SOLUTION: The length of a room exceeds the width by 5 feet, and the height is 3 feet less than the width. The area of the four walls exceeds the sum fo the areas of ceiling and floor by 240

Question 369984: The length of a room exceeds the width by 5 feet, and the height is 3 feet less than the width. The area of the four walls exceeds the sum fo the areas of ceiling and floor by 240 square feet. Find the dimensions of the room.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let width be x
length = x+5
height = x-3
..
Area of 4 walls = 2*LH + 2*WH
= 2*(x-3)(x+5)+2*x*(x-3)
=2(x^2+2x-15)+2x^2-6x
=2x^2+4x-30+2x^2-6x
=4x^2-2x-30
...
Area of ceiling & floor
Area = 2 * LW
Area = 2*x*(x+5)
Area = 2x^2+10x
..
4x^2-2x-30 =2x^2+10x+240
2x^2-12x-270=0
/2
x^2-6x-135=0
x^2-15x+9x-135=0
x(x-15)+9(x-15)=0
(x-15)(x+9)=0
x= 15 feet the width
length = x+5 = 15 +5 = 20 feet
height = x-3 = 15-3 = 12 feet
...
m.ananth@hotmail.ca
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