Question

Rectangle $ABCD$ has $AB=3$ and $BC=4.$ Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}.$ What is the area of $\triangle AED?$

$1$

$\dfrac{42}{25}$

$\dfrac{28}{15}$

$2$

$\dfrac{54}{25}$

Answer: E

Concepts:similarity area ratio right triangle

Difficulty rating: 1600

Solution:

The area of $\triangle CDA$ is $\dfrac{3\cdot4}{2}=6$ . Since $E$ lies on $AC$ , triangles $EAD$ and $CDA$ share the same altitude from $D$ , so $[EAD]=[CDA]\cdot \dfrac{AE}{AC}$ .

By the Pythagorean Theorem, $AC=5$ . Also $\triangle ABE\sim \triangle ACB$ , so $\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35$ , giving $AE=\dfrac95$ . Thus $\dfrac{AE}{AC}=\dfrac{9}{25}$ .

Therefore $[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}$ . Thus, E is the correct answer.

2017 AMC 10B Problem 15

Answer: E

Difficulty rating: 1600

Solution:

Problem 15 in Other Years