A geometry problem by Chung Kevin

Geometry Level 2

$ABCD$ is a $25 \times 12$ rectangle. $E$ is a point on $AB$ such that $AE = 9, EB = 16$ .

What is the measure of $\angle DEC$ ?

Note: Diagram not drawn to scale.

60 ^ \circ

120 ^ \circ

105 ^ \circ

90^ \circ

1 solution

Brian Charlesworth
Jan 30, 2017

By the Pythagorean Theorem, $|DE|^{2} = |AD|^{2} + |AE|^{2} = 9^{2} + 12^{2} = 225 = 15^{2} \Longrightarrow |DE| = 15$ .

Also, $|EC|^{2} = |EB|^{2} + |BC|^{2} = 16^{2} + 12^{2} = 400 = 20^{2} \Longrightarrow |EC| = 20$ .

But $|DE|^{2} + |EC|^{2} = 15^{2} + 20^{2} = 625 = 25^{2} = |DC|^{2}$ ,

and so $\Delta DEC$ is a right triangle with $\angle DEC = \boxed{90^{\circ}}$ .