Question

16. The length of a rectangle is 5 m greater than its width. If the perimeter of the rectangle is 50m find its length
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Answer 1

Given:

Length of a rectangle is 5 m greater than its width.

Perimeter of the rectangle is 50m

To Find:

It's Length

Solution:

Let length of rectangle be = x

And width of rectangle be = y

Now ATQ

Length of a rectangle is 5 m greater than its width, So.

x = y + 5 ...(1)

And

Perimeter of the rectangle is 50m, So.

2(x + y) = 50 ...(2)

Solving equation (1) and (2) we get :-

=> 2( (y+5) + y) = 50

=> 2( 2y + 5) = 50

=> 4y + 10 = 50

=> 4y = 40

=> y = 10m

=> x = 15m

Hence, length of rectangle is 15m.

Answer 2

• GIVEN:-

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

Perimeter of the rectangle is 50m.

• To Find:-

Find its length.

• SOLUTION:-

Let the width be x and length be

x + 5.

Perimeter = 50 m.

Formula of perimeter of rectangle,

$\boxed{\sf{2(length+breadth) }}$

According to the question,

$\large\Longrightarrow{\sf{Perimeter=2(length+breadth) }}$

$\large\Longrightarrow{\sf{50=2(x+x+5) }}$

$\large\Longrightarrow{\sf{50=2(2x+5) }}$

$\large\Longrightarrow{\sf{50=4x+10}}$

$\large\Longrightarrow{\sf{50-10=4x }}$

$\large\Longrightarrow{\sf{40=4x }}$

$\large\Longrightarrow{\sf{\huge\frac{40}{4}\large\:=x}}$

$\large\Longrightarrow{\sf{10=x }}$

$\large\therefore\boxed{\sf{x=10}}$

Therefore,

The width is 10 m.

Length = x + 5 = 10 + 5 = 15 m.

$\large\green\therefore\boxed{\sf{\green{Length \:of \:the \:rectangle\: is\: 15\: m. }}}$