2015 AMC 12A Problem 8

Attempt Problem 8 of the 2015 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

The ratio of the length to the width of a rectangle is 4:3.4 : 3. If the rectangle has diagonal of length d,d, then the area may be expressed as kd2kd^2 for some constant k.k. What is k?k?

27\dfrac{2}{7}

37\dfrac{3}{7}

1225\dfrac{12}{25}

1625\dfrac{16}{25}

34\dfrac{3}{4}

Answer: C
Concepts:Pythagorean Theoremratio and proportionarea
Difficulty rating: 1440
Solution:

Let the sides of the rectangle be 4a4a and 3a.3a. By the Pythagorean Theorem the diagonal is 5a=d,5a = d, so a=d5.a = \dfrac{d}{5}.

The area is 4a3a=12a24a\cdot 3a = 12a^2 =12(d5)2= 12\left(\dfrac{d}{5}\right)^2 =1225d2,= \dfrac{12}{25}d^2, so k=1225.k = \dfrac{12}{25}.

Thus, the correct answer is C.

← Problem 7#7
Full Exam

Problem 8 in Other Years