Free math worksheets

- 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

- 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 \(y=7\).

\(y=7\)

- 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 \(y=7\).

\(y=7\)

- 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 and y-intercepts of \(f\) and \(g\). - Use the stated x-coordinate to locate the required intersection point of \(f\) and \(h\). - Find every vertex of the bounded quadrilateral before computing its area. - Choose a coordinate-area method that works for a non-axis-aligned quadrilateral.

1. Lines \(f\) and \(g\) both have slope \(\frac{1}{2}\) and have different y-intercepts, so they are distinct parallel lines. 2. At \(x=2\), \(f(2)=\frac{1}{2}\cdot2+1=2\). Since \(h\) passes through \((2,2)\), \(2=-2+b\), so \(b=4\) and \(h(x)=-x+4\). 3. The y-axis meets \(f\) at \((0,1)\) and \(g\) at \((0,-2)\). Line \(h\) meets \(f\) at \((2,2)\). 4. To find the remaining vertex, solve \(-x+4=\frac{1}{2}x-2\). This gives \(x=4\) and \(y=0\), so the fourth vertex is \((4,0)\). 5. Using the vertices in order \((0,1)\), \((2,2)\), \((4,0)\), \((0,-2)\), the shoelace formula gives area \(\frac{1}{2}|-8-10|=9\) square units.

a) They are parallel because both have slope \(\frac{1}{2}\) and different y-intercepts. b) \(b=4\) c) \(9\) 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

- 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