Question

how do the areas of triangle abc and def compare? the area of △abc is 1 square unit greater than the area of △def. the area of △abc is 2 square units greater than the area of △def. the area of △abc is 1 square unit less than the area of △def. the area of △abc is equal to the area of △def.

Explanation:

Step1: Find area of △ABC

The base of △ABC is the horizontal distance from A(4, 2) to B(6, 2), so base $b = 6 - 4=2$. The height is the vertical distance from C(4, 6) to the line $y = 2$, so height $h=6 - 2 = 4$. Using the area formula $A=\frac{1}{2}bh$, we have $A_{ABC}=\frac{1}{2}\times2\times4 = 4$ square - units.

Step2: Find area of △DEF

The base of △DEF is the horizontal distance from D(-2,-1) to F(4,-1), so base $b = 4-(-2)=6$. The height is the vertical distance from E(4,-3) to the line $y=-1$, so height $h=-1-(-3)=2$. Using the area formula $A = \frac{1}{2}bh$, we have $A_{DEF}=\frac{1}{2}\times6\times2=6$ square - units.

Step3: Compare the areas

$A_{DEF}-A_{ABC}=6 - 4=2$. So the area of △ABC is 2 square units less than the area of △DEF, or the area of △DEF is 2 square units greater than the area of △ABC.

Answer:

The area of △ABC is 2 square units less than the area of △DEF (it seems there is an error in the provided options as the correct comparison based on our calculation is that the area of △DEF is 2 square units greater than the area of △ABC).