The middle line of the trapezoid is 15 cm, the sum of the angles at one of the bases is 90.

univerkov | September 11, 2021 | education

The middle line of the trapezoid is 15 cm, the sum of the angles at one of the bases is 90. Find the area of a trapezoid if one side is equal to the root of 10 m and the base difference is 10 m.

Let the base BC = X cm, then, by condition, the base AD = (X + 10) cm.

The middle line of the trapezoid is: KM = (BC + AD) / 2 = (X + X + 10) / 2 = 15.

2 * X + 10 = 30.

X = BC = 20/2 = 10 cm.

AD = 10 + 10 = 20 cm.

Let us extend the lateral lines of the trapezoid until they intersect at point K. Angle AKD = 90, since, by condition, the sum of the angles at the larger base of the trapezoid is 90. Then the triangle AKD is rectangular.

Triangles AKD and BKS are similar. Let the segment BK = X cm, then AK = √10 + X.

Then: AD / BC = AK / BK.

20 / (X + √10) = 10 / X.

10 * X + 10 * √10 = 20 * X.

10 * X = 10 * √10.

X = BK = √10 cm.

Then AK = 2 * √10.

Then, by the Pythagorean theorem, DK ^ 2 = AD ^ 2 – AK ^ 2 = 400 – 40 = 360.

DК = 6 * √10 cm.

KC ^ 2 = BC ^ 2 – BK ^ 2 = 100 – 10 = 90.

KC = 3 * √10.

Determine the area of the triangle AKD. Sacd = AK * DK / 2 = 2 * √10 * 6 * √10 / 2 = 60 cm2.

The area of the VKS triangle is equal to: Svcs = BK * KC / 2 = √10 * 3 * √10 / 2 = 30/2 = 15 cm2.

Determine the area of the trapezoid: Savsd = Sackd – Svcs = 60 – 15 = 45 cm2.

Answer: The area of the trapezoid is 45 cm2.

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