# In an isosceles trapezoid with mutually perpendicular diagonals, the side is 26 cm. The height drawn

**In an isosceles trapezoid with mutually perpendicular diagonals, the side is 26 cm. The height drawn from the top of an obtuse angle divides the larger base into segments, the smaller of which is 24 cm. Find the area of the trapezoid.**

The ABH triangle is rectangular, since BH is the height of the trapezoid. Then, by the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2 = 26 ^ 2 – 24 ^ 2 = 676 – 576 = 100.

BH = 10 cm.

Let’s draw the CM height through the point O, the point of intersection of the diagonals.

The BOC triangle is rectangular and isosceles since the diagonals of the isosceles trapezoid intersect at right angles, OB = OC.

We divide the height of the OK in half, the angle of the OBK in the triangle OBK is 45, then the triangle OBK is rectangular and equilateral, OK = BK = BC / 2.

Similarly, OM = AM = AD / 2.

Then KM = (OK + OM) = (BC + AD) / 2, which is the middle line of the trapezoid.

Then the area of the trapezoid is equal to: Savsd = KM ^ 2 = BH ^ 2 = 100 cm2.

Answer: The area of the trapezoid is 100 cm2.