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(b) of the room be 'x' metres.
then, length(l) of the room = x+2 metres.
Area(A) = $$l\times b$$ = x(x+2) $$m^2$$
Given, length is increased by 4 meters and the breadth decreased by 2 meters
Then, new length(l') of the room = x+6 metres
new breadth(b') of the room = x-2 metres
New Area(A') of the room = $$l'\times b'$$ = (x+6)(x-2) $$m^2$$
Also given that, A = A'
$$\Rightarrow x(x+2) = (x+6)(x-2)$$
$$\Rightarrow x^2+2x = x^2+4x-12$$
$$\Rightarrow 2x = 12$$
$$\Rightarrow x = 6$$
Therefore the length of the room (l) = 8 metres
and breadth of the room (b) = 6 metres
and given height of the room (h) = 3 metres
Since the room will be in the shape of a cuboid, Surface area = 2 ($$l\times b+b\times h+l\times h$$)
But the Surface area of Walls = Total Surface area - Area of Roof and Floor = 2 ($$l\times b+b\times h+l\times h) - 2 (l\times b) = 2 (8\times 3+6\times 3) = 84 m^2$$
Hence, Surface Area of walls = 84 $$m^2$$.
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