Question

Your mother wants to get a new television but she would like to have an older model with an aspect ratio of 4:3 she is looking at a 32 inch television. what is the width height and area of the television?

Answer 1

Answer:

Part A) The height of the television is  $25.6\ in$

Part B) The width of the television is  $19.2\ in$

Part C) The area of the television is  $491.52\ in^{2}$

Step-by-step explanation:

Let

x-----> the width of the television

y-----> the height of the television

we know that

The diagonal of the television is $32\ in$

$\frac{y}{x}=\frac{4}{3}$

$y=\frac{4}{3}x$ ----> equation A

Applying the Pythagoras Theorem

${y^{2} +x^{2}} =32^{2}$------> equation B

substitute the equation A in equation B

${(\frac{4}{3}x)^{2} +x^{2}} =32^{2}$

${\frac{16}{9}x^{2} +x^{2}} =1,024$

${\frac{25}{9}x^{2}=1,024$

$x=3*32/5=19.2\ in$

Find the value of y

$y=\frac{4}{3}x$-------> $y=\frac{4}{3}(19.2)=25.6\ in$

Find the area of the television

The area of rectangle is equal to

$A=bh$

substitute

$A=19.2*25.6=491.52\ in^{2}$