In trapezoid ABCD, the base lengths are 3: 1, and the diagonals intersect at point O. Find the area of the trapezoid if the area of the triangle BOC = 1
The BOC and AOD triangles are similar in two angles, and the ratio of their areas is equal to the square of the similarity coefficient.
Let the length of the base BC = X cm, then AD = 3 * X cm.
Then Svos = Sаod = 1/9.
Saod = 9 cm2.
Let’s draw the height of the KH.
The ratio of the heights of similar triangles is equal to the coefficient of similarity.
KO / HO = 1/3.
HO = 3 * KO.
Height KH = KO + HO = KO + 3 * KO = 4 * KO.
Svos = BC * KO / 2 = 1 cm2. (1).
The area of the trapezoid is:
Savsd = (BC + AD) * KH / 2 = (BC + 3 * BC) * 4 * KO / 2 = 16 * BC * KO / 2.
Substitute equation 1 in the last expression.
Savsd = 16 * Sbos = 16 cm2.
Answer: The area of the trapezoid is 16 cm2.
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