Find the area of an isosceles trapezoid if its bases are 5 cm and 17 cm, and the side is 10 cm.

Given:
ABCD – isosceles trapezoid
AB = CD = c = 10cm
BC = b = 5 cm
AD = a = 17 cm
BQ = CS = h

Since the angles at the base are equal in an isosceles trapezoid, the angle BAQ = angle CDS = alpha, angle ABC = angle BCD = beta
By theorem, triangle ABQ = triangle CDS
means AQ = SD = (AD – BC) / 2 = (a-b) / 2 = (17 – 5) / 2 = 6 cm

Consider a triangle ABQ. Since in an isosceles triangle, the height is perpendicular to the base, then the angle AQB is straight, which means that triangle ABQ is right-angled.
Let’s use the Pythagorean theorem:
AB ^ 2 = AQ ^ 2 + BQ ^ 2
10 ^ 2 = 6 ^ 2 + BQ ^ 2
10 ^ 2 – 6 ^ 2 = BQ ^ 2
(10-6) (10 + 6) = BQ ^ 2
64 = BQ ^ 2
BQ = root of 64
BQ = 8
8 cm – trapezoid height ABCD

S = (a + b) / 2 * h
S = (17 + 5) / 2 * 8

S = 22/2 * 8
S = 88
88 cm: 2 – trapezoid area

Answer: 88 cm ^ 2