Given: ABCD – parallelogram AB = 6 HD = 5 angle BHD = 90 degrees angle ABD = 90 degrees Find: S ABCD -?

Let the length of the segment AH be equal to X cm.

In a right-angled triangle ABH, we express the leg BH by the Pythagorean theorem.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 36 – X2.

By the property of the height of a right-angled triangle drawn from a right angle to the hypotenuse

BH ^ 2 = AH * DH = 5 * X.

Then 5 * X = 36 – X ^ 2.

X ^ 2 + 5 * X – 36 = 0.

Let’s solve the quadratic equation.

X1 = 4.

X2 = -9. (Doesn’t match because <0).

AH = 4 cm.

Then AD = AH + DH = 4 + 5 = 9 cm.

BH ^ 2 = 5 * 4 = 20.

BH = √20 = 2 * √5 cm.

Determine the area of the parallelogram.

Savsd = BH * AD = 2 * √5 * 9 = 18 * √5 cm2.

Answer: The area of the parallelogram is 18 * √5 cm2.