Question

The length of a rectangle exceeds its width by 6 m. If its perimeter is 44 m find its dimension. ​

Answer 1

✬ Length = 14 m ✬

✬ Breadth = 8 m ✬

Step-by-step explanation:

Given:

• Length of a rectangle exceeds width by 6 cm
• Perimeter of rectangle is 44 m.

To Find:

• What are the dimensions of rectangle?

Solution: Let the width of rectangle be x cm. Therefore,

➼ Length of rectangle = 6 m more than x

➼ Length = (x + 6) m

As we know that

★ Perimeter of Rectangle = 2(Length + Width) ★

A/q

$\implies{\rm }$ 44 = 2(Length + Width)

$\implies{\rm }$ 44 = 2(x + 6 + x)

$\implies{\rm }$ 44 = 2(2x + 6)

$\implies{\rm }$ 44 = 4x + 12

$\implies{\rm }$ 44 – 12 = 4x

$\implies{\rm }$ 32 = 4x

$\implies{\rm }$ 32/4 = x

$\implies{\rm }$ 8 = x

So,

➮ Width of rectangle is x = 8 m.

➮ Length of rectangle = x + 6

=> 8 + 6 = 14 m

Answer 2

GIVEN :-

•The length of a rectangle exceed its width by 6m.

• Perimeter of rectangle is 44m.

TO FIND :-

•The dimensions of rectangle.

SOLUTION :-

Let, the width of rectangle be R m , and length of rectangle is R + 6 m.

As we know that,

$\dagger \: { \boxed{ \red{ \tt{Perimeter \: of \: rectangle \: = \: 2( l + b) }}}}$

[ Put the values ]

$\tt \longrightarrow \: 44 \: = 2(R + R + 6 ) \\ \\ \tt \longrightarrow \: 44 = 2R \: + 2R \: + 12 \\ \\ \tt \longrightarrow \: 44 - 12 = 4R \\ \\ \longrightarrow \tt 32 \: = 4R \\ \\ \longrightarrow \tt R \: = \frac{ \cancel{32}}{ \cancel4} \\ \\ \longrightarrow \tt {\red{ R \: = 8}}$

Hence,

$\longrightarrow$ The length of rectangle is 14 m.

$\longrightarrow$ The width of rectangle is 8 m.