2017 AMC 10B Problem 15

Attempt Problem 15 of the 2017 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

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 ADE?

1

\dfrac{42}{25}

\dfrac{28}{15}

2

\dfrac{54}{25}

Answer: E

Concepts:similarity area ratio right triangle

Difficulty rating: 1600

Small hint:

Use similarity between

\triangle ABE

and

\triangle ACB

Big hint:

Then compare

\triangle EAD

to

\triangle CDA

using their bases on

AC

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.