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