2025 AMC 8 Problems/Problem 10

Problem

In the figure below, $ABCD$ is a rectangle with sides of length $AB = 5$ inches and $AD$ = 3 inches. Rectangle $ABCD$ is rotated $90^\circ$ clockwise around the midpoint of side $DC$ to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles?

$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 22.25 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 23.75 \qquad \textbf{(E)}\ 25$

Solution 1

The area of each rectangle is $5 \cdot 3 = 15$ . Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length $2.5$ (as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is $15+15-2.5^2=\boxed{\textbf{(D)}~23.75}$ .

~ cxsmi

~ Edited by Aoum

Video Solution by Cool Math Problems

https://youtu.be/BRnILzqVwHk?si=YIRjHgok5ifMsK_x&t=560

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=geTr-OnL-4wC94XG&t=814 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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