In trapezoid ABCD, angle A = 90 degrees. The height of the CK

| September 23, 2021 | education

In trapezoid ABCD, angle A = 90 degrees. The height of the CK with the diagonal AC and the lateral side of the CD is angles equal to 45 degrees, AK = 8cm. Find the area of the trapezoid.

Given:

ABCD is a trapezoid.

∠A = 90.

∠ASK = ∠КСD = 45.

AK = 8 cm.

Find: Sabcd.

Solution:

Consider triangles ACK and DCК, for which ∠AСK = ∠КСD = 45, and the angles ∠СКА = ∠СКD = 90, since SK is height, then triangles ACK and DCК are equal in terms of the common leg СK and an acute angle of 45, as well as isosceles. Then AK = KD = CK = BC = 8 cm.

АВСК – a square with a side of 6 cm, its area is: S1 = AK * CK = 8 * 8 = 64 cm2.

CKD is a right-angled isosceles triangle, its area is: S2 = (CK * KD) / 2 = 8 * 8/2 = 32 cm2.

Then the area of the trapezoid is:

S = S1 + S2 = 64 + 32 = 96 cm2.

Answer: S = 96 cm2.



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