Question

what is the area, in square feet, of an isosceles triangle whose vertex angle is $120^{\circ}$ and whose base is $20$ feet long? express your answer as a common fraction in simplest radical form.

Answer 1

(100√3)/3 square feet

Explanation:

You want the area of the isosceles triangle that has a vertex angle of 120° and a base of 20 feet.

Altitude

The altitude divides the triangle into two 30-60-90 right triangles with a long side of half the base, or 10 ft. That means the altitude is 10/√3 ft.

Area

The area is ...

A = 1/2bh

A = 1/2(20 ft)(10/√3 ft) = 100/√3 ft²

We usually want the radical in the numerator, so that would be ...

Area = (100√3)/3 ft²