Free math worksheets

- Can you read the diagonal lengths from the coordinates? - How are the diagonals positioned on the coordinate plane? - What formula uses the two diagonals to find the area of a rhombus? - Can you decompose the rhombus into two or four congruent triangles?

1. Diagonal \(AC\) is horizontal, and diagonal \(BD\) is vertical. 2. Their lengths are \(AC=10-2=8\) units and \(BD=8-2=6\) units. 3. Use the rhombus area formula: \(A=\frac{1}{2}d_1d_2\). 4. Therefore, \(A=\frac{1}{2}\times8\times6=24\) square units.

The area of the rhombus is \(24\) square units.

- How do you find the distance between two points on a horizontal line? - A height is perpendicular to its base. - Use the triangle area relationship after finding the base and height.

1. Segment \(AB\) is horizontal, so its length is \(7-1=6\,\text{cm}\). 2. The height is the vertical distance from \(C\) to line \(AB\): \(6-2=4\,\text{cm}\). 3. The area is \(A=\frac{1}{2}bh=\frac{1}{2}\times6\times4=12\,\text{cm}^2\).

a) \(AB=6\,\text{cm}\), \(h=4\,\text{cm}\) b) \(A=12\,\text{cm}^2\)

- Which sides are parallel? - How do you find the length of a vertical segment from its coordinates? - What is the distance between the two vertical lines? - Use the trapezoid area relationship.

1. Sides \(AB\) and \(CD\) are vertical and parallel. 2. Their lengths are \(AB=6-1=5\) units and \(CD=5-2=3\) units. 3. The height is the horizontal distance between \(x=2\) and \(x=5\), which is \(3\) units. 4. The area is \(A=\frac{1}{2}(5+3)\times3=12\) square units.

The area is \(12\) square units.

- Which sides of the quadrilateral are parallel? - Use the x-coordinates to find each horizontal side length. - Use the y-coordinates to find the perpendicular distance between the parallel sides.

1. The horizontal parallel sides have lengths \(10-2=8\) units and \(7-4=3\) units. 2. Their vertical distance is \(5-1=4\) units, so the height is \(4\) units. 3. The area is \(A=\frac{1}{2}(8+3)\times4=22\) square units.

The area of the trapezoid is \(22\) square units.

- Identify the horizontal and vertical sides. - The distance between one pair of opposite sides equals the length of the other pair. - Subtract the coordinates that differ.

1. Sides \(\overline{AB}\) and \(\overline{CD}\) lie on the horizontal lines \(y=1\) and \(y=6\). Their distance is \(6-1=5\) units. 2. Sides \(\overline{AD}\) and \(\overline{BC}\) lie on the vertical lines \(x=2\) and \(x=11\). Their distance is \(11-2=9\) units.

The distance between \(\overline{AB}\) and \(\overline{CD}\) is \(5\) units. The distance between \(\overline{AD}\) and \(\overline{BC}\) is \(9\) units.

- Adjacent vertices on a horizontal or vertical side share one coordinate. - Match the x-coordinate from one opposite vertex with the y-coordinate from the other. - For horizontal or vertical segments, subtract the changing coordinates.

1. Point \(B\) shares the x-coordinate of \(C\) and the y-coordinate of \(A\), so \(B=(9, 1)\). 2. Point \(D\) shares the x-coordinate of \(A\) and the y-coordinate of \(C\), so \(D=(2, 6)\). 3. The horizontal side length is \(9-2=7\) units. 4. The vertical side length is \(6-1=5\) units.

a) \(B=(9, 1)\) and \(D=(2, 6)\) b) The side lengths are \(7\) units and \(5\) units.

- Compare the coordinates of \(A\) and \(B\). - Opposite sides of a parallelogram are parallel and equal in length. - Apply the movement from \(A\) to \(B\) starting at \(D\).

1. Points \(A\) and \(B\) have the same y-coordinate, so \(\overline{AB}\) is horizontal. Its length is \(8-1=7\) units. 2. Opposite sides of a parallelogram are parallel and equal in length. Therefore, \(\overline{DC}\) is horizontal and has length \(7\). 3. Starting at \(D(3, 6)\), move \(7\) units right to get \(C=(10, 6)\).

a) \(7\) units b) \(C=(10, 6)\). Moving from \(A\) to \(B\) is \(7\) units right, so move \(7\) units right from \(D\).

