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.
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