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?

,

i

- $2$
- $2.25$
- $2.5$
- $2.7$
- $3$

**Key Concepts**

Geometry

Square

Pythagoras

## Check the Answer

Answer: $2.7$

AMC-10A (2008) Problem 14

Pre College Mathematics

## Try with Hints

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 ……….

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 $\frac{3 x-y}{2}=\frac{3 x-2 x}{2}=\frac{x}{2}=\frac{27}{10}=2.7$

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=gGT15ls_brU&t=2s