Question

Find the area of an isosceles trapezoid, if the lengths of its bases are 16 cm, and 30 cm, and the diagonals are perpendicular to each other.

Answer 1

For this case we have:
 a = 30 cm
 c = 16 cm
 We look for the length of the diagonal:
 d = x + y
 Where,
 For x:
 a ^ 2 = x ^ 2 + x ^ 2
 x = a / root (2) = 30 / root (2) = 21.2132 cm
 For y:
 c ^ 2 = y ^ 2 + y ^ 2
 y = c / root (2) = 16 / root (2) = 11.3137 cm
 The diagonal is:
 d = x + y
 d = 21.2132 + 11.3137
 d = 32.5269 cm
 Then, the height is:
 h = h1 + h2
 For h1:
 h1 = root (x ^ 2 - (a / 2) ^ 2) = root ((21.2132) ^ 2 - (30/2) ^ 2)
 h1 = 15 cm
 For h2:
 h2 = root (y ^ 2 - (c / 2) ^ 2) = root ((11.3137) ^ 2 - (16/2) ^ 2)
 h2 = 8 cm
 Finally:
 h = h1 + h2
 h = 15 + 8
 h = 23 cm
 Then, the area is:
 A = (1/2) * (a + c) * (h)
 A = (1/2) * (30 + 16) * (23)
 A = 529 cm ^ 2
 Answer:
 the area of an isosceles trapezoid is:
 A = 529 cm ^ 2