What is the area of the Square?

Geometry Level 4

A triangle DEF is inscribed in square ABCD such that E and F are on lines AB and BC respectively, and the lengths of DE and EF are 4 and 3 respectively. It is known that angle DEF is 90°.

The area of the square can be represented as a/b, where a and b are positive integers.

What is a + b?

The answer is 273.

1 solution

David Vreken
Sep 19, 2020

By the Pythagorean Theorem on $\triangle DEF$ , $DF = \sqrt{3^2 + 4^2} = 5$ .

Let the side of the square be $x$ . By the Pythagorean Theorem on $\triangle DCF$ , $CF = \sqrt{25 - x^2}$ , and by the Pythagorean Theorem on $\triangle DAE$ , $AE = \sqrt{16 - x^2}$ .

Also, $FB = BC - CF = x - \sqrt{25 - x^2}$ and $EB = AB - AE = x - \sqrt{16 - x^2}$ .

By the Pythagorean Theorem on $\triangle EBF$ , $EB^2 + FB^2 = EF^2$ , or $(x - \sqrt{16 - x^2})^2 + (x - \sqrt{25 - x^2})^2 = 3^2$ , which solves to $x = \frac{16}{\sqrt{17}}$ and $x = \frac{16}{\sqrt{65}}$ . However, $x = \frac{16}{\sqrt{65}}$ would make $EB = x - \sqrt{16 - x^2}$ negative, so $x = \frac{16}{\sqrt{17}}$ .

The area of the square is then $A = x^2 = (\frac{16}{\sqrt{17}})^2 = \frac{256}{17}$ , so $a = 256$ , $b = 17$ , and $a + b = \boxed{273}$ .

What is the area of the Square?

The answer is 273.

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