In an isosceles trapezoid ABCD with bases DC and AD, the diagonals AC and BD meet at point O.

univerkov | March 23, 2021 | education

In an isosceles trapezoid ABCD with bases DC and AD, the diagonals AC and BD meet at point O. AO: OC = 4: 3, and the area of triangle ABO is 6. Find the area of trapezoid ABCD.

Since the trapezoid is isosceles, then its diagonals are equal, AC = BD, and the point O divides them into equal segments. ОВ = OC, AO = ОD.

Let AO = OD = 4 * X cm, then OB = OC = 3 * X cm.

The area of the triangle AOB is equal to: Saov = OB * AO * SinAOB / 2.

3 * X * 4 * X * SinAOB / 2 = 6.

X2 * SinAOB = 1.

In the BOC triangle, the BOC angle = (180 – AOB), then SinBOC = SinAOB.

Then the area of the BOC triangle is: Svos = DO * OC * SinBOC / 2 = 3 * X * 3 * X * SinBOC / 2 = 4.5 * X2 * SinBOC = 4.5 cm2.

Determine the area of the triangle AOD.

Saod = AO * OD * SinAOD / 2 = 4 * X * 4 * X * SinAOD / 2 = 8 * X2 * SinAOB = 8 cm2.

The area of the triangle COD is equal to the area of the triangle AOB.

Then Savsd = Saov + Ssod + Svos + Saod = 6 + 6 + 4.5 + 8 = 24.5 cm2.

Answer: The area of the trapezoid is 24.5 cm2.

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