- Compare the coordinates of \(P\) and \(Q\), and then compare the coordinates of \(P\) and \(R\). - Equal y-coordinates make a horizontal segment. - Equal x-coordinates make a vertical segment.

1. Connecting three distinct points forms a triangle. 2. Segment \(\overline{PQ}\) is horizontal because \(P\) and \(Q\) have the same y-coordinate. 3. Segment \(\overline{PR}\) is vertical because \(P\) and \(R\) have the same x-coordinate. 4. A horizontal segment and a vertical segment are perpendicular, so they form a right angle at their common endpoint, \(P\). Therefore, the polygon is a right triangle.

a) A right triangle is formed. b) \(\overline{PQ}\) is horizontal and \(\overline{PR}\) is vertical, so they are perpendicular. The right angle is at \(P\).

- Which vertices must line up horizontally and vertically? - All four sides of a square have the same length. - Subtract the corresponding coordinates of adjacent vertices.

1. Point \(D\) must have the same x-coordinate as \(A\) and the same y-coordinate as \(C\). Therefore, \(D=(2, 5)\). 2. The horizontal side length is \(5.5-2=3.5\) units. 3. The vertical side length is \(5-1.5=3.5\) units, confirming that the figure is a square.

a) \(D=(2, 5)\) b) The side length is \(3.5\) grid units.

- Opposite sides of a rectangle are parallel. - Use coordinate differences to find the side lengths. - Recall the area relationship for a rectangle.

1. To complete the axis-aligned rectangle, \(D\) must share the x-coordinate of \(A\) and the y-coordinate of \(C\). Thus, \(D=(2, 4)\). 2. The length is \(7-2=5\,\text{cm}\), and the width is \(4-1=3\,\text{cm}\). 3. The area is \(5\times3=15\,\text{cm}^2\).

a) \(D=(2, 4)\) b) \(15\,\text{cm}^2\)

- Read the coordinates of the three labeled vertices from the grid. - Opposite sides of a square are parallel, and all four sides have equal length. - The missing vertex must line up vertically with one shown vertex and horizontally with another.

1. Read the coordinates from the graph: \(P(3, 2)\), \(Q(-2, 2)\), and \(R(-2, -3)\). 2. Segment \(PQ\) is horizontal and has length \(3-(-2)=5\) units. Segment \(QR\) is vertical and also has length \(2-(-3)=5\) units. 3. To complete the square, \(S\) has the same x-coordinate as \(P\) and the same y-coordinate as \(R\). 4. Therefore, \(S=(3, -3)\).

\(S=(3, -3)\)

- Count the sides of the closed outline. - A vertical side has endpoints with the same x-coordinate; use the y-coordinates to find its length. - The highest vertex has the greatest y-coordinate.

1. The closed outline has five sides, so it is a pentagon. 2. Sides \(BC\) and \(EA\) are vertical because each pair of endpoints has the same x-coordinate. Each extends from \(y=-2\) to \(y=1\), so each has length \(3\) units. 3. The highest vertex is \(D=(0, 3)\).

a) Pentagon b) \(BC\) and \(EA\); each is \(3\) units long. c) \((0, 3)\)

- Read the coordinates of the triangle's vertices and point \(P\) from the graph. - What point is halfway along the horizontal base \(\overline{AB}\)? - Check whether \(P\) lies on the segment from that midpoint to vertex \(C\).

1. Read the graph: \(A(1,1)\), \(B(7,1)\), \(C(4,5)\), and \(P(4,2)\). 2. The midpoint of \(\overline{AB}\) is \(M(4,1)\). 3. Segment \(\overline{MC}\) lies inside the triangle. Point \(P(4,2)\) lies on this segment because its x-coordinate is \(4\) and its y-coordinate lies between \(1\) and \(5\). 4. Therefore, \(P\) lies inside the triangle.

Point \(P\) lies inside the nature preserve.

- Compare the coordinate change from \(B\) to \(C\) with the change needed from \(A\) to \(D\). - The height to horizontal base \(AB\) is a vertical distance. - Find the base length from the x-coordinates before using the parallelogram area formula.

