Question

if the length of a rectangle is reduced by 50% and the breadth is increased by 25% it takes the shape of a square find the ratio of(I) the area of the rectangle and the square and (2)the perimeter of the rectangle and the square

Answer 1

Answer:

(1) 1.6

(2) 1.4

Step-by-step explanation:

for the rectangle let :

length =x     and     breadth=y

the area of the rectangle (A) : A=x*y

the perimeter of the rectangle (P): P=(x+y)*2

then for the square

length =(1-.5)*x = .5x        and  breadth=(1+.25)*y=1.25y

the area of the square (A')= .5x * 1.25y = .625 x*y

the perimeter of the square (P')=.5 *x*4 =2x

the ratio of the area of the rectangle and the square =$\frac{x*y}{.625*x*y}$ =$\frac{1}{.625}$=1.6

the ratio of the perimeter of the rectangle and the square =$\frac{(x+y)*2}{2x}$=$\frac{x+y}{x}$

∵ .5x=1.25 y

∴y = .4x

∴the ratio of the perimeter of the rectangle and the square =$\frac{x+.4x}{x}$ =1.4

Answer 2

Answer:

for the rectangle let :

length =x and breadth=y

the area of the rectangle (A) : A=x*y

the perimeter of the rectangle (P): P=(x+y)*2

then for the square

length =(1-.5)*x = .5x and breadth=(1+.25)*y=1.25y

the area of the square (A')= .5x * 1.25y = .625 x*y

the perimeter of the square (P')=.5 *x*4 =2x

the ratio of the area of the rectangle and the square =\frac{x*y}{.625*x*y}

.625∗x∗y

x∗y

=\frac{1}{.625}

.625

1

=1.6

the ratio of the perimeter of the rectangle and the square =\frac{(x+y)*2}{2x}

2x

(x+y)∗2

=\frac{x+y}{x}

x

x+y

∵ .5x=1.25 y

∴y = .4x

∴the ratio of the perimeter of the rectangle and the square =\frac{x+.4x}{x}

x

x+.4x

=1.4