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2015 AMC 10A Problem 11

2015 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 2 min · position P11-15

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 $k$ ?

A)
$\frac{2}{7}$
B)
$\frac{3}{7}$
C)
$\frac{12}{25}$
D)
$\frac{16}{25}$
E)
$\frac{3}{4}$

Answer

Key insight

Sides 4x and 3x make the diagonal 5x, so d = 5x and the area 12x^2 becomes 12(d/5)^2 = (12/25)d^2.

Solution

Let the sides be $4x$ and $3x$ . The diagonal is the hypotenuse of a $3$ - $4$ - $5$ right triangle scaled by $x$ , so
$d = 5x, \qquad x = \frac{d}{5}.$
The area is
$4x \cdot 3x = 12x^2 = 12\left(\frac{d}{5}\right)^2 = \frac{12}{25}\,d^2.$
Hence $k = \frac{12}{25}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{12}{25}}$ .

Why this works

A ratio condition means the shape is fixed up to scale, so everything can be written in one parameter $x$ . Since area scales with the square of length, any "area $= k \cdot (\text{length})^2$ " question reduces to computing the shape once at a convenient size and reading off the constant.

Alternative approach

Take the concrete $3 \times 4$ rectangle: $d = 5$ and the area is $12$ , so $k = \frac{12}{5^2} = \frac{12}{25}$ . The constant is scale-invariant, so one example suffices.

The trap

Using the diagonal as a side (area 12 with d = 5, ratio 12/5) or forgetting to square the 1/5 when substituting x = d/5.

Common mistakes

Using the diagonal as a side (area $12$ with $d = 5$ , ratio $\frac{12}{5}$ ) or forgetting to square the $\frac{1}{5}$ when substituting $x = d/5$ .
Reporting $\frac{16}{25}$ or $\frac{9}{25}$ , the squares of the individual side fractions, instead of their product.

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)