1. For a), moving from \(B(7, 1)\) to \(C(8, 4)\) changes the coordinates by \((1, 3)\). Apply the same change to \(A(2, 1)\): \(D=(3, 4)\). 2. For b), side \(AB\) lies on \(y=1\), and side \(DC\) lies on \(y=4\). Their perpendicular distance is \(4-1=3\) units, so \(h=3\) units. 3. For c), the base is \(7-2=5\) units. Therefore, \(A=bh=5\times3=15\) square units.

a) \(D(3, 4)\) b) \(h=3\) units c) \(A=15\) square units

- 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.

- How can you find the distance between two points on a horizontal line? - What does the height of a parallelogram represent? - Which measurements determine the area in \(A=bh\)?

1. Since \(A\) and \(B\) have the same y-coordinate, \(AB=7-1=6\,\text{cm}\). The height is the perpendicular distance between the horizontal lines \(y=1\) and \(y=5\), so \(h=5-1=4\,\text{cm}\). 2. The area is \(A=bh=6\times4=24\,\text{cm}^2\). 3. Sliding the upper side horizontally changes the slant of the parallelogram but not its base or height. Therefore, the area remains \(24\,\text{cm}^2\).

a) \(b=6\,\text{cm}\), \(h=4\,\text{cm}\) b) \(A=24\,\text{cm}^2\) c) The area stays \(24\,\text{cm}^2\) because the base and perpendicular height do not change.

- Which side lies exactly on a vertical grid line? - How do you measure the height when the base is vertical? - Compare the base lengths and the distances between the opposite sides.

1. Side \(AB\) is vertical, so its length is \(8-2=6\,\text{cm}\). 2. The corresponding height is the horizontal distance between \(x=2\) and \(x=6\), which is \(6-2=4\,\text{cm}\). 3. The area is \(A=bh=6\times4=24\,\text{cm}^2\). 4. The second parallelogram has the same base \(AB\) and the same horizontal distance of \(4\,\text{cm}\) between its parallel sides. Therefore, it has the same area.

a) \(A=24\,\text{cm}^2\) b) The second parallelogram has the same area, \(24\,\text{cm}^2\), because its base and corresponding height are unchanged.

- First find the area using the side whose length and height are easiest to read from the coordinates. - The area is the same no matter which side is chosen as the base. - Use the known area with side \(QR\) to recover its corresponding perpendicular height.

1. Use horizontal side \(PQ\) as the base. Its length is \(11-1=10\,\text{cm}\), and the vertical height is \(5-1=4\,\text{cm}\). 2. The area is \(A=10\times4=40\,\text{cm}^2\). 3. Using side \(QR\) as the base, \(40=5\times h_{QR}\). Therefore, \(h_{QR}=40\div5=8\,\text{cm}\).

a) \(A=40\,\text{cm}^2\) b) \(h_{QR}=8\,\text{cm}\)

- How do you find the length of a horizontal segment from its coordinates? - Recall the area formulas for a parallelogram and a triangle. - How do you find a height when the base is horizontal? - Compare the y-coordinates to find the perpendicular distances.

1. Segment \(AB\) has length \(10-2=8\) units. The parallelogram's height is \(5-2=3\) units, so its area is \(8\times3=24\) square units. 2. Triangle \(ABE\) has the same base, \(8\) units. Its height is \(8-2=6\) units, so its area is \(\frac{1}{2}\times8\times6=24\) square units. 3. The areas are equal. The triangle's height is twice the parallelogram's height, which offsets the factor \(\frac{1}{2}\) in the triangle area formula.

a) \(24\) square units b) \(24\) square units c) The areas are equal. The figures have the same base, and the triangle's height is twice the parallelogram's height.

- Use half the sum of the parallel side lengths times the height. - Which coordinate differences give the lengths of the parallel sides? - Does the height change when the two upper points move horizontally?

1. In \(ABCD\), the parallel sides have lengths \(AB=9-1=8\) units and \(CD=6-4=2\) units. The height is \(4-1=3\) units. Thus, \(A=\frac{1}{2}(8+2)\times3=15\) square units. 2. In \(ABC_1D_1\), the lower base is still \(8\) units, the upper base is \(7-3=4\) units, and the height is still \(3\) units. Thus, \(A_1=\frac{1}{2}(8+4)\times3=18\) square units. 3. The increase is \(18-15=3\) square units.

a) \(15\) square units b) \(18\) square units c) The area increased by \(3\) square units.

- Use the least and greatest x- and y-coordinates to define the surrounding rectangle. - Find the leg lengths of each corner triangle by subtracting coordinates. - Subtract the four triangle areas from the rectangle's area.

