Question

helpppp!!!!!!!!!!!!!​

Answer 1

Answer:

$A=37.5\ units^2$

Step-by-step explanation:

we know that

The area of a trapezoid is equal to

$A=\frac{1}{2}[b1+b2]h$

where

b1 and b2 are the parallel bases

h is the height of trapezoid (perpendicular distance between the two parallel bases)

In this problem the area is equal to

$A=\frac{1}{2}[BC+AD]AB$

we have the coordinates

$A(-3,2),B(1,5),C(7,-3),D(0,-2)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(-3,2),B(1,5$

substitute in the formula

$d=\sqrt{(5-2)^{2}+(1+3)^{2}}$

$d=\sqrt{(3)^{2}+(4)^{2}}$

$d_A_B=5\ units$

step 2

Find the distance BC

$B(1,5),C(7,-3)$

substitute in the formula

$d=\sqrt{(-3-5)^{2}+(7-1)^{2}}$

$d=\sqrt{(-8)^{2}+(6)^{2}}$

$d_B_C=10\ units$

step 3

Find the distance AD

$A(-3,2),D(0,-2)$

substitute in the formula

$d=\sqrt{(-2-2)^{2}+(0+3)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d_A_D=5\ units$

step 4

Find the area

$A=\frac{1}{2}[BC+AD]AB$

substitute the values

$A=\frac{1}{2}[10+5]5$

$A=37.5\ units^2$