Given a trapezoid ABCD. AB = 25cm, BC = 11cm, CD = 26cm, AD = 28cm. Find the area of the trapezoid.

| September 7, 2021 | education

From the tops of the trapezoid B and C, we lower the heights BH and CK to the base of AD.

The formed quadrangle BCKH is a rectangle, since BC is parallel to HC, and BH and CK are perpendicular to HC, then NC = BC = 11 cm, and BH = CK.

Let the segment AH = X cm, then the segment DH = AD – HK – X = 17 – X cm.

Let us express, according to the Pythagorean theorem, the height BH from the triangle ABH, and the height of the CK from the triangle CDK.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 25 ^ 2 – X ^ 2 = 625 – X ^ 2.

CK ^ 2 = CD ^ 2 – DK ^ 2 = 26 ^ 2 – (17 – X) ^ 2 = 676 – 289 + 34 * X – X ^ 2.

Let’s equate the equations.

625 – X ^ 2 = 676 – 289 + 34 * X – X ^ 2.

34 * X = 238.

X = 238/34 = 7 cm.

AH = 7 cm, DK = 17 – 7 = 10 cm.

Then BH ^ 2 = 625 – X ^ 2 = 625 – 49 = 576.

BH = 24 cm.

Determine the area of ​​the trapezoid.

S = (АD + ВС) * ВН / 2 = (11 + 28) * 24/2 = 468 cm2.

Answer: The area of ​​the trapezoid is 468 cm2.



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