1. The surrounding rectangle has vertices \((1, 1)\), \((8, 1)\), \((8, 6)\), and \((1, 6)\). 2. Its width is \(8-1=7\) units, and its height is \(6-1=5\) units. Its area is \(7\times5=35\) square units. 3. The four outside triangles have areas \(\frac{1}{2}\times4\times1=2\), \(\frac{1}{2}\times3\times3=4.5\), \(\frac{1}{2}\times5\times2=5\), and \(\frac{1}{2}\times2\times4=4\) square units. 4. Their total area is \(2+4.5+5+4=15.5\) square units, so the quadrilateral's area is \(35-15.5=19.5\) square units.

1. The rectangle's vertices are \((1, 1)\), \((8, 1)\), \((8, 6)\), and \((1, 6)\). 2. The rectangle's area is \(35\) square units. 3. The four triangle areas are \(2\), \(4.5\), \(5\), and \(4\) square units. 4. The area of \(ABCD\) is \(19.5\) square units.

- Compare the x-coordinates of the endpoints of the two horizontal sides. - Imagine vertical segments from the upper vertices to the lower base. - Add the areas of the resulting rectangle and two triangles.

1. The lower base \(AB\) has length \(10-2=8\) units, the upper base \(CD\) has length \(8-4=4\) units, and the height is \(5-1=4\) units. 2. Vertical segments from \(C\) and \(D\) to \(\overline{AB}\) form a central \(4\times4\) rectangle and two congruent side triangles. 3. The rectangle's area is \(4\times4=16\) square units. 4. Each side triangle has base \(2\) units and height \(4\) units, so each has area \(\frac{1}{2}\times2\times4=4\) square units. 5. The total area is \(16+4+4=24\) square units.

Split the quadrilateral into a central \(4\times4\) rectangle and two congruent triangles, each with base \(2\) units and height \(4\) units. The area is \(24\) square units.

- Look for two vertices that lie on the same horizontal line. - Use a horizontal segment to divide the pentagon. - Identify the triangle and trapezoid formed by the segment.

1. Draw horizontal segment \(\overline{TR}\) to split the pentagon into triangle \(TSR\) and trapezoid \(PQRT\). 2. For \(\triangle TSR\), the base is \(9-(-1)=10\) units and the height is \(8-5=3\) units. Its area is \(\frac{1}{2}\times10\times3=15\) square units. 3. The trapezoid has bases \(7-1=6\) units and \(10\) units and height \(5-2=3\) units. Its area is \(\frac{1}{2}(6+10)\times3=24\) square units. 4. The pentagon's area is \(15+24=39\) square units.

The area of pentagon \(PQRST\) is \(39\) square units.

- Use the least and greatest coordinates to draw a surrounding rectangle. - Find the legs of each outside right triangle from coordinate differences. - Subtract the outside areas from the rectangle's area.

1. The surrounding rectangle extends from \(x=2\) to \(x=8\) and from \(y=2\) to \(y=7\). Its dimensions are \(6\times5\), so its area is \(30\) square units. 2. The lower-right triangle has area \(\frac{1}{2}\times6\times1=3\) square units. 3. The upper-right triangle has area \(\frac{1}{2}\times2\times4=4\) square units, and the upper-left triangle has area \(\frac{1}{2}\times4\times2=4\) square units. 4. No triangle must be removed at the lower-left corner because \(K\) is a corner of the rectangle and \(KN\) lies on its left side. 5. The quadrilateral's area is \(30-3-4-4=19\) square units.

The area of quadrilateral \(KLMN\) is \(19\) square units.

- How do you find the distance between two vertically aligned points? - Use the triangle area formula to find the height. - If the base is vertical, in which direction is its height measured? - Move the required distance to either side of the base line.

1. Since \(P\) and \(Q\) have the same x-coordinate, \(PQ=7-1=6\,\text{cm}\). 2. Use \(12=\frac{1}{2}\times6\times h\). This gives \(12=3h\), so \(h=4\,\text{cm}\). 3. The base lies on \(x=1\), so point \(R\) must have a horizontal distance of \(4\) units from that line. 4. Therefore, \(R\) lies on \(x=5\) or \(x=-3\). One possible point is \(R(5, 4)\).

a) \(PQ=6\,\text{cm}\) b) \(h=4\,\text{cm}\) c) \(R\) must lie on \(x=5\) or \(x=-3\). One possible point is \(R(5, 4)\).

