Answer: (2)
Let us consider the room with
Length (l) = 3x
Breadth (b) = 2x
Height = 1x
Area of the walls (A) = 2(l + b) × h
A = 2(3x + 2x) × 1x
A = 10x2 sq. unit.
Now
l1 = 6x
b1 = x
h1 = x/2
Nwe area (A1) = 2(6x + x) (x/2)
A1 = 7x2
\(\begin{array}{l}% decrease in the area of walls = \frac{10x^{2}-7x^{2}}{10x^{2}}\end{array} \)
% decrease in the area of walls = 30%
Suppose we have a square ABCD measuring sides be (x+1). If we draw a diagonal BC calculate the diagonal by using Pythagoras theorem.