2022 AMC 10B Problem 16

Question

The diagram below shows a rectangle with side lengths

4

and

8

and a square with side length

5.

Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

15\dfrac{1}{8}

15\dfrac{3}{8}

15\dfrac{1}{2}

15\dfrac{5}{8}

15\dfrac{7}{8}

Answer: D

Concepts:congruence (geometry)similarity trapezoid

Difficulty rating: 2150

Small hint:

The side length

5

gives a

3

–

4

–

5

slope relation

Big hint:

Use coordinates or similar triangles for the tilted square

Solution:

Label the points as shown:

Because

AB=4

and the square side

BC=5,

right triangle

ABC

gives

AC=3.

Also

BC\perp CE

and

A,C,D

are collinear, so

\angle ABC=\angle DCE.

The right triangles

ABC

and

CDE

have equal hypotenuses

BC=CE=5,

so they are congruent. Thus

CD=4, DE=3,

and

EF=4-DE=1.

Right triangles

EFG

and

CDE

are similar, so

\frac{EG}{EF}=\frac{EC}{CD}=\frac54.

Hence

EG=\frac{5}{4}.

The shaded region

BCEG

is a trapezoid whose parallel sides are

BC=5

and

EG=\frac{5}{4},

and whose height is the perpendicular side

CE=5.

Its area is

\frac12\left(5+\frac54\right)5=\frac{125}{8}=15\frac58.

Thus, the answer is D .