- What is the length of base \(AB\)? - Use the triangle area formula. - What happens when a triangle's vertex moves parallel to its base? - Does a horizontal movement change the vertical distance to the x-axis?

1. Base \(AB\) has length \(3-(-3)=6\) units. 2. Use \(18=\frac{1}{2}\times6\times h\). This gives \(18=3h\), so \(h=6\) units. 3. Since \(AB\) lies on \(y=0\) and \(y>0\), the y-coordinate of \(D\) is \(6\). 4. Changing the x-coordinate moves \(D\) parallel to the horizontal base. The perpendicular distance to the base stays \(6\), so the area does not change.

a) \(y=6\) b) The x-coordinate only moves \(D\) parallel to the base. It does not change the height, so it does not change the area.

- Compare the area formulas for triangles and parallelograms. - What factor appears only in the triangle formula? - After finding the parallelogram's height, add it to the base line's y-coordinate.

1. The base length is \(6-2=4\) units. 2. For the triangle, \(10=\frac{1}{2}\times4\times h_T=2h_T\), so \(h_T=5\) units. 3. For the parallelogram, \(10=4h_P\), so \(h_P=2.5\) units. 4. Since \(AB\) lies on \(y=2\), side \(DE\) must lie on \(y=2+2.5=4.5\). 5. The triangle formula includes \(\frac{1}{2}\), so with the same base and area, the triangle must be twice as high as the parallelogram.

a) \(5\) units b) \(y=4.5\) c) The triangle needs twice the height because its area formula includes the factor \(\frac{1}{2}\).

- Compare the horizontal and vertical changes from \(A\) to \(D\), then apply them from \(B\). - Which base and perpendicular height are easiest to read from the coordinates? - A rectangle with the same base and area must have the same height.

1. For a), from \(A(1, 1)\) to \(D(3, 5)\), move \(2\) units right and \(4\) units up. Make the same move from \(B(8, 1)\), giving \(C(10, 5)\). 2. Base \(AB\) has length \(8-1=7\) units, and the perpendicular height is \(5-1=4\) units. The area is \(7\times4=28\) square units. 3. For b), a rectangle with the same \(7\)-unit base and area \(28\) square units also needs height \(4\) units. Its upper vertices lie directly above \(A\) and \(B\), so \(D_1(1, 5)\) and \(C_1(8, 5)\).

a) \(C(10, 5)\); area \(28\) square units b) \(C_1(8, 5)\) and \(D_1(1, 5)\)

- Choose the side parallel to an axis as the base. - Find the vertical distance from the third vertex to the horizontal base. - What changes when the vertex moves parallel to the base? - With the base fixed, how must the height change to double the area?

1. Base \(AB\) has length \(10-2=8\,\text{cm}\), and the height is \(7-1=6\,\text{cm}\). 2. The area is \(A=\frac{1}{2}\times8\times6=24\,\text{cm}^2\). 3. Moving \(C\) to \(C_1\) does not change its y-coordinate, so the height and area stay the same. 4. Twice the original area is \(48\,\text{cm}^2\). With the same base, the required height is \(12\,\text{cm}\). 5. Since \(AB\) lies on \(y=1\), point \(D\) may lie on \(y=13\) or \(y=-11\). One possible point is \(D(5, 13)\).

a) \(24\,\text{cm}^2\) b) The area stays \(24\,\text{cm}^2\). c) One possible point is \(D(5, 13)\). Any point on \(y=13\) or \(y=-11\) works.

- Compare the area formulas for a parallelogram and a triangle. - What base and height do the two figures share? - Does moving \(P\) along \(CD\) change its perpendicular distance from \(AB\)?

1. Base \(AB\) has length \(7-1=6\) units, and the parallelogram's height is \(5-1=4\) units. Its area is \(6\times4=24\) square units. 2. Triangle \(ABP\) has the same \(6\)-unit base and the same \(4\)-unit height. Its area is \(\frac{1}{2}\times6\times4=12\) square units. 3. The triangle has one-half the parallelogram's area. Every point on \(CD\) is the same perpendicular distance from \(AB\), so the relationship holds for every such point.

a) \(24\) square units b) \(12\) square units c) The triangle has one-half the parallelogram's area. This is true for every point \(P\) on \(CD\).

