- Read the coordinates of the rectangle's vertices from the graph. - Find the horizontal and vertical side lengths using coordinate differences. - Use the rectangle area formula once the side lengths are known.

1. The horizontal side length is \(6-1=5\) units. 2. The vertical side length is \(4-1=3\) units. 3. The area is \(5\cdot 3=15\) square units.

\(15\) square units

- Identify a base and a perpendicular height from the coordinate graph. - Horizontal and vertical coordinate differences give those two lengths directly. - Use the triangle area formula rather than the rectangle area formula.

1. \(\overline{AB}\) is a horizontal leg of length \(4\) units. 2. \(\overline{AC}\) is a vertical leg of length \(3\) units. 3. The area is \(\frac{1}{2}\cdot 4\cdot 3=6\) square units.

\(6\) square units

- Which measurement adds the lengths around the outside? - Which measurement counts the unit squares inside the rectangle?

1. The rectangle is \(5\) units wide and \(2\) units tall. 2. Its perimeter is \(2(5+2)=14\) units, the distance around its edges. 3. Its area is \(5\cdot2=10\) square units, the space inside it.

Perimeter: \(14\) units. Area: \(10\) square units. Area uses square units because it measures a two-dimensional region.

- What do you notice about the endpoint y-coordinates of \(AB\)? - Which coordinate difference gives the length of a horizontal side? - For the height, look in the direction perpendicular to the base.

1. Points \(A=(2, 1)\) and \(B=(8, 1)\) have the same y-coordinate, so \(AB\) is horizontal and has length \(8-2=6\) units. 2. Point \(C=(2, 5)\) is directly above \(A\). The perpendicular height to \(AB\) is \(5-1=4\) units.

Base \(6\) units; perpendicular height \(4\) units. The base uses the horizontal coordinate difference; the height uses the vertical difference.

- Match x- and y-coordinates to complete the axis-aligned rectangle. - Find the two side lengths from coordinate differences before using either measurement formula. - Keep linear units for perimeter and square units for area.

1. Point \(S\) has the same x-coordinate as \(P\) and the same y-coordinate as \(R\), so \(S=(1, 5)\). 2. The side lengths are \(9-1=8\,\text{cm}\) and \(5-1=4\,\text{cm}\). The perimeter is \(2(8+4)=24\,\text{cm}\). 3. The area is \(8\,\text{cm}\cdot4\,\text{cm}=32\,\text{cm}^2\).

1. \(S=(1, 5)\) 2. \(24\,\text{cm}\) 3. \(32\,\text{cm}^2\)

- Read the horizontal and vertical side lengths from the coordinate bounds. - Use the rectangle formulas for area and perimeter. - A set is bounded when it cannot extend indefinitely in any direction.

1. The x-values run from \(0\) to \(4\), so the rectangle has width \(4\). 2. The y-values run from \(0\) to \(3\), so the rectangle has height \(3\). 3. Its area is \(4\cdot3=12\) square units, and its perimeter is \(2(4+3)=14\) units. 4. Both coordinates have lower and upper bounds, so the solution set is bounded.

The area is \(12\) square units, the perimeter is \(14\) units, and the solution set is bounded.

- Choose a side that makes the base length easy to read from the coordinates. - The height must be perpendicular to the chosen base. - Use the triangle area formula after finding the base and height.

1. From the graph, \(A=(1, 1)\), \(B=(7, 1)\), and \(C=(4, 5)\). 2. Segment \(\overline{AB}\) is a horizontal base of length \(7-1=6\) units. 3. The perpendicular height from \(C\) to line \(AB\) is \(5-1=4\) units. 4. The area is \(\frac{1}{2}\cdot 6\cdot 4=12\) square units.

\(12\) square units

- Read each horizontal and vertical side length from coordinate differences on the graph. - Be sure the perimeter follows the inward notch as well as the outside boundary. - For area, compare the L-shape with a surrounding rectangle and identify the missing rectangular piece.

1. From the graph, the side lengths are \(AB=6\), \(BC=2\), \(CD=3\), \(DE=3\), \(EF=3\), and \(FA=5\) units. 2. The perimeter is \(6+2+3+3+3+5=22\) units. 3. For the area, view the figure as a \(6\times5\) outer rectangle with a \(3\times3\) rectangle removed. 4. The area is \(6\cdot5-3\cdot3=21\) square units.

Perimeter: \(22\) units Area: \(21\) square units

- Use the known x-coordinates to determine one side length. - Express the other side length using \(y\) and the given condition \(y>2\). - Set the product of the two side lengths equal to the stated area.

