2003 AMC 10B Problems/Problem 6

Revision as of 19:06, 5 June 2011 by Djmathman (talk | contribs) (Created page with "==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 ...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

Because the legs are in the ratio of $4 : 3$, let one leg be $4x$ and the other be $3x$.

Then we use the Pythagorean Theorem:

$(4x)^2+(3x)^2=27^2$

$16x^2+9x^2=729$

$25x^2=729$

$x^2=\frac{729}{25}$

$x=\frac{27}{5}$

$x=5.4$

The horizontal length is $5.4\cdot4=21.6$, which is closest to $\boxed{21.5 \text{ (D)}}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
Invalid username
Login to AoPS