2022 AMC 10B Problems/Problem 16

The following problem is from both the 2022 AMC 10B #16 and 2022 AMC 12B #13, so both problems redirect to this page.

Problem

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?

$[asy] import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)); label("8",C,S); label("5",(3, 5.5),NW); label("4",(8, 2), E); [/asy]$

$\textbf{(A) }15\frac{1}{8}\qquad \textbf{(B) }15\frac{3}{8}\qquad \textbf{(C) }15\frac{1}{2}\qquad \textbf{(D) }15\frac{5}{8}\qquad \textbf{(E) }15\frac{7}{8}$

Solution 1

Let us label the points on the diagram.

$[asy] import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A", F, NW); label("B", B, S); label("C", C, S); label("D", D, SE); label("E", I, E); label("F", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); [/asy]$

By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC \cong \angle DCE \cong \angle FEG$ . Similarly, $\angle ACB \cong \angle CED \cong \angle EGF$ . Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$ .

As we are given a rectangle and a square, $AB = 4$ and $AC = 5$ . Therefore, $\triangle ABC$ is a 3-4-5 right triangle and $BC = 3$ .

$CE$ is also $5$ . So, using the similar triangles, $CD = 4$ and $DE = 3$ .

$EF = DF - DE = 4 - 3 = 1$ . Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$ . So,

$\begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*}$

Finally, we have

$\begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D)}\ 15 \frac{5}{8}}. \end{align*}$

~Connor132435

Solution 2 (Clever)

(Refer to the diagram above) Proceed the same way as solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of $\triangle ABC$ , $\triangle CDE$ , and the rectangle to the far left of the diagram. The area of $EFG$ is $\frac{3}{8}$ and thus the fractional part of the answer is $\frac{5}{8}$ . Our answer is $\boxed{\textbf{(D)}\ 15 \frac{5}{8}}$

~mathboy100

Solution 3 (Area of trapezoid)

Proceed similar to solution 1 and use similar triangles to find side length of GE. Then use area of a trapezoid to solve for the area of ACEG.

Video Solution by OmegaLearn Using Similar Triangles

https://youtu.be/mv2tYNhbAfk

~ pi_is_3.14

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 AMC 12 Problems and Solutions

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