# The bases of the trapezoid are 25 and 4 cm, and the sides are 20 and 13 cm. Find the area of the trapezoid.

Draw two heights BH and CK from the vertices B and C and denote the segment AH by X, then the segment KD = (25 – X).

In a right-angled triangle ABH, according to the Pythagorean theorem, the height BH ^ 2 = AB ^ 2 – AH ^ 2 = 400 – X ^ 2.

In a right-angled triangle CDK, according to the Pythagorean theorem, the height is CK ^ 2 = CD ^ 2 – KD ^ 2 = 169 – (25 – X – 4) ^ 2.

Since BH = CK, then 400 – X ^ 2 = 169 – (21 – X) ^ 2.

400 – X ^ 2 = 169 – 441 + 42 * X – X ^ 2).

42 * X = 672.

X = 672/42 = 16 cm.

AH = 16 cm.

Then BH ^ 2 = 400 – 16 ^ 2 = 400 – 256 = 144.

BH = √144 = 12 cm.

Determine the area of the trapezoid.

S = (АD + ВС) * ВН / 2 = (25 + 4) * 12/2 = 174 cm2.

Answer: The area of the trapezoid is 174 cm2.

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