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 $$m^2$$.