ProblemBC of trapezoid ABCD is tangent at any point on circular arc DE whose center is O. Find the length of BC so that the area of ABCD is maximum.

Solution

From the figure: $z^2 + 5^2 = 15^2$

$z = 10\sqrt{2}$

For the angle theta: $\cos \theta = \dfrac{z}{15} = \dfrac{20}{BC}$

$BC = \dfrac{300}{z} = \dfrac{300}{10\sqrt{2}}$

$BC = 15\sqrt{2} \, \text{ cm} = 21.21 \, \text{ cm}$ answer

A solution to the same problem has been carried out with the aid of Calculus. See this link to to see the solution by maxima and minima: Solution by Calculus

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