1. The horizontal side length is \(5-(-1)=6\) units. 2. Because \(y>2\), the vertical side length is \(y-2\) units. 3. Use the rectangle area: \(6(y-2)=30\). 4. Then \(y-2=5\), so \(y=7\).

\(y=7\)

- How can you find a rectangle's width and length from the corners? - Which calculation describes paving the inside, and which describes edging the outside? - Does the gate opening need edging?

1. From the corner coordinates, the rectangle is \(7-1=6\) feet wide and \(11-2=9\) feet long. 2. The paving area is \(6\cdot9=54\) square feet. 3. The full perimeter is \(2(6+9)=30\) feet. The gate opening needs no edging, so the edging length is \(30-3=27\) feet.

a) \(54\,\text{ft}^2\) b) \(27\,\text{ft}\)

- What are the largest and smallest x- and y-coordinates of the four vertices? - What are the base and height of the shapes outside the trapezoid?

1. The enclosing rectangle spans \(7\) units horizontally and \(3\) vertically, so its area is \(21\) square units. 2. The region outside the trapezoid consists of two right triangles, each with legs \(2\) and \(3\). Each triangle has area \(\frac{1}{2}\cdot2\cdot3=3\). 3. Subtracting both triangles gives \(21-3-3=15\) square units.

The enclosing rectangle has area \(21\). Subtract two triangles of area \(3\) each: \(21-3-3=15\) square units.

- Which sides meet at a right angle in the coordinates? - Write the vertical leg's length using \(h\) and the stated inequality. - Use the triangle's area to find the missing leg.

1. The horizontal leg \(AB\) is \(4-(-2)=6\) units long. 2. The vertical leg \(BC\) has length \(h-1\), since \(h>1\). 3. The area equation is \(\frac{1}{2}\cdot6(h-1)=12\), so \(3(h-1)=12\), \(h-1=4\), and \(h=5\).

\(h=5\)

- Express each vertex in terms of \(c\). - Which two vertices lie on the horizontal line \(y = c\)? - What happens to the square of a quantity when the quantity is doubled?

1. The lines \(f\) and \(g\) intersect where \(x = -x\), so \(x = 0\) and \(y = 0\). One vertex is \((0, 0)\). 2. The intersection of \(f\) and \(h\) satisfies \(x = c\), giving \((c, c)\). The intersection of \(g\) and \(h\) satisfies \(-x = c\), giving \((-c, c)\). 3. The horizontal base has length \(c - (-c) = 2c\), and the height from the origin to \(y = c\) is \(c\). 4. Therefore, \(A = \frac{1}{2} \cdot 2c \cdot c = c^2\). 5. If \(c\) is doubled, the new area is \((2c)^2 = 4c^2\), so the area is multiplied by \(4\).

a) The vertices are \((0, 0)\), \((c, c)\), and \((-c, c)\). The base is \(2c\) and the height is \(c\), so \(A = c^2\). b) The area quadruples because \((2c)^2 = 4c^2\).

- Find where the two lines have the same y-value. - The two y-intercepts form one side of the triangle. - What is the horizontal distance from the intersection point to the y-axis? - How does reflection affect area?

a) Set the line equations equal: \(x+1=-0.5x+4\). Then \(1.5x=3\), so \(x=2\). Substitution gives \(y=3\), so \(S=(2, 3)\). b) The y-intercepts are \((0, 1)\) and \((0, 4)\), so the vertical base has length \(3\). The perpendicular distance from \(S(2, 3)\) to the y-axis is \(2\), so the area is \(A=\frac{1}{2}\cdot 3\cdot 2=3\) square units. c) Reflection preserves area. The kite consists of the original triangle and its reflected copy, so its area is \(2\cdot 3=6\) square units.

a) \(S=(2, 3)\) b) \(3\) square units c) \(6\) square units

- Identify the pairs of parallel lines first. - Find the vertices by intersecting one line from each parallel pair. - What is the vertical distance between the two horizontal lines? - Use the parallelogram area formula once you know a base and its corresponding height.

1. Find the vertices from pairwise intersections. From \(2 = x + 1\), \(f\) and \(h\) meet at \((1, 2)\). From \(2 = x - 4\), \(f\) and \(k\) meet at \((6, 2)\). From \(-3 = x - 4\), \(g\) and \(k\) meet at \((1, -3)\). From \(-3 = x + 1\), \(g\) and \(h\) meet at \((-4, -3)\). 2. The horizontal side from \((1, 2)\) to \((6, 2)\) has length \(6 - 1 = 5\). 3. The perpendicular height is the vertical distance between \(y = 2\) and \(y = -3\), which is \(2 - (-3) = 5\). 4. Therefore, the area is \(A = 5 \cdot 5 = 25\) square units.

