# In trapezoid ABCD, angle A = 90 degrees, side CD and diagonal AC intersect at right

**In trapezoid ABCD, angle A = 90 degrees, side CD and diagonal AC intersect at right angles, AD = 5 cm, CD = 3 cm. Find the area of the trapezoid.**

Since AC is perpendicular to CD, the ACD triangle is rectangular, from which, according to the Pythagorean theorem, we determine the length of the AC leg.

AC ^ 2 = AD ^ 2 – CD ^ 2 = 25 – 9 = 16.

AC = 4 cm.

Let’s draw the height of CH, which is the height of the trapezoid and the height of the triangle ACD.

The area of the triangle ACD is equal to: Sacd = AC * CD / 2 = 4 * 3/2 = 6 cm2.

On the other hand, the area of the ACD triangle is equal to:

Savs = AD * CH / 2.

Then AD * CH / 2 = 6 cm2.

CH = 2 * 6/5 = 12/5 = 2.4 cm.

From a right-angled triangle ACH, AH ^ 2 = AC ^ 2 – CH ^ 2 = 16 – 5.76 = 10.24. AH = BC = 3.2 cm.

Determine the area of the trapezoid.

Savsd = (BC + AD) * CH / 2 = (3.2 + 5) * 2.4 / 2 = 9.84 cm2.

Answer: The area of the trapezoid is 9.84 cm2.