Question

find the breadth of the rectangle whose perimeter is 16cm and whose length is 60cm

Answer 1

Solution :-

There is a mistake in this question. The perimeter cannot be less than its length or breadth.

The perimeter should be 60 cm and breadth of the rectangle should be 16 cm.

I have assumed the perimeter as 60 cm and breadth as 16 cm and then solved it.

Perimeter of rectangle = 2(Length + Breadth)

Let the length of rectangle be 'x' cm

⇒ 60 = 2(x + 16)

⇒ 60 = 2x + 32

⇒ 2x = 60 - 32

⇒ 2x = 28

⇒ x = 28/2

⇒ x = 14 cm

So, the length of the given rectangle is 14 cm.

Answer.

Answer 2

Correct Question :

Find the breadth of the rectangle whose perimeter is 60cm and whose length is 16cm.

Given :

• Length $\leadsto$ 16 cm
• Perimeter $\leadsto$ 60 cm

To find :

• The Breadth of Rectangle

Solution :

$\;\boxed{\sf{\purple{Perimeter_{\:(rectangle)} = 2 \times Length + Breadth}}}$

Let the Breadth be "x" for now

$: \implies \sf 60 \: cm \: = 2 \: \times (x \: + 16 \: cm) \\ \\ \\ \: : \implies \sf \dfrac{ \cancel{60}}{ \cancel{2}} \: \hookrightarrow \: 30 \: cm \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: : \implies \sf (30 \: - 16) \: cm \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \sf { \underline{ \boxed{ \pink{ \mathfrak{x \leadsto 14 \: cm}}}}} \\ \\ \\ \: \: \: \: \: {\underline {\sf { \dag \:Verification }}} \\ \\ \\ : \implies \sf 2 \times (14 + 16) \: cm \\ \\ \\ : \implies \sf 2 \times 30 \: cm \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: : \implies { \underline{ \boxed{ \pink{ \mathfrak{P \leadsto 60 \: cm}}}}}$

•°• Hence, verified! that the Breadth is 14 cm

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