In the trapezoid ABCD with the bases of AD and BC, the diagonals intersect at point O. The area of the triangle BOC

In the trapezoid ABCD with the bases of AD and BC, the diagonals intersect at point O. The area of the triangle BOC is 4, the area of the triangle AOD is 9. Find the area of the trapezoid.

Let’s draw the heights of the triangles OK – the height of the triangle BOC, ON – the height of the triangle AOD. NK is the height of the trapezoid.
Consider a triangle BOC:
S1 = 1/2 * BC * OK;
Express OK, we get:
OK = 2 * S1 / BC;
Consider triangle AOD:
S2 = 1/2 * AD * NO;
NO = 2 * S2 / AD;
Triangles AOD and BOC, <A = <C, <D = <B – as internal criss-crossing. <AOD = <BOC. So the triangles are similar in three corners.
AD ^ 2 / BC ^ 2 = S1 / S2.
AD = 2 * BC / 3
Substitute this into the expression for NO:
NO = 2 * S2 / (2 * BC / 3) = 3 * S2 / BC;
NK = NO + OK = (3 * S2 / BC) + (2 * S1 / BC) = (3 * S2 + 2S1) / BC;
Find the area of the trapezoid:
S = (a + b) * h / 2;
S = (AD + BC) * NK / 2 = (((2 * BC + 3BC) / 3) * (3 * S2 + 2S1) / BC) / 2 = (2 * 5 * BC / 3) * (12 +18) / BC = 100 cm ^ 2.
Answer: S = 100 cm ^ 2.