Question 151151: A restaurant's aquarium is destroyed in an earthquake. the insurance company is willing to replace the aquarium with one of the same dimensions. The owner can only remember that the diagonal of the front was 45 inches, the length was 9 inches more than the height of the front, and the depths was the same as the height of the front. Find the dimensions of the aquarium.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A restaurant's aquarium is destroyed in an earthquake. the insurance company is willing to replace the aquarium with one of the same dimensions. The owner can only remember that the diagonal of the front was 45 inches, the length was 9 inches more than the height of the front, and the depths was the same as the height of the front. Find the dimensions of the aquarium.
:
Let x = height
then
(x+9) = length
:
Using Pythagorean a^2 + b^2 = c^2
Assign as follows:
a = x; b = (x+9); c = 45
:
x^2 + (x+9)^2 = 45^2
:
x^2 + x^2 + 18x + 81 = 2025
:
2x^2 + 18x + 81 - 2025 = 0
:
2x^2 + 18x - 1944 = 0
Simplify, divide by 2
x^2 + 9x - 972 = 0
Factors to:
(x+36)(x-27) = 0
Positive solution:
x = +27 inches is the height (and the width)
;
27 + 9 = 36 inches is the length
:
:
Check: 27^2 + 36^2 = 2025 which is 45^2
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