2022 AMC 10B Problems/Problem 16

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

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

Label the points on the diagram.

By doing some angle chasing, using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC$ , $\angle DCE$ , and $\angle FEG$ are congruent. Similarly, $\angle ACB$ , $\angle CED$ , and $\angle EGF$ are congruent as well.

Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$ .

Since the triangles are created by a rectangle and a square, $AC = 5$ and $AB = 4$ . $\angle ABC$ is a right angle, so $BC = 3$ .

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

$EF = 1$ by subtracting $DE$ from $DF$ . The area of a 3-4-5 right triangle is $6$ , so both $\triangle ABC$ and $\triangle CDE$ have an area of 6. The ratio of $EF$ to $CD$ = $1:4$ , so by similarity, the area of of $\triangle EFG$ is $6/4^2 = 3/8$ .

$BD = BC + CD = 3 + 4 = 7$ and $DF = 4$ , so the area of rectangle $ABDF$ is $28$ . Subtracting the area of $\triangle ABC, \triangle CDE,$ and $\triangle EFG$ , gets the shaded area, $28 - 6 - 6 - \tfrac{3}{8} = \boxed{\textbf{(D)} 15 \tfrac{5}{8}}$

~Connor132435

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2022 AMC 10B Problems/Problem 16

Problem

Solution