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,

$AspectRatio=\frac{\text{width}}{\text{height}}=\frac{4}{3}$

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

$\frac{\text{width}}{\text{height}}=\frac{4c}{3c}=\frac{4}{3}$

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=\frac{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

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