- First determine the direction of \(\overline{AB}\). - A parallel line has the same direction. - Use the horizontal and vertical changes from \(B\) to \(C\), then apply them from \(A\).

1. Line \(\overleftrightarrow{AB}\) is horizontal with equation \(y=2\). The distance from \(C(5, 6)\) to this line is \(6-2=4\) units. 2. The line through \(C\) parallel to \(\overline{AB}\) has equation \(y=6\). 3. From \(B\) to \(C\), the change is \(3\) units left and \(4\) units up. Applying the same change to \(A(3, 2)\) gives \((3-3, 2+4)=(0, 6)\). 4. The point \((0, 6)\) lies on both requested lines, so it is their intersection.

a) \(4\) units b) \((0, 6)\)

- Compare the x- and y-coordinates of adjacent rectangle vertices. - Use matching x- and y-coordinates to complete the rectangle. - Find the length of the horizontal side. - A square has four congruent sides.

1. For \(ABCD\) to be a rectangle, \(D\) must have the same x-coordinate as \(A\) and the same y-coordinate as \(C\). Therefore, \(D(2, 5)\). 2. The horizontal side length is \(AB=7-2=5\) units. 3. For a square, the vertical side length must also be \(5\) units. Starting from \(y=2\), the upper vertices must have y-coordinate \(2+5=7\). 4. Therefore, the new points are \(C(7, 7)\) and \(D(2, 7)\).

a) \(D(2, 5)\) b) The y-coordinate of both \(C\) and \(D\) must be \(7\).

- Compare the direction and length of \(\overline{EF}\) with those of \(\overline{HG}\). - Use coordinate differences to find a horizontal segment's length. - A trapezoid needs at least one pair of parallel sides. - Part b has more than one possible answer.

1. Segment \(\overline{EF}\) is horizontal and has length \(6-1=5\) units. 2. For a parallelogram, \(\overline{HG}\) must be parallel and congruent to \(\overline{EF}\). Moving \(5\) units left from \(G(8, 4)\) gives \(H(3, 4)\). 3. For a trapezoid that is not a parallelogram, one pair of opposite sides can be parallel without being congruent. One choice is \(H_2(5, 4)\). Then \(\overline{H_2G}\) is horizontal and parallel to \(\overline{EF}\), but the other pair of opposite sides is not parallel.

a) \(H(3, 4)\) b) One possible answer is \(H_2(5, 4)\).

- Vertical segments have endpoints with the same x-coordinate. - A vertical segment of length \(4\) can extend up or down. - Apply the same vertical change to both given points.

1. Because \(\overline{PS}\) and \(\overline{QR}\) are vertical, \(S\) has x-coordinate \(3\), and \(R\) has x-coordinate \(10\). 2. One possibility is to move both points up \(4\) units. Then \(S=(3, 9)\) and \(R=(10, 12)\). 3. A second possibility is to move both points down \(4\) units. Then \(S=(3, 1)\) and \(R=(10, 4)\).

One possibility: \(R=(10, 12)\) and \(S=(3, 9)\) Another possibility: \(R=(10, 4)\) and \(S=(3, 1)\)

- Read the coordinates of \(A\), \(B\), and \(C\) from the graph. - Compare the horizontal and vertical changes from \(B\) to \(C\). - Apply the same change from \(A\) to reach the missing vertex.

1. Read the graph: \(A(2, 1)\), \(B(6, 3)\), and \(C(5, 5)\). 2. From \(B\) to \(C\), move \(1\) unit left and \(2\) units up. 3. Opposite sides of a rectangle are parallel and equal, so make the same move from \(A\): \(D=(2-1,1+2)=(1,3)\).

\(D=(1, 3)\)

- Read the x-coordinate before the y-coordinate. - Opposite sides of a rectangle are parallel and equal. - Find the two side lengths before using the perimeter formula.

1. Reading the graph gives \(A = (-2, -3)\), \(B = (4, -3)\), and \(C = (4, 1)\). 2. Point \(D\) must share the x-coordinate of \(A\) and the y-coordinate of \(C\), so \(D = (-2, 1)\). 3. The side lengths are \(4 - (-2) = 6\,\text{cm}\) and \(1 - (-3) = 4\,\text{cm}\). 4. The perimeter is \(2(6 + 4) = 20\,\text{cm}\).

a) \(A = (-2, -3)\), \(B = (4, -3)\), \(C = (4, 1)\) b) \(D = (-2, 1)\) c) \(20\,\text{cm}\)

- Read the coordinates of the three shown vertices from the grid. - Determine the horizontal and vertical change along one completed side. - Use the equal, parallel opposite side to locate the missing vertex.

