Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 10B Problems/Problem 6"

(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 ...")
 
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
Because the legs are in the ratio of <math>4 : 3</math>, let one leg be <math>4x</math> and the other be <math>3x</math>.
+
If you divide the television screen into two right triangles, the legs are in the ratio of <math>4 : 3</math>, and we can let one leg be <math>4x</math> and the other be <math>3x</math>. Then we can use the Pythagorean Theorem.
  
−
Then we use the Pythagorean Theorem:
+
<cmath>\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*}</cmath>
  
−
<math>(4x)^2+(3x)^2=27^2</math>
+
The horizontal length is <math>5.4\times4=21.6</math>, which is closest to <math>\boxed{\mathrm{(D) \ } 21.5}</math>.
−
 
 
−
<math>16x^2+9x^2=729</math>
 
−
 
 
−
<math>25x^2=729</math>
 
−
 
 
−
<math>x^2=\frac{729}{25}</math>
 
−
 
 
−
<math>x=\frac{27}{5}</math>
 
−
 
 
−
<math>x=5.4</math>
 
−
 
 
−
The horizontal length is <math>5.4\cdot4=21.6</math>, which is closest to <math>\boxed{21.5 \text{ (D)}}</math>.
 
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2003|ab=B|num-b=5|num-a=7}}

Revision as of 03:15, 10 June 2011

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

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 $\boxed{\mathrm{(D) \ } 21.5}$.

See Also

2003 AMC 10B (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_6&oldid=39447"