Question

James bought a 55" TV. If it's aspect ratio is 4:3 what is its area?

Answer 1

The TV length is measured diagonally, meaning a 55 inch TV has its diagonal of 55 inches as shown in the sketch below.

An aspect ratio is the ratio of width to height

`AspectRatio=\frac{\text{width}}{\text{height}}`

and we are told that it is 4: 3; therefore,

`AspectRatio=\frac{\text{width}}{\text{height}}=(4)/(3)`

Now, from Pythagoras's theorem, we know that

`55^2=(\text{width)}^2+(\text{height)}^2`

But we do not know the height and width, rather, we only know the ratio. But let us solve this by multiplying the ratio by a constant c such that

`\begin{gathered} \text{width = 4c} \\ \text{height}=3c \end{gathered}`

Notice that multiplying by c does not change the ratio because

`\frac{\text{width}}{\text{height}}=(4c)/(3c)=(4)/(3)`

which is the same thing.

Now the Pythagorean theorem gives

`(4c)^2+(3c)^2=55^2`
`\rightarrow(4c)^2+(3c)^2=3025`
`16c^2+9c^2=3025`
`25c^2=3025`
`c^2=(3025)/(25)`

`\therefore c=11.`

We have the value of c! The only thing remaining now is finding the width and height to calculate the area.

`\begin{gathered} \text{width =4c=4}\cdot11=44 \\ \text{height}=3c=3\cdot11=33 \end{gathered}`

Hence, the area of the TV is

`\text{area}=44\cdot33=1452in^2\text{.}`