Question

The area of the rectangle gets reduced by 9 sq units if the length is reduced by 5 units and the breath is increased by 3 units. The area is increased by 63 sq units if length is increased by 3 units and breath is increased by 2 units . Find the perimeter of the rectangle

Answer 1

Given :

• The area of the rectangle gets reduced by 9 sq units if the length is reduced by 5 units and the breath is increased by 3 units.
• The area is increased by $\bold{\red{63}}$ sq units if length is increased by 3 units and breath is increased by 2 units.

To Find :

• Perimeter of the rectangle.

Solution :

Let the length of the rectangle be x units.

Let the breadth of the rectangle be y units.

Area of rectangle = xy units²

Case 1:

The area of the rectangle is reduced by 9 sq units, when the length is reduced by 5 units and breadth is increased by 3 units.

Length = (x - 5) units.

Breadth = (y + 3) units.

Equation :

$\implies$ $\sf{xy-9\:=\:(x-5)(y+3)}$

$\implies$ $\sf{xy-9=x(y+3)-5(y+3)}$

$\implies$ $\sf{xy-9=xy+3x-5y-15}$

$\implies$ $\sf{xy-9-xy=3x-5y-15}$

$\implies$ $\sf{-9=3x-5y-15}$

$\implies$ $\sf{-9+15=3x-5y}$

$\implies$ $\sf{6=3x-5y\:\:\:\:\:(1)}$

Case 2 :

The area of the rectangle is increased by 63 sq units when the length is increased by 3 units and breath is increased by 2 units.

Length = (x + 3) units.

Breadth = (y + 2) units.

Equation :

$\implies$ $\sf{xy+63=(x+3)(y+2)}$

$\implies$ $\sf{xy+63=x(y+2)+3(y+2)}$

$\implies$ $\sf{xy+63=xy+2x+3y+6}$

$\implies$ $\sf{xy+63-xy=2x+3y+6}$

$\implies$ $\sf{63=2x+3y+6}$

$\implies$ $\sf{63-6=2x+3y}$

$\implies$ $\sf{57=2x+3y}$

$\implies$ $\sf{57-3y=2x}$

$\implies$ $\sf{\dfrac{57-3y}{2}=x\:\:\:\:(2)}$

Now, consider equation (1),

$\implies$ $\sf{3x-5y=6}$

$\implies$ $\sf{3\Big(\dfrac{57-3y}{2}\Big)\:-5y=6}$

$\bold{\big[From\:equation\:(1)\:x\:=\:\dfrac{57-3y}{2}\big]}$

$\implies$ $\sf{\Big(\dfrac{171-9y}{2}\Big)\:-\:5y\:=6}$

$\implies$ $\sf{\dfrac{171-9y}{2}-5y=6}$

$\implies$ $\sf{\dfrac{171-9y-10y}{2}=6}$

$\implies$ $\sf{171-19y=12}$

$\implies$ $\sf{-19y=12-171}$

$\implies$ $\sf{-19y=12-171}$

$\implies$ $\sf{y=\dfrac{-159}{-19}}$

$\implies$ $\sf{y=\dfrac{159}{19}}$

$\implies$ $\sf{y=8.36}$

Substitute, y = 8.36 in equation (1),

$\implies$ $\sf{3x-5y=6}$

$\implies$ $\sf{3x-5(8.36)=6}$

$\implies$ $\sf{3x-41.8=6}$

$\implies$ $\sf{3x=6+41.8}$

$\implies$ $\sf{3x=47.8}$

$\implies$ $\sf{x=\dfrac{47.8}{3}}$

$\implies$ $\sf{x=15.93}$

We have length and breadth of the rectangle.

Length = x = 15.93 units

Breadth = y = 8.36 units

We know,

$\large{\boxed{\bold{Perimeter_{rectangle}\:=\:2(length\:+\:breadth)}}}$

Block in the values,

$\sf{Perimeter\:=\:2(x+y)}$

$\implies$ $\sf{Perimeter\:=\:2(15.93+8.36)}$

$\implies$ $\sf{Perimeter\:=\:31.86+16.72}$

$\implies$ $\sf{Perimeter\:=\:48.58}$

$\large{\boxed{\bold{\purple{Perimeter_{rectangle}\:=\:48.58\:units}}}}$

Answer 2

Answer:

$\boxed{\red{\bf \therefore Perimeter = 48.58 \: units. }}$

Step-by-step explanation:

Let the length of the rectangle be x units. And, the breadth of the rectangle be y units.

We know that,

Area of rectangle = length * Breadth

= xy units².

It is given that,

The area of the rectangle gets reduced by 9 sq units, if the length is reduced by 5 units and the breadth is increased by 3 units.

Thus,

$\implies \sf (x-5)(y+3)= xy - 9\\\\\implies \sf xy + 3x - 5y - 15 = xy - 9\\\\\implies \bf 3x - 5y = 15-9\\\\\implies \sf 3x - 5y = 6 \: \: \dots (i)$

Again, it is given that -

The area is increased by 63 sq units, if length is increased by 3 units and breadth is increased by 2 units .

So,

$\implies \sf (x+3)(y+2) = xy + 63\\\\\implies \sf xy + 2x + 3y + 6 = xy + 63 \\\\\implies \sf 2x + 3y = 63-6\\\\\implies \sf 2x + 3y = 57 \: \: \dots (ii)$

Multiplying equation (i) by 2 and equation (ii) by 3. We get ;

$\sf \implies 6x - 10y = 12 \: \:\dots (iii) \\\\\implies \sf 6x + 9y = 171 \:\:\dots (iv)$

Subtracting equation (iii) from (iv),

$\implies \sf 6x - 10y - (6x + 9y)= 12 - 171\\\\\implies \sf 6x - 10y - 6x - 9y = - 159 \\\\\implies \sf - 19y = - 159 \\\\\implies \sf y = \frac{-159}{-19} \\\\\boxed{\red{\bf \therefore y = 8.36}}$

Now, substituting the value of y in (i),

$\implies \sf 3x-5y=6\\\\\implies \sf 3x - 5(8.36)=6\\\\\implies \sf 3x - 41.8 = 6 \\\\\implies \sf 3x = 6+41.8\\\\\implies \sf 3x = 47.8 \\\\\implies \sf x = \frac{47.8}{3} \\\\\boxed{\red{\bf \therefore x= 15.93}}$

Now, we know that :

Perimeter of rectangle = 2 ( length + breadth )

                                      = 2 ( x + y )

                                      = 2 ( 15.93 + 8.36 )

                                      = 2 ( 24.29 )

                                      = 48.58 units.

Hence, the perimeter of the given rectangle is 48.58 units.