# 2003 AMC 10B Problems/Problem 6

The following problem is from both the 2003 AMC 12B #5 and 2003 AMC 10B #6, so both problems redirect to this page.

## Problem

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? $\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$

## Solution 1

If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$, and we can let one leg be $4x$ and the other be $3x$. Then we can use the Pythagorean Theorem. \begin{align*}(4x)^2+(3x)^2&=27^2\\ 16x^2+9x^2&=729\\ 25x^2&=729\\ x^2&=\frac{729}{25}\\ x&=\frac{27}{5}\\ x&=5.4\end{align*}

The horizontal length is $5.4\times4=21.6$, which is closest to $\boxed{\textbf{(D) \ } 21.5}$.

## Solution 2

One can realize that the diagonal, vertical, and horizontal lengths all make up a $3,4,5$ triangle. Therefore, the horizontal length, being the $4$ in the $4 : 3$ ratio, is simply $\frac{4}{5}$ times the hypotenuse. $\frac{4}{5}\cdot27=21.6 \approx \boxed{\textbf{(D) } 21.5}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 