2015 AMC 10A Problem 11

Question

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}

A

3

-

4

-

5

rectangle has diagonal scale

5x

Express the area in terms of

d=5x

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.

Problem 11 in other years