2013 AMC 8 Problem 24
Attempt Problem 24 of the 2013 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
24.
Squares and are equal in area. Points and are the midpoints of sides and respectively. What is the ratio of the area of the shaded pentagon to the sum of the areas of the three squares?
Answer: C
Small hint:
Let each square have side length .
Big hint:
Subtract the unshaded rectangle and triangle from the total area .
Video solution:
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Written solution:
Let each square have side length , so the total area of the three squares is . It is easier to subtract the unshaded region to the left of diagonal from this total.
Using the solution diagram, rectangle has area , and triangle has area Thus the unshaded area is , so the shaded pentagon has area .
The requested ratio is .
Thus, C is the correct answer.
Problem 24 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
- 1996 AMC 8
- 1997 AMC 8
- 1998 AMC 8
- 1999 AMC 8
- 2000 AMC 8
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