Question

Length of a rectangle is 8 m less
than twice its width. If the
perimeter of rectangle is 56 m.
Find its length and width.​

Answer 1

Answer :

• Length and breadth of Rectangle is 16m and 12m

Given:

• Length of a rectangle is 8 m less than twice its width.
• If the perimeter of rectangle is 56 m.

To Find:

• Find its length and width.

Solution:

• Length of Rectangle = (2x - 8)m
• Let Breadth be x m
• Perimeter = 56m

Now,

Perimeter = 2 (L + B)

➠ 56 = 2 (2x - 8 + x)

➠ 56 = 2 (3x - 8)

➠ 56 = 6x - 16

➠ 56 + 16 = 6x

➠ 72 = 6x

➠ x = 72/6

➠ x = 12

Putting values of x in Length and breadth

• Breadth = 12m
• Length = 2 × 12 - 8 = 16m

Answer 2

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Given :-

• Length of a rectangle is 8 m less than twice its width.

• The perimeter of rectangle is 56 m.

To Find :-

• Length of a rectangle.

• Breadth of a rectangle.

We know that,

${\underline {\boxed {\bf {\pink {★~Perimeter~ of~ a~ rectangle = 2 (l + b)}}}}}$

Where,

• l = Length
• b = Breadth

Solution :-

Let,

• The length of a rectangle = (2x -8) m.

• The breadth of a rectangle = x m.

Given,

• The Perimeter of a rectangle = 56 m.

Therefore,

$\qquad \longmapsto {\sf {2(l+b) = 56}}$

$\qquad \longmapsto {\sf {2(2x - 8 + x) = 56}}$

$\qquad \longmapsto {\sf {2(3x - 8) = 56}}$

$\qquad \longmapsto {\sf {6x - 16 = 56}}$

$\qquad \longmapsto {\sf {6x = 56 + 16}}$

$\qquad \longmapsto {\sf {6x = 72}}$

$\qquad \longmapsto {\sf {x = {\dfrac{{\cancel{{72}}^{12}}}{{\cancel{{6}}^{1}}}}}}$

$\qquad\red \longmapsto \red {\sf {x = 12}}$

Hence,

$✰~ {\sf {\purple{Length~ of~ a~ rectangle = (2x - 8) m. }}}$

$\qquad \qquad ~~~~~~~\qquad \qquad$ ${\sf {\purple{= 2 \times 12 - 8 }}}$

$\qquad \qquad ~~~~~~~\qquad \qquad$${\sf {\purple{= 24 - 8 = 16~ m.}}}$

$✰~ {\sf {\purple{Width~ of~ a~ rectangle = 12~ m. }}}$

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