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2003 AMC 10B Problem 6

2003 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 1 of 5 · about 1 min · position P1-10

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4:3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?

A)
$20$
B)
$20.5$
C)
$21$
D)
$21.5$
E)
$22$

Answer

Key insight

A 4:3 rectangle has a 5-unit diagonal in the same units, so the horizontal length is 4/5 of the 27-inch diagonal, about 21.6.

Solution

Write the width as $4k$ and the height as $3k$ . The diagonal splits the screen into right triangles with legs $4k$ and $3k$ , so the diagonal is $\sqrt{(4k)^2 + (3k)^2} = 5k$ : the sides and the diagonal are in the ratio $4 : 3 : 5$ .

A $27$ -inch screen has diagonal $5k = 27$ , so $k = 5.4$ and the horizontal length is $4k = 21.6$ inches.

Among the choices, $21.6$ is closest to $21.5$ , so the answer is $\boxed{\textbf{(D)}\ 21.5}$ .

Why this works

Any rectangle with sides in ratio $4:3$ is a scaled $3$ - $4$ - $5$ right triangle doubled, so all three lengths scale together. Recognizing the Pythagorean triple removes the need to solve $25k^2 = 729$ from scratch: the width is always $\frac{4}{5}$ of the diagonal.

The trap

Taking the height (3/5 of 27 = 16.2) or treating 27 inches as the horizontal length itself.

Common mistakes

Taking the height ( $\frac{3}{5} \cdot 27 = 16.2$ ) or treating $27$ inches as the horizontal length itself.
Rounding $21.6$ to $22$ because it "rounds up," when the question asks for the closest listed value and $21.5$ is only $0.1$ away.

Techniques

Set up the equation/formula and compute; no special trick needed