In trapezoid ABCD with bases BC and AD, the diagonals intersect at point O. The areas of triangles BOC and AOD are equal to S1 and S2. Find the area of the trapezoid.
Triangles AOB and COD are similar in two angles. The angle AOB is equal to the angle OCD as the vertical angles at the intersection of the diagonals. The OBA angle is equal to the ODC angle as criss-crossing angles. The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
S1 / S2 = (OC / OA) ^ 2.
OC / OA = √ (S1 / S2).
Consider the triangles AOD and COD in which one can draw the same height to the bases of OA and OC.
Then the ratio of the areas of these triangles is equal to the ratio of the bases.
Sod / S2 = OC / OA = √ (S1 / S2).
Sod = S2 * √ (S1 / S2) = √ (S1 * S2).
Since the diagonals of the trapezoid cut off two equal triangles at the lateral sides, then Saov = S sod = √ (S1 * S2).
Then the area of the trapezoid will be equal to:
Savsd = S1 + S2 + √ (S1 * S2) + √ (S1 * S2) = S1 + 2 * √ (S1 * S2) + S2 = (√S1 + √S2) 2 cm2.
Answer: The area of the trapezoid is (√S1 + √S2) ^ 2 cm2.
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