Geometry Trapezoid Word Problem

Question

What is the area of an isosceles trapezoid if the length of its shorter base is 18 cm, the length of an altitude is 9 cm, and the measure of the acute angle is 45º.

 

I know that the bottom two angles are equal, and the legs are also equal. I am just having a hard time figuring out the lower base. once I find that, I can use the formula 1/2h(b1+b2) to find the area.

Answer 1

draw the isosceles trapezoid

draw the altitude from the left top vertex and also draw the altitude from the top right vertex

this forms two right triangles, one on the left and one on the right

the bottom left acute angle is 45º (as is the other acute angle)

one leg is the altitude of 9 cm

the other leg is not known

the hypotenuse is also not known but it is not needed to solve the problem

use the tan function

tan=opposite/adjacent

tan45º=9/x where x=the unknown leg

x=9/tan45º

tan45º=1

x=9/1

x=9 cm

the same reasoning applies to the triangle on the right side

you have a rectangle 9 by 18 and a right triangle on each end

the bottom base is 18 cm + 9cm + 9cm=36 cm (you are adding the two bottom bases from the two right triangles to the length of the top trapezoidal base to get the bottom trapezoidal base)

A=(1/2)(h)(B+b)

A=(1/2)(9)(36+18)

A=(1/2)(9)(54)

A=(1/2)(486)

A=243 square cm is the area of the trapezoid

 

Answer 2

Draw a picture and look at the triangle formed by one of the non-parallel sides and the altitude. The 45° angle will also appear in this triangle. Since the altitude is perpendicular to the base, the triangle is a right triangle. Thus, the third angle is also 45°, which makes an isosceles triangle.

So that means the two congruent sides each equal 9. Now draw the same triangle on the other side of your trapezoid. So the longer base = 9 + 18 + 9 = 36. Now you can use the area formula.