2015 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2015 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10A solutions, or check the answer key.

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Concepts:rectangle Pythagorean Theorem ratio and proportion

Difficulty rating: 1070

11.

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 side lengths be $4x$ and $3x.$ Then the diagonal has length $\sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x.$

The area of the rectangle is $4x \cdot 3x = 12x^2.$ Then we get that $kd^2 = 12x^2$ $k = \dfrac{12x^2}{25x^2} = \dfrac{12}{25}.$

Thus, C is the correct answer.

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