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).

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}

Let the sides be

4a

and

3a,

so the diagonal is

5a

Write the area

12a^2

in terms of

d = 5a

Let the sides of the rectangle be

4a

and

3a.

By the Pythagorean Theorem the diagonal is

5a = d,

a = \dfrac{d}{5}.

The area is

4a\cdot 3a = 12a^2

= 12\left(\dfrac{d}{5}\right)^2

= \dfrac{12}{25}d^2,

k = \dfrac{12}{25}.

Thus, the correct answer is C.

Problem 8 in other years