In an isosceles trapezoid ABCD, side AD is parallel to BC, angle A = 30 degrees, height BK

univerkov | May 28, 2021 | education

In an isosceles trapezoid ABCD, side AD is parallel to BC, angle A = 30 degrees, height BK = 1 cm, BC = 2√3. You need to find the area of the trapezoid and the area of the triangle KMD – where M is the midpoint of the segment BD.

In a right-angled triangle ABK, the leg BK lies opposite the angle 30, then the length of the hypotenuse AB will be: AB = 2 * BK = 2 * 1 = 2 cm. The leg AK is determined by the Pythagorean theorem. AK ^ 2 = AB ^ 2 – BK ^ 2 = 4 – 1 = 3. AK = √3 cm.

Let’s draw the height of the CH. Since the trapezoid is isosceles, the triangles ABK and CDH are equal in hypotenuse and acute angle, then DH = AK = √3 cm.

Quadrilateral BCHK is a rectangle, then KН = BC = 2 * √3 cm.

Then the segment DК = DН + КН = √3 + 2 * √3 = 3 * √3 cm, and AD = DК + AK = 3 * √3 + √3 = 4 * √3 cm.

Determine the area of the trapezoid. Savsd = (BC + AD) * BK / 2 = (2 * √3 + 4 * √3) * 1/2 = 3 * √3 cm2.

Let’s draw the height МР, the length of which is equal to half of the height BК, since the point М is the middle of the diagonal, and therefore, lies on the middle line of the trapezoid. MR = BK / 2 = 1/2 cm.

Determine the area of the triangle KMD.

Sкмд = КD * МР / 2 = 3 * √3 * (1/2) / 2 = 3 * √3 / 4 cm2.

Answer: The area of the trapezoid is 3 * √3 cm2, the area of the KMD triangle is 3 * √3 / 4 cm2.

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