Question

ABCD is a trapezoid. segment AB = 1 cm and segment CD = 5 cm, and the area = 7.5 cm2. What is the altitude of the trapezoid?

Answer 1

Answer: 2.5 cm

Step-by-step explanation:

We know that area of trapezoid A = $\frac{1}{2}h(a+b)$, where a and b are the opposite sides of trapezoid and h is the altitude of the trapezoid.

Given: AB = 1 cm and  CD = 5 cm, and the area = 7.5 square cm.

By the name of trapezoid we can see that AB and CD are opposite sides of trapezoid, then

Area of trapezoid=$\frac{1}{2}h(AB+CD)=7.5\ cm^2$

$\Rightarroe\frac{1}{2}h(1+5)=7.5\\\Rightarrow\ h(6)=2\times7.5\\\Rightarrow\ h=\frac{2\times7.5}{6}\\\Rightarrow\ h=2.5$

Hence, the altitude of the trapezoid = 2.5 cm

Answer 2

Area of a trapezoid = 1/(a + b)h; where a and b are the two parallel sides and h is the height (altitude).
7.5 = 1/2(1 + 5)h = 1/2(6)h = 3h
h = 7.5/3 = 2.5 cm.