Question

Length of a rectangle is 20 centimetres and its breadth is 4 centimetres. When a new rectangle is formed by changing the length and breadth, its perimeter decreased by 8 centimetres and area increased by 16 square centimetres. Find
the change in its length and breadth.

DONT SPAM​

Answer 1

The perimeter of the original rectangle is,

$\longrightarrow\sf{P=2(20+4)}$

$\longrightarrow\sf{P=48\ cm}$

And its area is,

$\longrightarrow\sf{A=20\times4}$

$\longrightarrow\sf{A=80\ cm^2}$

Let the length and breadth be decreased by $\sf{x}$ and $\sf{y}$ respectively.

The new perimeter of the rectangle is,

$\longrightarrow\sf{2(20-x+4-y)=48-8}$

$\longrightarrow\sf{2(24-x-y)=40}$

$\longrightarrow\sf{24-x-y=20}$

$\longrightarrow\sf{x+y=4}$

$\longrightarrow\sf{y=4-x\quad\quad\dots(1)}$

And the new area of the rectangle is,

$\longrightarrow\sf{(20-x)(4-y)=80+16}$

$\longrightarrow\sf{80-4x-20y+xy=80+16}$

$\longrightarrow\sf{xy-4x-20y=16}$

From (1),

$\longrightarrow\sf{x(4-x)-4x-20(4-x)=16}$

$\longrightarrow\sf{4x-x^2-4x-80+20x=16}$

$\longrightarrow\sf{20x-80-x^2=16}$

$\longrightarrow\sf{x^2-20x+96=0}$

$\longrightarrow\sf{x^2-12x-8x+96=0}$

$\longrightarrow\sf{x(x-12)-8(x-12)=0}$

$\longrightarrow\sf{(x-12)(x-8)=0}$

$\Longrightarrow\sf{\underline{\underline{x=12\ cm}}\quad OR\quad \underline{\underline{x=8\ cm}}}$

And from (1),

$\Longrightarrow\sf{\underline{\underline{y=-8\ cm}}\quad OR\quad \underline{\underline{y=-4\ cm}}}$

Thus there are two cases:

1.  We can decrease length by 12 cm and increase breadth by 8 cm.

2. We can decrease length by 8 cm and increase breadth by 4 cm.

However, the new dimensions of the rectangle are 12 cm and 8 cm.

Answer 2

Answer:

perimeter of original rectangle

P= 2(20+4)

P= 48cm2

area 20 ✖ 4=80

let the length any breadth be decreased by x any y

new perimeter 2(20-x+4-y) =48-8

2(24-x-y)=40

x+x=4

y=4-x

new area of rectangle (20-x) (4-y) =80+16

xy -4x -20=16

from ( I)

x(4-x)-4x-20(4-x)=16

x^2-20x +96