The aspect ratio of a film means the ratio at which one dimmension is bigger than the other. In this case the movie has an aspect ratio of "1.25:1", which means that the width is 1.25 times greater than the height. The same is true for the TV. When we say that a TV has 36 inches, we are talking about its diagonal, the ratio from width to height in this case is 4:3, which means that for every 3 inches on the height there are 4 in the width.
For the TV the relationship between the height and width of the TV is:
[tex]\text{width}=\frac{4}{3}\cdot height[/tex]We can notice that the dyagonal forms a right triangle with the sides. So we can use Pythagora's to find the dimmensions of the TV.
[tex]\begin{gathered} 36^2=width^2+height^2 \\ 1296=(\frac{4}{3}\cdot height)^2+height^2 \\ 1296=\frac{16}{9}\cdot height^2+height^2 \\ \frac{25}{9}\cdot height^2=1296 \\ \text{height}^2=\frac{1296\cdot9}{25}=466.56 \\ \text{height}=\sqrt[]{466.56=21.6} \end{gathered}[/tex]With the value of the height we can calculate the width.
[tex]\text{width}=\frac{4}{3}\cdot21.6=28.8[/tex]The area of the TV is given by:
[tex]\begin{gathered} \text{area = height}\cdot\text{ width} \\ \text{area}=21.6\cdot28.8=622.08 \end{gathered}[/tex]The area of the TV is 622.08 square inches.
We now need to calculate the width, height and the area of the image. We have:
[tex]\begin{gathered} \text{height}_{\text{image}}=\text{height}_{tv}=21.6 \\ \text{ with}_{\text{image}}=height_{tv}\cdot1.25=21.6\cdot1.25=27 \end{gathered}[/tex]The height of the image is the same as the TV, while the width of the image gets adjusted to 27" instead of 28.8". The area of the image is:
[tex]\text{area}_{\text{image}}=27\cdot21.6=583.2[/tex]The area of the image is 583.2 square inches.
The percentage of image area is given by:
[tex]\begin{gathered} \text{percent = }\frac{583.2}{622.08}\cdot100 \\ \text{percent}=93.75\text{ \%} \end{gathered}[/tex]The bar is on the left and right side of the image, the image will be centralized so we need to divide the difference between the TV's width and image width and divide it by 2.
[tex]\text{width }_{blackbar}=\frac{28.8-27}{2}=\frac{1.8}{2}=0.9[/tex]While the height of the blackbar is the same as the height of the TV.
[tex]\text{height}_{\text{blackbar}}=21.6[/tex]The area is the product between height and width.
[tex]A_{blackbar}=0.9\cdot21.6=19.44[/tex]The area of one blackbar is 19.44 square inches.
