2015 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2015 AMC 12A, 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.

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Concepts:Pythagorean Theorem ratio and proportion area

Difficulty rating: 1440

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

$\dfrac{2}{7}$

$\dfrac{3}{7}$

$\dfrac{12}{25}$

$\dfrac{16}{25}$

$\dfrac{3}{4}$

Solution:

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

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

Thus, the correct answer is C.

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