# The large diagonal of the rectangular trapezoid divides the height drawn from the top of the obtuse angle into segments

**The large diagonal of the rectangular trapezoid divides the height drawn from the top of the obtuse angle into segments of 20 and 12 cm. The large side of the trapezoid is equal to its smaller base. Find the area of the trapezoid.**

Since BC = CD by condition, AH = BC by construction, we denote AH = BC = CD = X cm, and the segment DH = Y cm.

Lateral side AB = CK + НK = 20 + 12 = 32 cm.

Rectangular triangles ABD and DKН are similar in acute angle D, then:

AD / AB = DH / KН.

(X + Y) / 32 = Y / 12.

32 * Y = 12 * X + 12 * Y.

12 * X = 20 * Y.

X = 5 * Y / 3. (1).

In a right-angled triangle CDH X ^ 2 = 32 ^ 2 + Y ^ 2.

X ^ 2 = 1024 + Y ^ 2. (2).

Let’s solve the system of equations 1 and 2.

(5 * Y / 3) ^ 2 – Y ^ 2 = 1024.

25 * Y ^ 2/9 – 9 * Y ^ 2/9 = 1024.

16 * Y ^ 2 = 9216.

Y ^ 2 = 9216/16 = 567.

Y = DH = 24 cm.

Then BC = AH = CD = 24 * 5/3 = 40 cm.

AD = AH + DH = 40 + 24 = 64 cm.

Determine the area of the trapezoid.

Savsd = (ВС + АD) * СН / 2 = (40 + 64) * 32/2 = 1664 cm2.

Answer: The area of the trapezoid is 1664 cm2.