Free Math Worksheets

- What is true about opposite sides of a parallelogram? - What is the vertical distance from the lower side to the upper side? - What formula gives the area of a parallelogram?

1. Segment \(AB\) moves \(5\) units to the right. In a parallelogram, the opposite side must be parallel and equal in length, so move \(5\) units left from \(C(8, 4)\). This gives \(D(3, 4)\). 2. Side \(AB\) lies on \(y=1\), and side \(CD\) lies on \(y=4\). Their perpendicular distance is \(4-1=3\) units, so \(h=3\) units. 3. 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

- 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. Plot \(A(1, 1)\), \(B(7, 1)\), \(C(9, 5)\), and \(D(3, 5)\), then connect the points in order to graph the parallelogram. 2. Since \(A\) and \(B\) have the same y-coordinate, \(AB=7-1=6\,\text{cm}\). 3. The height is the perpendicular distance between the horizontal lines \(y=1\) and \(y=5\): \(h=5-1=4\,\text{cm}\). 4. The area is \(A=bh=6\times4=24\,\text{cm}^2\). 5. 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) A graph of the parallelogram with the given vertices b) \(b=6\,\text{cm}\), \(h=4\,\text{cm}\) c) \(A=24\,\text{cm}^2\) d) The area stays \(24\,\text{cm}^2\) because the base and perpendicular height do not change.

- 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.

- 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.

- How do you find the distance between two points on a horizontal line? - A height is perpendicular to its base. - What is the area formula for a triangle?

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

a) A graph of triangle \(ABC\) with the given vertices b) \(AB=6\,\text{cm}\), \(h=4\,\text{cm}\) c) \(A=12\,\text{cm}^2\)

- 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 formula.

1. Plot \(A(2, 1)\), \(B(2, 6)\), \(C(5, 5)\), and \(D(5, 2)\), then connect the points in order to graph the trapezoid. 2. Sides \(AB\) and \(CD\) are vertical and parallel. 3. Their lengths are \(AB=6-1=5\) units and \(CD=5-2=3\) units. 4. The height is the horizontal distance between \(x=2\) and \(x=5\), which is \(3\) units. 5. The area is \(A=\frac{1}{2}(5+3)\times3=12\) square units.

A graph of trapezoid \(ABCD\) with the given vertices; the area is \(12\) square units.

- Opposite sides of a parallelogram are parallel and congruent. - Which base and height are easiest to read from the coordinates? - A rectangle with the same base and area must have what height? - Its upper vertices lie directly above the base endpoints.

1. The vector from \(A\) to \(D\) is \((2, 4)\). Apply the same vector to \(B\): \(C=(8+2, 1+4)=(10, 5)\). 2. Base \(AB\) has length \(8-1=7\) units, and the height is \(5-1=4\) units. The area is \(7\times4=28\) square units. 3. The rectangle must also have height \(4\) units. Its upper vertices lie directly above \(A\) and \(B\) on \(y=5\), so \(D_1(1, 5)\) and \(C_1(8, 5)\).

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

- 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.

- 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 \(P\).

- Count the grid units along the two drawn sides. - Opposite sides of a square are parallel. - The missing vertex lies directly below \(P\) and directly to the right of \(R\).

1. Segment \(\overline{PQ}\) is horizontal and has length \(3 - (-2) = 5\) units. 2. Segment \(\overline{QR}\) is vertical and has length \(2 - (-3) = 5\) units. 3. To complete the square, \(S\) must have the same x-coordinate as \(P\) and the same y-coordinate as \(R\). 4. Therefore, \(S = (3, -3)\).

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

- Compare the point’s location with all three sides of the triangle. - Find the midpoint of the horizontal base. - Look for a segment from the base to vertex \(C\) that contains \(P\).

1. The midpoint of \(\overline{AB}\) is \(M(4, 1)\). 2. Segment \(\overline{MC}\) lies inside the triangle. 3. Point \(P(4, 2)\) lies on \(\overline{MC}\) because it has x-coordinate \(4\) and its y-coordinate is between \(1\) and \(5\). 4. Therefore, \(P\) lies inside the triangle.

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

- 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. - The area is the same no matter which side is chosen as the base. - Rearrange \(A=bh\) to solve for the height. - Think about the limits of a pencil, ruler, and right-angle construction.

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, \(h_{QR}=40\div5=8\,\text{cm}\). 4. A hand-drawn diagram has limited precision because of line thickness, ruler markings, and small errors when constructing a perpendicular segment.

a) \(A=40\,\text{cm}^2\) b) \(h_{QR}=8\,\text{cm}\) c) Line thickness, ruler precision, and small construction errors can make a measured value slightly inaccurate.

- 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 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.

