2022 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2022 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:congruence (geometry)similaritytrapezoid

Difficulty rating: 2150

16.

The diagram below shows a rectangle with side lengths 44 and 88 and a square with side length 5.5. 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?

1518 15\dfrac{1}{8}

1538 15\dfrac{3}{8}

1512 15\dfrac{1}{2}

1558 15\dfrac{5}{8}

1578 15\dfrac{7}{8}

Solution:

Firstly, let's label the points as follows:

Since we have a rectangle, AB=4.AB = 4. By the Pythagorean Theorem, we have AC=3.AC = 3. Then, since ACB\angle ACB and DCE\angle DCE are complementary, BAC=CDE=90,\angle BAC = \angle CDE = 90^\circ, and BC=CEBC = CE we know ABCCDE.ABC \cong CDE. Therefore, ED=3ED = 3 and EF=1.EF = 1.

Since CED\angle CED and GEF\angle GEF are complementary and GEF=CDE=90,\angle GEF = \angle CDE = 90^\circ, we know EFGEFG and CDECDE are similar. This means ECCD=EGEF, \dfrac{EC}{CD} = \dfrac{EG}{EF}, so EG=1.25.EG = 1.25. Since the shaded region is a trapezoid, we can get the area as (GE+BC)EC2=5(5+1.25)2\dfrac{(GE+BC)EC}2 = \dfrac{ 5(5+1.25)}2 =56.252=15.625.= \dfrac{5\cdot 6.25}{2} = 15.625.

This is equal to 1558. 15 \dfrac 58.

Thus, the answer is D .

Problem 16 in Other Years