In a rectangular trapezoid ABCD (BAD = 90 degrees) with bases AD = 12cm BC = 4cm large diagonal BD

| March 28, 2021 | education

In a rectangular trapezoid ABCD (BAD = 90 degrees) with bases AD = 12cm BC = 4cm large diagonal BD = 15cm intersects the height CK at point M. a) Prove that triangles BMC and DMK are similar b) Find the area of trapezoid ABCD

Since the CК is the height of the trapezoid, the triangles BMC and DMC are rectangular. In triangles, the angle of the BMC is equal to the angle DMC as the vertical angles at the intersection of the straight lines ВD and СK, then the right-angled triangles BMC and DMC are similar in acute angle. Q.E.D.

From the right-angled triangle ABD, by the Pythagorean theorem, we determine the length of the leg AB.

AB ^ 2 = ВD ^ 2 – AD ^ 2 = 15 ^ 2 – 12 ^ 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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