Question

5. The length of a rectangle is 8 m less than the
twice of its breadth. If the perimeter of the
rectangle is 56 m, find the length and breadth.​

Answer 1

Answer:

l= 2b-8

perimeter= 2(l+b) = 2(2b-8+b)

= 6b-16=56

b= 12 , l=16

Answer 2

➤ Given :-

Length ${\sf {}_{(Rectangle)}}$ :- 8m less than twice the breadth

Perimeter ${\sf {}_{(Rectangle)}}$ :- 56 metres

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➤ To Find :-

The length and breadth of the rectangle.

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★ How to do :-

Here, we are given the perimeter of the Rectangle. Also, The length of the Rectangle is 8 meters less than twice the breadth. So, we should take the variables value in the place of length or breadth and also when required, and we should find the value of length and breadth of that rectangle by using the formula which is used to calculate the perimeter. The transportation of numbers from one side to other is also used here. So, let's solve!!

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➤ Solution :-

Breadth of the rectangle :-

${\tt \leadsto \underline{\boxed{\tt 2 \: (Length + Breadth)}}}$

Substitute the given values.

Take the breadth as 'x'.

${\tt \leadsto 2 \: (2x - 8 + x) = 56}$

Multiply the number two with all numbers in bracket.

${\tt \leadsto 4x - 16 + 2x = 56}$

Add the values having same variables.

${\tt \leadsto 6x - 16 = 56}$

Shift the number 16 from LHS to RHS, changing it's sign.

${\tt \leadsto 6x = 56 + 16}$

Subtract the values to get the value of 6x.

${\tt \leadsto 6x = 72}$

Shift the number 6 from LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{72}{6}}$

Simplify the fraction to get the value of breadth.

${\tt \leadsto x = 12}$

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We have found the value of the breadth as 12 metres. Now, let's solve for the length.

Length of the rectangle :-

${\tt \leadsto 2x - 8}$

Substitute the value of 'x'.

${\tt \leadsto (2 \times 12) - 8}$

Multiply the numbers first.

${\tt \leadsto 24 - 8}$

Subtract the values to get the answer.

${\tt \leadsto 16 \: \: metres}$

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Final answers :-

${\tt \leadsto \pink{\underline{\boxed{\sf Length :- \: \: 16 \: \: metres}}}}$

${\tt \leadsto \pink{\underline{\boxed{\sf Breadth :- \: \: 12 \: \: metres}}}}$

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Hence solved !!