1. Read the graph: \(A(3, 2)\), \(B(8, 3)\), and \(C(7, 8)\). 2. From \(A\) to \(B\), move \(5\) units right and \(1\) unit up. 3. Opposite sides of a square are parallel and equal, so make the same move from \(D\) to \(C\). Working backward from \(C\), move \(5\) units left and \(1\) unit down. 4. Thus \(D=(7-5,8-1)=(2,7)\).

\(D=(2, 7)\)

- Read the coordinates of the three shown posts from the grid. - Find the horizontal and vertical change from \(E\) to \(F\). - Apply the same change starting at \(H\).

1. Read the graph: \(E(1, 1)\), \(F(6, 2)\), and \(H(2, 5)\). 2. From \(E\) to \(F\), move \(5\) units right and \(1\) unit up. 3. Opposite sides of a parallelogram are parallel and equal, so make the same move from \(H\): \(G=(2+5,5+1)=(7,6)\).

\(G=(7, 6)\)

- Read each ordered pair as x-coordinate first, y-coordinate second. - Which pairs of vertices have the same y-coordinate? Which have the same x-coordinate? - Use those coordinate relationships to justify parallel and perpendicular sides.

1. Reading the graph gives \(A(-4, -1)\), \(B(3, -1)\), \(C(3, 2)\), and \(D(-4, 2)\). 2. The horizontal sides each have length \(3 - (-4) = 7\) units, and the vertical sides each have length \(2 - (-1) = 3\) units. 3. Adjacent sides are horizontal and vertical, so all four angles are right angles. Therefore, \(ABCD\) is a rectangle.

a) \(A(-4, -1)\), \(B(3, -1)\), \(C(3, 2)\), and \(D(-4, 2)\) b) \(ABCD\) is a rectangle. \(AB\) and \(CD\) are horizontal, \(BC\) and \(DA\) are vertical, so opposite sides are parallel and adjacent horizontal and vertical sides meet at right angles.

- Read all four vertex coordinates from the graph. - Compare the horizontal and vertical changes along one pair of opposite sides. - Repeat the comparison for the other pair before deciding.

1. Read the graph: \(A(1, 1)\), \(B(6, 1)\), \(C(8, 4)\), and \(D(3, 4)\). 2. From \(A\) to \(B\), move \(5\) units right and \(0\) units vertically. From \(D\) to \(C\), the change is also \((5,0)\). 3. From \(A\) to \(D\), move \(2\) units right and \(3\) units up. From \(B\) to \(C\), the change is also \((2,3)\). 4. Both pairs of opposite sides have matching coordinate changes, so opposite sides are parallel and equal in length. Therefore, \(ABCD\) is a parallelogram.

Yes. \(\overline{AB}\) and \(\overline{DC}\) both have change \((5,0)\), while \(\overline{AD}\) and \(\overline{BC}\) both have change \((2,3)\).

- Read point \(Q\) and the endpoints of the slanted right boundary from the graph. - At \(y=3\), where is the point on \(\overline{BC}\)? - Compare that boundary x-coordinate with the x-coordinate of \(Q\).

1. Read the graph: \(B(8,1)\), \(C(7,5)\), and \(Q(8,3)\). 2. The right boundary is segment \(\overline{BC}\). The y-coordinate \(3\) is halfway from \(1\) to \(5\), so the corresponding x-coordinate on \(\overline{BC}\) is halfway from \(8\) to \(7\), which is \(7.5\). 3. Point \(Q\) has x-coordinate \(8\), so at the same y-coordinate it lies to the right of the boundary. 4. Therefore, \(Q\) lies outside the quadrilateral.

Point \(Q\) lies outside the quadrilateral.

- Use the vertical alignment with \(B\) to determine one coordinate of \(D\). - Use the horizontal alignment with \(C\) to determine the other coordinate. - After locating \(D\), read side lengths from coordinate differences.

