Question

The perimeter of a rectangle is 240cm.its length is increased by 10% and its breadth is decreased by 20% but the perimeter remains unchanged. find the length and breadth of the rectangle

Answer 1

Solution:-

Let the dimensions of the

rectangle are

i) length = x cm

breadth = y cm

Perimeter = 240 cm ( given )

2( x + y ) = 240

x + y = 120 ----( 1 )

If length is decreased by 10% and

breadth is increased by 20% then

the new dimensions are

Length = x ( 100-10)/100

= 90x /100

= 9x /10

Breadth = y× ( 100 +20 )/100

= 120y /100

= 12y /10

Perimeter = 240 cm

2 [ 9x /10 + 12y /10 ] = 240

9x + 12y = 1200

Divide each term with 3

3x + 4y = 400-----( 2 )

Multiply equation ( 1 ) with 3 and

Subtract from ( 2 )

y = 40

Put y = 40 in ( 1 )

x = 80

Therefore ,

Required rectangle dimensions are

Length = x = 80 cm

Length = x = 80 cmBreadth = y = 40 cm

=================

@GauravSaxena01

Answer 2

Answer:

Question:-

the length of a rectangle is 8m more than its breadth if its perimeter is 128m, find its length , breadth and Area

Answer:-

The length of Rectangle is 36 m

The breadth of rectangle is 28 m

The area of Given rectangle is 1008 m².

To find:-

Length and breadth of rectangle

Area of rectangle

Solution:-

Let the breadth be x

Length = 8 + x

Perimeter = 128 m

$\boxed{ \large{ \mathfrak{perimeter = 2(l + b)}}}$

According to question,

$\large{ \tt: \implies \: \: \: \: \: 2(8 + x + x) = 128}$

$\begin{gathered} \large{ \tt: \implies \: \: \: \: \: 8 + 2x = \frac{128}{2} } \\ \end{gathered}:$

$\large{ \tt: \implies \: \: \: \: \: 8 + 2x = 64}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 64 - 8}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 56}$

$\large{ \tt: \implies \: \: \: \: \: x = 28}$

The breadth of rectangle is 28 m

Length = 8 + x = 28 + 8 = 36 m

$\large{ \boxed{ \mathfrak{area = l \times b}}}$

$\large{ \tt: \implies \: \: \: \: \: area = 28\times 36}$

$\large{ \tt: \implies \: \: \: \: \: area = 1008 \: {m}^{2} }$

The area of Given rectangle is 1008 m².

Step-by-step explanation:

Question:-

the length of a rectangle is 8m more than its breadth if its perimeter is 128m, find its length , breadth and Area

Answer:-

The length of Rectangle is 36 m

The breadth of rectangle is 28 m

The area of Given rectangle is 1008 m².

To find:-

Length and breadth of rectangle

Area of rectangle

Solution:-

Let the breadth be x

Length = 8 + x

Perimeter = 128 m

$\boxed{ \large{ \mathfrak{perimeter = 2(l + b)}}}$

According to question,

$\large{ \tt: \implies \: \: \: \: \: 2(8 + x + x) = 128}$

$\begin{gathered} \large{ \tt: \implies \: \: \: \: \: 8 + 2x = \frac{128}{2} } \\ \end{gathered}:$

$\large{ \tt: \implies \: \: \: \: \: 8 + 2x = 64}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 64 - 8}$

$\large{ \tt: \implies \: \: \: \: \: 2x = 56}$

$\large{ \tt: \implies \: \: \: \: \: x = 28}$

The breadth of rectangle is 28 m

Length = 8 + x = 28 + 8 = 36 m

$\large{ \boxed{ \mathfrak{area = l \times b}}}$

$\large{ \tt: \implies \: \: \: \: \: area = 28\times 36}$

$\large{ \tt: \implies \: \: \: \: \: area = 1008 \: {m}^{2} }$

The area of Given rectangle is 1008 m².