# In an isosceles trapezoid ABCD of the base AD = 20 and BC = 10, ∠BAD = 45 °. Find the area of the trapezoid ABCD.

1. The area of the trapezoid is equal to the product of the half-sum of its bases by the height.

2. Find the height h.

To do this, from the vertex B we lower the perpendicular DO to the base of the trapezoid AD – we get a right-angled triangle ABO, in which, according to the condition of the problem, the angle A is 45 *, leg AO = (AD – BC): 2 = (20 – 10): 2 = 5.

The leg BО, which is the height of the trapezoid, is calculated by the tangent 45 *.

h: AO = tg 45 * = 1;

h = AO * 1 = 5.

3. Let’s calculate what is the area S of the trapezoid.

S = (BC + AD) * h: 2 = (10 + 20) * 5: 2 = 75.

Answer: The area of a trapezoid is 75 square units.

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