In trapezoid ABCD, angle BAD = 90 ° with base AD = 12 cm and BC = 4 cm large diagonal BD = 15 cm crosses

| May 24, 2021 | education

In trapezoid ABCD, angle BAD = 90 ° with base AD = 12 cm and BC = 4 cm large diagonal BD = 15 cm crosses the height CK at point M prove that triangles BMC and DMK are similar find the area of trapezoid ABCD.

Since CK is the height of the trapezoid, the triangles BMC and DMC are rectangular, in which the angle BMC = DMC as the vertical angles at the intersection of the height of the CK and the diagonal BD, then the triangles BMC and DMC are similar in acute angle, which was required to be proved.

In a right-angled triangle ABD, according to the Pythagorean theorem, AB ^ 2 = BD ^ 2 – AD ^ 2 = 225 – 144 = 81.

AB = 9 cm.

Determine the area of the trapezoid.

Savsd = (BC + AD) * AB / 2 = (4 + 12) * 9/2 = 72 cm2.

Answer: The area of the trapezoid is 72 cm2.



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