# In a trapezoid with a diagonal of 20, a height of 12 and an area of 150, the second diagonal is equal to?

Let us draw the height of the trapezium CH and, by the Pythagorean theorem, determine the leg AH.

AH ^ 2 = AC ^ 2 – CH ^ 2 = 400 – 144 = 256.

AH = 16 cm.

Determine the area of the triangle ACH. Ssn = AH * CH / 2 = 16 * 12/2 = 96 cm2.

Let us draw a line CE parallel to the diagonal BD.

BCED parallelogram, since its opposite sides are pairwise parallel, then DE = BC.

In the ACE triangle, the area will be equal to: Sace = AE * CH / 2 = (AD + DE) * CH / 2 = (AD + BC) * CH / 2, which is equal to the area of the trapezoid. Sace = Savsd = 150 cm2.

Then Ssnd = Sace – Sasn = 150 – 96 = 54 cm2.

Then Ssnd = HE * CH / 2 = HE * 12/2 = 54 cm.

HЕ = 54/6 = 9 cm.

Then in the triangle CHE, by the Pythagorean theorem, CE ^ 2 = HE ^ 2 + CH ^ 2 = 81 + 144 = 225.

CE = BD = 15 cm.

Answer: The length of the second diagonal is 15 cm.