Question

lets try: find the area of rectangle efgh. round to the nearest tenth if necessary. e(-2, 6), f(4, 3), g(2, -1), h(-4, 2) 1. graph quadrilateral efgh. 2. find the length of ef or gh. 3. find the length of eh or fg.

Explanation:

Step1: Recall distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

Step2: Find length of EF

For points \(E(-2,6)\) and \(F(4,3)\):
\(x_1=-2,y_1 = 6,x_2 = 4,y_2=3\)
\(EF=\sqrt{(4 - (-2))^2+(3 - 6)^2}=\sqrt{(6)^2+(-3)^2}=\sqrt{36 + 9}=\sqrt{45}\approx6.7\)

Step3: Find length of EH

For points \(E(-2,6)\) and \(H(-4,2)\):
\(x_1=-2,y_1 = 6,x_2=-4,y_2 = 2\)
\(EH=\sqrt{(-4-(-2))^2+(2 - 6)^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.5\)

Step4: Calculate area of rectangle

Area of rectangle \(A=\text{length}\times\text{width}\), so \(A = EF\times EH\approx6.7\times4.5 = 30.15\approx30.2\) (or using exact values: \(\sqrt{45}\times\sqrt{20}=\sqrt{45\times20}=\sqrt{900} = 30\)). Wait, let's recalculate with exact distance:

Wait, \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{20}=2\sqrt{5}\), so \(3\sqrt{5}\times2\sqrt{5}=6\times5 = 30\). So the exact area is 30.

Answer:

The area of rectangle EFGH is \(\boxed{30}\) (or 30.0 when rounded to nearest tenth).