24 + 16 = 40\) units for each octagon. 4. Changing the order of the side lengths does not change their sum.

Yes. Both octagons have a perimeter of \(40\) units.

- First find the sum of one length and one width. - List the whole-number pairs systematically so you do not repeat a pair in reverse order.

1. The perimeter has two lengths and two widths, so one length plus one width is \(24 \div 2 = 12\) meters. 2. The whole-number pairs with a sum of \(12\) are \(1\) and \(11\), \(2\) and \(10\), \(3\) and \(9\), \(4\) and \(8\), \(5\) and \(7\), and \(6\) and \(6\). 3. Reversing a pair only turns the same rectangle, so there are six different pairs.

The possible dimensions are \(1\,\text{m} \times 11\,\text{m}\), \(2\,\text{m} \times 10\,\text{m}\), \(3\,\text{m} \times 9\,\text{m}\), \(4\,\text{m} \times 8\,\text{m}\), \(5\,\text{m} \times 7\,\text{m}\), and \(6\,\text{m} \times 6\,\text{m}\).

- List whole-number side-length pairs whose product gives the area. - Check the perimeter of each possible rectangle rather than choosing from area alone.

1. Whole-number factor pairs of \(18\) with both factors greater than \(1\) are \(2\) and \(9\), and \(3\) and \(6\). 2. The \(2\)-by-\(9\) rectangle has perimeter \(2 + 9 + 2 + 9 = 22\) units. 3. The \(3\)-by-\(6\) rectangle has perimeter \(3 + 6 + 3 + 6 = 18\) units, so it does not fit.

The side lengths are \(2\) units and \(9\) units.

- Read the rectangle and corner-cut lengths from the figure. - Compare the boundary segments removed by the cut with the new boundary segments it creates. - Check whether the total boundary length changes.

1. The original rectangle is \(10\) units by \(3\) units, so its perimeter is \(10 + 3 + 10 + 3 = 26\) units. 2. The corner cut removes two \(1\)-unit pieces from the old boundary. 3. The cut also creates two new \(1\)-unit boundary pieces. 4. The removed and added lengths are equal, so the perimeter remains \(26\) units.

The new figure has a perimeter of \(26\) units.

- Calculate both complete boundaries rather than comparing only one side. - Track how many sides increase and how many sides decrease.

1. Rectangle A's perimeter is \(9 + 5 + 9 + 5 = 28\) units. 2. Rectangle B's perimeter is \(10 + 4 + 10 + 4 = 28\) units. 3. Increasing each length side by \(1\) adds \(2\) units total, while decreasing each width side by \(1\) removes \(2\) units total.

Both perimeters are \(28\) units. The added boundary length and removed boundary length are equal, so the perimeter does not change.

- Read the square side length and all detour lengths from the figure. - Start with the original square's perimeter. - Replace the removed straight segment with the three detour segments.

1. The original square has side length \(9\,\text{ft}\), so its perimeter is \(4 \times 9 = 36\,\text{ft}\). 2. The old \(2\)-foot boundary section is replaced. 3. The detour adds \(1 + 2 + 1 = 4\) feet of boundary in its place. 4. The new perimeter is \(36 - 2 + 4 = 38\,\text{ft}\).

The new perimeter is \(38\,\text{ft}\).

- Reconstruct the distance of two complete laps first. - The skipped side belongs to the second complete perimeter. - Split the two-lap distance into two equal boundary lengths.

1. Add the skipped side to the traveled distance: \(53 + 7 = 60\) units. 2. This total represents two complete laps around the polygon. 3. Divide by \(2\): \(60 \div 2 = 30\) units.

The polygon's perimeter is \(30\) units.

- Find how much perimeter is already accounted for by the known side groups. - Use the remaining perimeter to determine how many equal sides are missing.

1. The known sides total \(3 \times 2 + 3 \times 5 = 6 + 15 = 21\) units. 2. The remaining perimeter is \(42 - 21 = 21\) units. 3. The number of \(7\)-unit sides is \(21 \div 7 = 3\). 4. The polygon has \(3 + 3 + 3 = 9\) sides altogether.

There are \(3\) sides of length \(7\) units, and the polygon has \(9\) sides altogether.

- Track which boundary segments belong to one smaller polygon and which belong to both. - The dividing segment appears once in each new perimeter.

1. The original outside boundary still contributes \(48\) units altogether. 2. The new \(7\)-unit segment is part of the boundary of each smaller polygon, so it is counted twice. 3. The sum is \(48 + 7 + 7 = 62\) units.

The two smaller perimeters have a sum of \(62\) units.