Question

the length of a rectangle exceed its width by 4 metre if its perimeter is 40 metre find its dimensions​

Answer 1

Answer:

let the width be x

length=x+4

perimeter=40m

perimeter of rectangle is 2(l+b)

solution:

2(x+ x+4) =40m

2(2x+4) =40m

2x + 4=40/2

2x+4=20

2x=20-4=16

x=16/2

x=8

so length =8+4=12

Width=8

Answer 2

Answer:

Answer:-

$\red{\bigstar}$ Length $\large\leadsto\boxed{\rm\purple{12 m}}$

$\red{\bigstar}$ Width $\large\leadsto\boxed{\rm\purple{8 m}}$

• Given:-

Length of a rectangle exceed its width by 4 metre.

Perimeter of the rectangle is 40 metre.

• To Find:-

The dimensions

• Solution:-

Let the width of the rectangle be 'x'.

According to the question:-

Length exceeds the width by 4 metre.

Hence,

Length = x + 4

• Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x+4 m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large x m}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\blue{Perimeter = 2(l+b)}}}$

→ $\sf 40 = 2(x+4 + x)$

→ $\sf 40 = 2(2x + 4)$

→ $\sf \dfrac{40}{2} = 2x + 4$

→ $\sf 2x + 4 = 20$

→ $\sf 2x = 20 - 4$

→ $\sf 2x = 16$

→ $\sf x = \dfrac{16}{2}$

→ $\bf\green{x = 8}$

Now,

Width = 8 m

→ Length = x + 4

→ Length = 8 + 4

Length = 12 m

Therefore, the length of the rectangle is 12m and width is 8 m.