In an isosceles trapezoid of the base 7 and 19, the angle B is 135. Find the area of the trapezoid.

Let’s build the heights of the trapezoid BH and СK.

Since the quadrangle ВСКH is a rectangle, then HК = ВС = 7 cm.

Triangles ABH and CDK are rectangular, in which AB = CD as lateral sides of an isosceles trapezoid, angle BAH = CDK as angles at the base of an isosceles trapezoid. Then the triangles ABH and CDK are equal in hypotenuse and acute angle, which means AH = DK.

Then AH = (AD – HK) / 2 = (19 – 7) / 2 = 12/2 = 6 cm.

In a right-angled triangle ABH, the angle ABH = (ABC – CBH) = (135 – 90) = 45.

Then the triangle ABH is rectangular and isosceles, and therefore BH = AH = 6 cm.

Determine the area of ​​the trapezoid.

Savsd = (ВС + АD) * ВН / 2 = (7 + 19) * 6/2 = 78 cm2.

Answer: The area of ​​the trapezoid is 78 cm2.

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