In an isosceles described trapezoid, the lengths of the bases are 9 and 4. what is its area?

One of the features of the described quadrangle is that the sums of the opposite sides are equal. Thus, a circle can be inscribed into an isosceles trapezoid when the sum of its bases is equal to the sum of the sides.

Given a trapezoid ABCD: AD = a = 9, BC = b = 4, AB = CD = c.

1. On the basis of the described trapezoid:

a + b = c + c;

a + b = 2 * c.

Let’s match the data by the value condition:

2 * c = 9 + 4;

2 * c = 13;

c = 13/2;

c = 6.5.

2. The area of ​​an isosceles trapezoid is equal to:

S = (p – c) * √ ((p – a) * (p – b)),

where p is a semi-perimeter.

The semi-perimeter of an isosceles trapezoid is:

p = (a + b + 2 * c) / 2.

Semi-perimeter ABCD is:

p = (9 + 4 + 2 * 6.5) / 2 = (13 + 13) / 2 = 26/2 = 13.

Find the area of ​​the trapezoid ABCD:

S = (13 – 6.5) * √ ((13 – 9) * (13 – 4)) = 6.5 * √ (4 * 9) = 6.5 * √36 = 6.5 * 6 = 39 …

Answer: S = 39.