Question

. The length of a rectangle is 5 m greater than its width. If the perimeter of the rectangle is 50 m, find its length .​

Answer 1

$\sf\huge{\underline{\underline\pink{AnsWeR}}}$

let

,

breadth = x

so ,the length = x + 5

perimeter = 2(l+b)

50 = 2(x+5+x)

25 =2x + 5

x = 10

so the length is x + 5 = 10 + 5 = 15m

Answer 2

Answer:

Answer:-

$\red{\bigstar}$ Length of rectangle $\large\leadsto\boxed{\rm\pink{15 \: m}}$

• Given:-

Length of rectangle is 5m greater than its width.

Perimeter of rectangle is 50m

• To Find:-

Length of the rectangle

• Solution:-

Let the width of the rectangle be 'x' m.

• According to the question:-

The length of rectangle is 5m greater than its width.

Hence,

The length will be x+5 m

★ Diagram:-

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

We know, perimeter of rectangle:-

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

• Substituting in the Formula:-

→ $\sf 50 = 2(x+5 \: + x)$

→ $\sf 50 = 2(2x + 5)$

→ $\sf 50 = 4x + 10$

→ $\sf 4x = 50 - 10$

→ $\sf 4x = 40$

→ $\sf x = \dfrac{40}{4}$

$\green{\bigstar}$ $\large{\tt\red{x = 10 }}$

Now,

Width is 10 m and length will be

→ x + 5

→ 10 + 5

→ 15 m

Therefore, the the length of the rectangle is 15m.