\(25\) square units

- Compare the slopes before doing any intersections. - Use the point on \(f\) with the stated x-coordinate to find \(b\). - Which familiar shapes could make up the region between the lines? - Work out the height and base of each shape in coordinate units.

1. The graphs of \(f\) and \(g\) have the same slope \(\frac{1}{2}\) and different y-intercepts, so they are distinct parallel lines. 2. At \(x=2\), \(f(2)=2\). The point \((2, 2)\) lies on \(h\), so \(2=-2+b\) and \(b=4\). 3. The y-axis meets \(f\) at \((0, 1)\) and \(g\) at \((0, -2)\). Line \(h\) meets \(g\) at \((4, 0)\). 4. Draw the vertical segment at \(x=2\) between \(f\) and \(g\). The part from \(x=0\) to \(x=2\) is a parallelogram with a vertical side of \(3\) and horizontal height \(2\), so its area is \(6\). 5. The remaining triangle has a vertical base of \(3\) at \(x=2\) and horizontal height \(2\), so its area is \(\frac{1}{2}\cdot3\cdot2=3\). 6. The quadrilateral's area is \(6+3=9\) square units.

a) Both slopes are \(\frac{1}{2}\), with different intercepts. b) \(b=4\). c) The parallelogram has area \(6\) and the triangle has area \(3\), giving \(6+3=9\) square units.

- One of the given points already lies on a coordinate axis. Which one? - How can you find a line equation from two points? - What shape do the two intercepts and the origin form? - How do you find the area of a right triangle?

1. The slope through \(A\) and \(B\) is \(\frac{0 - 3}{4 - (-2)} = -\frac{1}{2}\). 2. Since \(B(4, 0)\) lies on the x-axis, the x-intercept is \((4, 0)\). 3. Write the line as \(y = -0.5x + b\). Using \((4, 0)\), \(0 = -0.5 \cdot 4 + b\), so \(b = 2\). The y-intercept is \((0, 2)\). 4. The axes and the line form a right triangle with legs of lengths \(4\) and \(2\). Its area is \(\frac{1}{2} \cdot 4 \cdot 2 = 4\) square units.

The intercepts are \((4, 0)\) and \((0, 2)\). The area is \(4\) square units.

- Read the four vertex coordinates in order around the quadrilateral. - Use the distance formula because none of the sides is horizontal or vertical. - Add the four side lengths only after finding each one.

1. From the graph, \(A=(0, 0)\), \(B=(4, 3)\), \(C=(-2, 11)\), and \(D=(-6, 8)\). 2. \(AB=\sqrt{4^2+3^2}=5\) and \(BC=\sqrt{(-6)^2+8^2}=10\). 3. Likewise, \(CD=\sqrt{(-4)^2+(-3)^2}=5\) and \(DA=\sqrt{6^2+(-8)^2}=10\). 4. The perimeter is \(5+10+5+10=30\) units.

\(30\) units

- Read the coordinates of \(A\), \(B\), and \(C\) from the graph. - Apply the distance formula separately to the two adjacent oblique sides requested in part a). - Use the exact side lengths from part a) in the rectangle area formula.

1. From the graph, \(A=(0,0)\), \(B=(3,2)\), \(C=(-1,8)\), and \(D=(-4,6)\). 2. \(AB=\sqrt{(3-0)^2+(2-0)^2}=\sqrt{13}\). 3. \(BC=\sqrt{(-1-3)^2+(8-2)^2}=\sqrt{52}=2\sqrt{13}\). 4. The rectangle area is \(AB\cdot BC=\sqrt{13}\cdot2\sqrt{13}=26\) square units.

a) \(AB=\sqrt{13}\) units; \(BC=2\sqrt{13}\) units b) \(26\) square units

- How does the garden differ from the smallest rectangle that contains it? - Which side length cannot be found by subtracting coordinates along just one axis? - Match each price to the kind of measurement it describes before adding costs.

1. The bounding rectangle is \(12\) by \(9\), so its area is \(108\) square feet. 2. The missing upper-right corner triangle has legs \(3\) and \(4\), with area \(\frac{1}{2}\cdot3\cdot4=6\) square feet. The garden's area is \(108-6=102\) square feet. 3. The straight horizontal and vertical sides measure \(12\), \(5\), \(9\), and \(9\) feet. The slanted side is \(\sqrt{3^2+4^2}=5\) feet. The perimeter is \(12+5+5+9+9=40\) feet. 4. The plants cost \(102\cdot\$4=\$408\), and edging costs \(40\cdot\$8=\$320\). The total is \(\$728\).

a) \(102\) square feet b) \(40\,\text{ft}\) c) \(\$728\)