In an isosceles trapezoid, the height = 5, the larger base = 15, the angle at the base is 45 degrees. Find a smaller base.

Let’s build the heights of the ВН and СK of the trapezoid, then in the right-angled triangles ABН and CDK the acute angle is 45, then the lengths of the legs of these triangles are equal to each other.

AH = BH = СK = DK = 5 cm.

The length of the segment НК = (АD – АН – DК) = (15 – 5 – 5) = 5 cm.

Quadrangle ВСКН is a rectangle, then НK = BC = 5 cm.

Second way.

The BH height forms an equilateral right-angled triangle, since its acute angle is 45. Then AH = BH = 5 cm.

The height BH divides the base AD into two segments, the length of the smaller of which is equal to the half-difference of the lengths of the bases.

AH = (AD – BC) / 2.

BC = (AD – AH) / 2 = (15 – 5) / 2 = 5 cm.

Answer: The length of the smaller base is 5 cm.