Question

In trapezoid, two triangles are formed by the bases and the intersecting diagonal segments. The areas of these triangles are S1 and S2. Find area of the trapezoid in terms of S1 and S2.

Answer 1

`S_1+S_2+2√(S_1S_2)`

Explanation:

Consider trapezoid ABCD. Let triangle BEC be of area
`S_1` and triangle AED be of area
`S_2.`

Triangles BEC and AED are similar with the coefficient of similarity
`(A_(\triangle BEC))/(A_(\triangle AED))=\sqrt{(S_1)/(S_2)}.`

Consider triangles ABC and ACD. The area of these triangles are

`A_(\triangle ABC)=(1)/(2)\cdot BC\cdot H,\\ \\A_(\triangle ACD)=(1)/(2)\cdot AD\cdot H.`

Note that

`H=h_(BC)+h_(AD),` where
`h_(BC)` is the height of the triangle BEC,
`h_(AD)` is the height of the triangle AED and

`(h_(BC))/(h_(AD))=\sqrt{(S_1)/(S_2)}.`

Thus,

`h_(BC)=\sqrt{(S_1)/(S_2)}\cdot h_(AD)` and

`H=\left(1+\sqrt{(S_1)/(S_2)}\right)h_(AD)` or

`H=\left(1+\sqrt{(S_2)/(S_1)}\right)h_(BC).`

Then

`A_(\triangle ACD)=(1)/(2)\cdot AD\cdot H=(1)/(2)\cdot AD\cdot\left(1+\sqrt{(S_1)/(S_2)}\right)h_(AD)=\left(1+\sqrt{(S_1)/(S_2)}\right)S_2,`

`A_(\triangle ABC)=(1)/(2)\cdot BC\cdot H=(1)/(2)\cdot BC\cdot\left(1+\sqrt{(S_2)/(S_1)}\right)h_(BC)=\left(1+\sqrt{(S_2)/(S_1)}\right)S_1.`

Thus, the area of trapezoid ABCD is

`A_(ABCD)=A_(\triangle ABC)+A_(\triangle ACD)=\left(1+\sqrt{(S_1)/(S_2)}\right)S_2+\left(1+\sqrt{(S_2)/(S_1)}\right)S_1=S_2+√(S_1S_2)+S_1+√(S_1S_2)=S_1+S_2+2√(S_1S_2).`