SOLUTION: You are planning on putting a new digital TV on a wall that is 12 feet long and 9 feet high. The digital TV has a diagonal of 72in. The length of the TV is twice the width of the

Algebra ->  Pythagorean-theorem -> SOLUTION: You are planning on putting a new digital TV on a wall that is 12 feet long and 9 feet high. The digital TV has a diagonal of 72in. The length of the TV is twice the width of the      Log On


   

Question 240504: You are planning on putting a new digital TV on a wall that is 12 feet long and 9 feet high. The digital TV has a diagonal of 72in. The length of the TV is twice the width of the TV. How much of the wall will still need to be decorated around the TV?
We have tried to input doubling the length. We used the numbers 48 and 24. But when we put the numbers into the therom, they did not match up.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
You are planning on putting a new digital TV on a wall that is 12 feet long and 9 feet high.
The digital TV has a diagonal of 72in.
The length of the TV is twice the width of the TV.
How much of the wall will still need to be decorated around the TV?
:
The area of the wall: 12 * 9 = 108 sq/ft
:
Find the dimensions of the TV using a^2 + b^2 = c^2
where
a = x
b = 2x (twice the width)
c = 72
:
x^2 + (2x)^2 = 72^2
:
x^2 + 4x^2 = 5184
:
5x^2 = 5184
:
x^2 = 5184%2F5
:
x^2 = 1036.8
x = sqrt%281036.8%29
x = 32.2 inches is the width of the TV
:
The dimensions of the TV; 32.2 by 64.4 inches (length is twice the width)
The area: 32.2 * 64.4 = 2073.68 sq/inches
Convert to sq/ft 2073.68%2F144 = 14.4 sq/ft
:
Find the remaining area on the wall
108 - 14.4 = 93.6 sq/ft