The bases of an isosceles trapezoid are 13 and 25, and its sides are 10. Find the area of the trapezoid.

Given:

isosceles trapezoid ABCE,

BC = 13,

AE = 25,

AB = CE = 10.

Find the area of ​​the trapezoid ABCE, that is, S ABCE -?

Solution:

1. Consider an isosceles trapezoid ABCE. Let’s draw the heights of the HV and CO. We get the HBCO rectangle. He has BH = CO and BC = HC = 13.

2. Right-angled triangle ABH = right-angled triangle COE along the hypotenuse and acute angle, since angle A = angle E and CE = AB. Then OE = AH = (25 – 13): 2 = 12: 2 = 6.

3. Consider the BHA triangle. By the Pythagorean theorem: BH ^ 2 = AB ^ 2 – AH ^ 2 = 100 – 36 = 64. Then BH = 8.

4. The area of ​​the trapezoid ABCE, that is, S ABCE = 1/2 * (BC + AE) * BH = 1/2 * (13 + 25) * 8 = 4 * 38 = 152 cm ^ 2.

Answer: 152 cm ^ 2.