# In an isosceles trapezoid, the bases are 6 cm and 10 cm, and the overshadowing angle is 45 °

In an isosceles trapezoid, the bases are 6 cm and 10 cm, and the overshadowing angle is 45 ° which is equal to the area of the trapezoid.

Consider an isosceles trapezoid ABCD with bases AD, BC and sides AB = CD.

By the condition of the problem, we have:

AD = 10 cm, BC = 6 cm.

Let us lower the heights BM and CN from the vertices B and C to the large base AD.

Triangles ABM and CDN are equal because are rectangular with equal hypotenuses and legs. Hence,

AM = DN. We have:

AD = AM + MN + DN = 2 * AM + BC,

10 = 2 * AM + 6,

AM = 2.

Consider a triangle ABM. Since BM is height, triangle ABM is right-angled.

By condition, it is known that the angle BAM = 45 °. Hence,

ABM = 180 ° – BAM – BMA = 180 ° – 45 ° – 90 ° = 45 ° and ABM = BAM.

This implies that triangle ABM is isosceles and BM = AM = 2.

Therefore, the area S of a given trapezoid is:

S = 1/2 * (AD + BC) * BM = 1/2 * (10 + 6) * 2 = 16.