# The computational area of an isosceles trapezoid, the bases of which are 5 and 7 cm

**The computational area of an isosceles trapezoid, the bases of which are 5 and 7 cm, and one of the angles at the base is 45 °.**

Let ABCD be an isosceles trapezoid, AD = 7 cm – a larger base, BC = 5 cm – a smaller base, angle A = 45 degrees.

1. Draw from vertices B and C heights ВН and СK to the base AD (ВН = СK). NK = BC = 5 cm (since НВСK is a rectangle), AH = KD = (AD – НK) / 2 = (7 – 5) / 2 = 2/2 = 1 (cm).

2. Consider a triangle ABН: angle BHA = 90 degrees (since BH is the height), angle HAB (angle A) = 45 degrees, AH = 1 cm – leg, AB – hypotenuse, ВН – second leg.

The cosine of the angle is the ratio of the adjacent leg to the hypotenuse. Then:

cosHAB = AH / AB;

cos45 = 1 / AB;

√2 / 2 = 1 / AB;

AB = 2 / √2 = 2√2 / 2 = √2 (cm).

By the Pythagorean theorem:

BH = √ (AB ^ 2 – AH ^ 2);

BH = √ ((√2) ^ 2 – 1 ^ 2) = √ (2 – 1) = √1 = 1 (cm).

3. The area of an isosceles trapezoid is equal to:

S = mh,

where m is the middle line, h is the height.

m = (a + b) / 2 = (AD + BC) / 2 = (7 + 5) / 2 = 12/2 = 6 (cm).

S = 6 * 1 = 6 (cm square).

Answer: S = 6 cm square.