Mensuration Part 6

Problem 1: If the length of a rectangular field is increased by 20 % and the breadth is reduced by 20 %. The area of the rectangle will be 192 m2. What is the area of the original rectangle?

Solution:

10                       10      ————->100

12                        8       ————->96

96——->192

100——->?$\frac{192\times&space;100}{96}=200$

The original area of the rectangle is 200

Problem 2: An order was placed for the supply of carpet of breadth 3 meters. The length of carpet was 1.44 times of breadth. Subsequently, the breadth and length were increased by 25 and 40 percent respectively. At the rate of Rs. 45 per square meter, what would be the increase in the cost of the carpet?

Solution:

Length =3*1.44 = 4.32

Breadth=$3\times&space;\frac{125}{100}=3.75$

Length = $4.32\times&space;\frac{140}{100}=6.048$

=3.75*6.048*45 = 1020.6

Problem 3: The surface area of water in a swimming pool forms a rectangle with the length of 40 m and breadth 15m. The depth of water increases uniformly from 1.2m to 2.4m at the other end. The volume ( in m3) of water in the pool is?

Solution:

Volume = Length * Breadth * $(\frac{h_{1}+h_{2}}{2})$ = 40 * 15 *( $\frac{1.2+2.4}{2}$)

= 600*1.8 =  1080