1. Being directly above \(B(7, 2)\) gives \(D\) x-coordinate \(7\). Being directly right of \(C(2, 6)\) gives \(D\) y-coordinate \(6\). Thus, \(D=(7, 6)\). 2. The horizontal side length is \(7-2=5\) units. The vertical side length is \(6-2=4\) units. 3. The perimeter is \(5+4+5+4=18\) units.

a) \(D(7, 6)\) b) \(5\) units by \(4\) units c) \(18\) units

- Use coordinate differences to read horizontal and vertical side lengths from each grid. - The concave polygon can be split into two non-overlapping rectangles. - Area measures the region inside a polygon, while perimeter measures its boundary length.

1. Figure a is a \(6\)-by-\(3\) rectangle, so its area is \(6\times3=18\) square units and its perimeter is \(2(6+3)=18\) units. 2. Split figure b into a lower \(6\)-by-\(2\) rectangle and an upper-left \(3\)-by-\(2\) rectangle. Its area is \(12+6=18\) square units. 3. The perimeter of figure b is \(6+2+3+2+3+4=20\) units. 4. The figures have equal areas but different perimeters, so equal area does not determine perimeter.

Figure a: area \(18\) square units, perimeter \(18\) units. Figure b: area \(18\) square units, perimeter \(20\) units. Equal area does not imply equal perimeter.

- First find the horizontal side length from the x-coordinates. - Use the rectangle's area to determine the required vertical distance. - A fixed distance from a horizontal line can be reached in two vertical directions.

1. Segment \(\overline{AB}\) is horizontal and has length \(2-(-4)=6\) units. 2. A rectangle of area \(42\) with base \(6\) must have height \(42\div6=7\) units. 3. The opposite horizontal side can be \(7\) units above \(y=3\), at \(y=10\), or \(7\) units below \(y=3\), at \(y=-4\). 4. Above \(\overline{AB}\), the other vertices are \(C=(2,10)\) and \(D=(-4,10)\). 5. Below \(\overline{AB}\), the other vertices are \(C=(2,-4)\) and \(D=(-4,-4)\).

The other vertices are either \((2,10)\) and \((-4,10)\), or \((2,-4)\) and \((-4,-4)\).

- Identify the horizontal and vertical legs of the whole triangle from the coordinates. - Determine where the slanted side crosses the axes. - Find the total area and the Quadrant I area separately before forming a percent.

1. Segments \(AB\) and \(AC\) are perpendicular and each has length \(10\) units. Therefore, the triangle's total area is \(\frac{1}{2}\times10\times10=50\) square units. 2. Segment \(BC\) lies on \(y=x\), so it passes through \((0, 0)\). The part in Quadrant I is a right triangle with vertices \((0, 0)\), \((0, 5)\), and \((5, 5)\). 3. That smaller triangle has area \(\frac{1}{2}\times5\times5=12.5\) square units. 4. The percent is \(\frac{12.5}{50}\times100\%=25\%\).

a) \(50\) square units b) \(25\%\)

- Enclose the triangle in a rectangle whose sides are parallel to the axes. - Identify the three right triangles outside \(XYZ\). - Use coordinate differences to find the legs of those triangles.

1. Enclose the triangle in the rectangle with vertices \((2, 2)\), \((10, 2)\), \((10, 9)\), and \((2, 9)\). 2. The rectangle has area \((10-2)\times(9-2)=8\times7=56\) square units. 3. The three outside right triangles have areas \(\frac{1}{2}\times8\times2=8\), \(\frac{1}{2}\times5\times5=12.5\), and \(\frac{1}{2}\times3\times7=10.5\) square units. 4. Subtract their areas from the rectangle: \(56-(8+12.5+10.5)=25\) square units.

The exact area of triangle \(XYZ\) is \(25\) square units.

- Read the horizontal base length and the vertical distance between the parallel lines from the graph. - Think of trapezoid area as average base length multiplied by height. - Once the required upper-base length is known, use a horizontal coordinate difference from point \(C\).

1. From the graph, \(A=(-4,-2)\), \(B=(4,-2)\), and \(C=(2,3)\). 2. The lower base has length \(8\) units, and the height between the horizontal lines is \(5\) units. 3. The trapezoid needs average base length \(30\div5=6\) units. 4. Therefore the two base lengths must sum to \(12\) units. Since the lower base is \(8\), the upper base must be \(4\) units long. 5. Starting at \(C=(2,3)\) and moving \(4\) units left gives \(D=(-2,3)\).

\(D=(-2,3)\)