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Explanation
The television has a diagonal that measures 36 inches:
And the ratio is 4:3
[tex]\begin{gathered} \frac{w}{h}=\frac{4}{3} \\ w=\frac{4}{3}h \end{gathered}[/tex]We can use the Pythagorean theorem to find the height of the TV:
[tex]\begin{gathered} 36^2=h^2+w^2 \\ 36^2=h^2+(\frac{4}{3}h)^2 \\ 36^2=h^2(1+\frac{4^2}{3^{2}}) \\ 1296=h^2(1+\frac{16}{9}) \\ 1296=h^2\times\frac{25}{9} \\ h^2=1296\times\frac{9}{25} \\ h=\sqrt[]{1296\times\frac{9}{25}}=21.6 \end{gathered}[/tex]The height of the TV is 60 inches. It's width is:
[tex]w=\frac{4}{3}h=\frac{4}{3}\times21.6=28.8[/tex]w=80 inches
Therefore the area of the TV is
[tex]A_{TV}=w\times h=28.8\times21.6=622.08in^2[/tex]The move has an aspec ratio of 25:1 shown as a pillarboxed image. This means that this is what we see:
So we know that the image height is the same as the TV's, 21.6 inches.
The relation between it's height and it's width is:
[tex]\begin{gathered} \frac{w}{h}=\frac{1.25}{1} \\ w=1.25h \\ \text{if h = 21.6 in} \\ w=27in \end{gathered}[/tex]The area of the image is:
[tex]A_{\text{image}}=w_{\text{image}}\times h=27\times21.6=583.2[/tex]The area of the two blackbars is the difference between the area of the TV and the area of the image:
[tex]A_{2-blackbars}=A_{TV}-A_{image}=622.08-583.2=38.88in^{2}[/tex]Since we need to find the area of just one blackbar, we just have to divide the area of both blackbars by 2:
[tex]A_{1-blackbar}=\frac{A_{2-blackbars}}{2}=\frac{38.88}{2}=19.44in^{2}[/tex]Answer
• Area of the TV: ,622.08 in²
,• Area of the image: ,583.2 in²
,• Area of one blackbar: ,19.44 in²
