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.

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
 
 the area of an isosceles trapezoid is:
 A = 529 cm ^ 2

I’m pretty sure the slope would be 1 since the x and y values increase by 1