- Graph the points and look for horizontal sides. - Draw vertical segments from the endpoints of the upper base to the lower base. - Add the areas of the resulting rectangle and two triangles.

1. Plot \(A(2, 1)\), \(B(10, 1)\), \(C(8, 5)\), and \(D(4, 5)\) and connect them in order. 2. 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. 3. Draw vertical segments from \(C\) and \(D\) to \(\overline{AB}\). This forms a central \(4\times4\) rectangle and two congruent side triangles. 4. The rectangle's area is \(4\times4=16\) square units. 5. Each side triangle has base \(2\) units and height \(4\) units, so each has area \(\frac{1}{2}\times2\times4=4\) square units. 6. The total area is \(16+4+4=24\) square units.

The graph can be split into a central \(4\times4\) rectangle and two congruent triangles, each with base \(2\) units and height \(4\) units. The area of quadrilateral \(ABCD\) 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 \(\frac{1}{2}\).

- 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 \(H_2(5, 4)\).

- 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.

- 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 \(S = (3, 1)\)

- 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 \(D\).

- 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 formula 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 \(5\,\text{cm} \times 3\,\text{cm} = 15\,\text{cm}^2\).

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

- Opposite sides of a rectangle have the same direction and length. - Determine the horizontal and vertical change from \(B\) to \(C\). - Apply that same change to \(A\).

1. From \(B(6, 3)\) to \(C(5, 5)\), the coordinate change is \((-1, 2)\). 2. Opposite sides of a rectangle are parallel and equal, so the change from \(A\) to \(D\) is also \((-1, 2)\). 3. Apply this change to \(A(2, 1)\): \(D = (2 - 1, 1 + 2) = (1, 3)\).

\(D = (1, 3)\)

- 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) \(20\,\text{cm}\)

- Determine the coordinate change from one endpoint of a side to the other. - Opposite sides of a square are parallel and equal. - Apply the same coordinate change to the opposite vertex.

1. The coordinate change from \(B\) to \(A\) is \((-5, -1)\). 2. Opposite sides of a square are parallel and equal, so the same coordinate change takes \(C\) to \(D\). 3. Apply the change to \(C(7, 8)\): \(D = (7 - 5, 8 - 1) = (2, 7)\).

\(D = (2, 7)\)

- Opposite sides of a parallelogram are parallel and equal. - Find the horizontal and vertical change from \(E\) to \(F\). - Apply that change starting at \(H\).

1. From \(E(1, 1)\) to \(F(6, 2)\), the coordinate change is \((5, 1)\). 2. Opposite sides of a parallelogram are parallel and equal, so the same change takes \(H\) to \(G\). 3. Apply the change to \(H(2, 5)\): \(G = (2 + 5, 5 + 1) = (7, 6)\).

\(G = (7, 6)\)

- Read the x-coordinate before the y-coordinate. - Compare the horizontal and vertical side lengths. - Use the side directions and angle types to classify the quadrilateral.

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) A rectangle

- Compare the coordinate changes from \(A\) to \(B\) with those from \(D\) to \(C\). - One pair of opposite sides that is both parallel and congruent is enough to prove a quadrilateral is a parallelogram.

1. From \(A(1, 1)\) to \(B(6, 1)\), the horizontal change is \(5\) and the vertical change is \(0\). 2. From \(D(3, 4)\) to \(C(8, 4)\), the horizontal change is also \(5\) and the vertical change is \(0\). 3. Therefore, \(\overline{AB}\) and \(\overline{DC}\) have the same length and direction. One pair of opposite sides is both parallel and congruent, so \(ABCD\) is a parallelogram.

Yes. Both \(\overline{AB}\) and \(\overline{DC}\) move \(5\) units right and \(0\) units vertically, so the quadrilateral is a parallelogram.

- Examine the slanted right side \(\overline{BC}\). - Find the point on \(\overline{BC}\) whose y-coordinate is \(3\). - Compare its x-coordinate with the x-coordinate of \(Q\).

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

Point \(Q\) lies outside the quadrilateral.

- Graph the axes and identify Quadrant I. - Determine where the triangle meets the x-axis and y-axis. - Identify the shape of the portion in Quadrant I. - Find the total area and the Quadrant I area separately.

1. Plot \(A(-5, 5)\), \(B(5, 5)\), and \(C(-5, -5)\) and connect the points to form the triangle. 2. 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. 3. 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 \((5, 5)\). 4. That smaller triangle has area \(\frac{1}{2}\times5\times5=12.5\) square units. 5. The percent is \(\frac{12.5}{50}\times100\%=25\%\).

a) Plot \(A(-5, 5)\), \(B(5, 5)\), and \(C(-5, -5)\) and connect them. The total area is \(50\) square units. b) \(25\%\) of the area lies in Quadrant I.

- 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.