The sides and the smaller base of the trapezoid have the same length – 50 cm each.

| September 8, 2021 | education

The sides and the smaller base of the trapezoid have the same length – 50 cm each. Find the size of its larger base, at which the area of the trapezoid would be the largest.

Let’s build the height BH of the trapezoid ABCD.

Let the angle BAD = α0, and the side AB = BC = CD = a cm.

In a right-angled triangle ABN, BH = AB * Sinα = a * Sinα.

AH = AB * Cosα = a * Cosα.

Then AD = 2 * Cosα + a.

The area of ​​the trapezoid will be equal to: Savsd = (BC + AD) * BH / 2 = (a + 2 * Cosα + a) * a * Sinα / 2 =

2 * a * (1 + Cosα) * Sinα / 2 = a2 * (1 + Cosα) * Sinα = a2 * (Sinα + Sinα * Cosα) = a2 * (Sinα + Sin (2 * α) / 2).

Take the derivative with respect to the angle α and equate it to zero.

Cosα + Cos (2 * α) = 0.

Cosα + 2 * Cos2α – 1 = 0.

We introduce a new variable, let Cosα = X.

Then 2 * X2 + X – 1 = 0.

Let’s solve the quadratic equation.

X1 = -1. (Not suitable, since the angle α is acute).

X2 = 1/2.

Cosα = 1/2.

Angle α = arcos (1/2) = 600.

Then AH = 50 * 1/2 = 25 cm.

AD = 25 + 50 + 25 = 100 cm.

Answer: With AD equal to 100 cm, the area of ​​the trapezoid is the largest.



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