The area of greatest circle included in a right trapezoid

Question

A trapezium ABCD for which AD //BC and AD = 18 cm , BC = 33 cm , CD = 25 cm and angle ABC = 90 degree . Whats the area of the greatest circle which can be drawn inside the trapezium ? My turn : The greatest circle is that touches the side of the rectangle ABED where DE is perpendicular to BC meets it at E and AE = AB = CE = $\sqrt{CD^2 - CE^2 }=\sqrt{ 25^2 - 15^2 }= 20 $cm Then compute the radius and calculating the area of the circle. Is the solution correct ? Note : Calculus is not allowed

Answer 1

Now, you need to show that if we'll take a circle with a radius is greater than $10$,

so we can not place this circle inside the trapezoid.

Can you end it now?

The hint:

Let $\Phi$ be a circle with radius is greater than $10$, $O$ be a center of the circle and $\Phi$ is placed inside the trapezoid.

Now, take an altitude of the trapezoid such that $O$ is placed on this altitude.

Answer 2

The trapezium is part of a right triangle $BCF,$ see figure. In similar triangles $\triangle FAD$ and $\triangle FBC$ $$\frac{18}{33}=\frac{d}{d+25},$$ hence $d=30.$ Thus $\triangle FBC$ is Pythagorean with the lengths of sides $$FC=55, FB=44, BC=33.$$ In the similar triangle $\triangle FAD$ are the sides $30, 24, 18.$ Therefore, $AB=44-24=20.$

As shows the figure, the largest circle has diameter $AB.$