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The length of a room exceeds its breadth by 2 meters. If the length be increased by 4 meters and the breadth decreased by 2 meters, the area remains the same. Find the surface area of its walls if the height is 3 meters?
Let the breadth of room be 'b' meters. So, the length will be 'b+2' meters. So, Area = b(b+2)
If the length isΒ increased by 4 meters, then length='b+6' metersΒ andΒ the breadth decreased by 2 meters, then breadth='b-2' meters
So, initial area = new area.Β
b(b+2) = (b+6)(b-2)
$$b^2+2b\ =\ b^2+4b-12$$
On solving,Β b = 6 meters. So, Breadth = 6 meters and length=8 meters.
Surface area of walls = 2*h(l+b) = 2*3(6+8) = 84m^2
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