# The bases of an isosceles trapezoid are 12 cm and 16 cm, and its diagonals are mutually perpendicular.

**The bases of an isosceles trapezoid are 12 cm and 16 cm, and its diagonals are mutually perpendicular. Find the area of the trapezoid.**

Consider a right-angled triangle AOD, which, by the property of the diagonals of an isosceles trapezoid, AO = OD, that is, an isosceles trapezoid.

Then, by the Pythagorean theorem, AD ^ 2 = AO ^ 2 + DO ^ 2 = 2 * AO ^ 2.

AO ^ 2 = AD ^ 2/2 = 256/2 = 128.

Then from the triangle AOH, OH ^ 2 = AO ^ 2 – AH ^ 2 = 128 – 82 = 64.

OH = 8 cm.

Likewise in the BOC triangle.

ВO ^ 2 = BC ^ 2/2 = 144 / = 72.

KO ^ 2 = ВO ^ 2 = ВK ^ 2 = 72 – 62 = 72 – 36 = 36.

KO = 6 cm.

Then the height KH = OH + KO = 8 + 6 = 14 cm.

Let’s define the middle line of the trapezoid.

PE = (AD + BC) / 2 = (16 + 12) / 2 = 28/2 = 14 cm.

Then the area of the trapezoid is: S = PE * KH = 14 * 14 = 196 cm2.

Answer: The area of the trapezoid is 196 cm2.