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Television Problem | AMC 10A, 2008 | Problem 14

Try this beautiful problem from Geometry:Squarefrom AMC-10A (2008) You may use sequential hints to solve the problem.

Try this beautiful Television Problem from AMC – 10A, 2008.

Television Problem – AMC-10A, 2008- Problem 14


Older television screens have an aspect ratio of 4: 3 . That is, the ratio of the width to the height is 4: 3 . The aspect ratio of many movies is not $4: 3,$ so they are sometimes shown on a television screen by “letterboxing” – darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2: 1 and is shown on an older television screen with a 27 -inch diagonal. What is the height, in inches, of each darkened strip?

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Television Problem
  • $2$
  • $2.25$
  • $2.5$
  • $2.7$
  • $3$

Key Concepts


Geometry

Square

Pythagoras

Check the Answer


Answer: $2.7$

AMC-10A (2008) Problem 14

Pre College Mathematics

Try with Hints


Television Problem

The above diagram is a diagram of Television set whose aspect ratio of $4: 3$.Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$-inch diagonal. Then we have to find the height, in inches, of each darkened strip.

we assume that the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively. If we can find out the value of \(x\) and \(y\) then  the height of each strip can be calculate eassily

Can you now finish the problem ……….

Television Problem

By the Pythagorean Theorem, the diagonal is $\sqrt{(3 x)^{2}+(4 x)^{2}}=5 x=27 .$ So $x=\frac{27}{5}$

Now the movie and the screen have the same width, $2 y=4 x \Rightarrow y=2 x$

can you finish the problem……..

Thus, the height of each strip is $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

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