Question

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

Answer 1

Answer:

Let the width be x.

So, the length be x+4.

The perimeter of rectangle is 40 m.

perimeter = 2(length + width)

40 = 2( x + x + 4)

40/2 = 2x + 4

20 - 4 = 2x

2x = 16

x = 8 m

Therefore,

length = 12 m

width = 8 m

Answer 2

$\sf{\underline{Answer}}$

$\bold{\sf{\red{Length\:=\:12\:m\:\:\:;\:\:Breadth\:=\:8\:m}}}$

$\sf{\underline{Step\:-\:by\:-\:step\:explanation:}}$

GIVEN :

• Length exceeds the breadth by 4 metres
• Perimeter → 40 m

TO FIND :

• Dimensions of the rectangle

SOLUTION :

Forthrightly 'm not gud in deciphering such kinda sums but I'll attempt. Apologies, if my solution is erroneous.

Let the breadth be called as x.

Let the length be called as y.

Then as per the question,

→ length = breadth + 4

→ y = x + 4 (1)

Sum of all sides of the rectangle i.e perimeter equal 40 m.

→ Perimeter = 2 ( length + breadth)

→ 40 = 2 (y + x)

→ 40 = 2 ( x + 4 + x)

→ 40 = 2x + 8 + 2x

→ 40 = 4x + 8

→ 40 - 8 = 4x

→ 32 = 4x

→ $\dfrac{32}{4}$ = x

→ 8 = x

We called x as breadth.

So we extrapolate, that the rectangle has breadth of 8 m.

To solve for y i.e length,

→ y = x + 4

→ y = 8 + 4

→ y = 12 m

Our rectangle has length equal to 12 m.

Dimensions :

• Length → 12 m
• Breadth → 8 m

I reckon my solution do helps you.