Question

The length of a rectangle is 5 cm. more than it's breadth. If the perimeter of the
rectangle is 46 cm
Find its length and breadth ?




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Answer 1

To Find:-

• Find its length and breadth.

Given:-

• Length of a rectangle is 5 cm more than it's breadth.
• If the perimeter of the  rectangle is 46 cm.

Solution:-

Let the length be " x + 5 "

The breadth be " x "

$\tt \implies { \: Perimeter \: \: of \: rectangle \: = \: 2 ( \: l \: + \: b \: ) \: }$

$\tt \implies { \: 46 \: = \: 2 ( \: x \: + \: 5 \: + \: x \: ) \: }$

$\tt \implies { \: 46 \: = \: 2 ( \: 2x \: + \: 5 \: ) \: }$

$\tt \implies { \: 46 \: = \: 4x \: + \: 10 \: }$

$\tt \implies { \: 46 \: - \: 10 \: = \: 4x \: }$

$\tt \implies { \: 4x \: = \: 36 \: }$

$\tt \implies { \: x \: = \: \cancel\frac{36}{4} \: }$

$\tt \implies \boxed{{ \: x \: = \: 9cm \: }}$

So,

$\tt \implies { \: Breadth \: = \: 9cm \: }$

$\tt \implies { \: Length \: = \: 14cm \: }$

Answer 2

Step-by-step explanation:

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$\large\bf{\underbrace{\underline{Question:-}}}$

:$\Longrightarrow$ ● The length of a rectangle is 5 cm. more than it's breadth. If the perimeter of the rectangle is 46 cm. Find its length and breadth ?

$\sf To\ find = \begin{cases} \sf{The\ length\ of\ the\ rectangle.} \\ \\ \sf{The\ breadth\ of\ the\ rectangle.} \end{cases}$

$\sf Given = \begin{cases} \sf{The\ length\ of\ rectangle\ is\ 5cm\ more\ than\ it's\ breadth.} \\ \\ \sf{The\ perimeter\ of\ the\ rectangle\ is\ 46cm.} \end{cases}$

$\large\bf{\underbrace{\underline{Solution:-}}}$

$\sf We\ have = \begin{cases} \sf{Let\ the\ breadth\ of\ the\ rectangle\ be\ x.} \\ \\ \sf{Then,\ the\ length\ of\ the\ rectangle\ be\ x\ +\ 5cm.} \end{cases}$

$\large\bf{\underbrace{\underline{According\ to\ the\ question:-}}}$

$\sf Value\ of\ x = \begin{cases} \sf{●\ Perimeter\ of\ rectangle\ =\ 2(length\ +\ breadth)} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ 46\ =\ 2(x\ +\ 5\ +\ x)} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ 46\ =\ 2(2x\ +\ 5)} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ 46\ =\ 4x\ +\ 10} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ 46\ -\ 10\ =\ 4x} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ 4x\ =\ 36} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ x\ =\ \dfrac{36}{4}} \\ \\ \sf{●\ Perimeter_{(rectangle)}\ =\ x\ =\ 9cm.} \end{cases}$

$\large\bf{\underbrace{\underline{Hence:-}}}$

$\sf\purple{●\ Breadth_{(rectangle)}} \sf{=\ x\ =\ 9cm.}$

$\sf\purple{●\ Length_{(rectangle)}} \sf{=\ x\ +\ 5\ =\ 9\ +\ 5\ =\ 14cm.}$

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