The area of a trapezoid is 160 square inches. One base of the trapezoid is 8 inches longer than the other base. The sum of the bases is 32 inches. What is the height? What are the lengths of the bases? Explain how you found your answers.

Height = 10 inches

Length of the bases are

base1 = 20 inches

base2 = 12 inches

Step-by-step explanation:

Given that the area of trapezoid = 160 inches²

Let base1 and base2 represent the trapezoid bases respectively.

Sum of base1 and base2 = 32 inches.

base1 + base2 = 32 inches

One base of the trapezoid is 8 inches longer than the other base.

Let assume base1 is the longer base:

base1 = base2 + 8

1) what is the height?

From the formula of area of trapezoid

Area = 1/2 x height x sum of bases.

Therefore

160 = 1/2 x height x 32

160 = 16 x height

Height = 160/16 = 10 inches.

2) What are the lengths of the bases?

Given that, sum of bases = 32 inches

:. base1 + base2 = 32

And base1 = base2 + 8

base1 + base2 = 32 will be equivalent to

base2 + 8 + base2 = 32

2(base2) + 8 = 32

base2 = (32-8)/2

base2 = 12 inches

And

base1 = base2 + 8

base1 = 12+8

base1 = 20 inches

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3