Question

and outer perimeter is 154 m
The inner circumference of circular park is 132 my and outer perimeter is 154 mt. what is the length of a equal width path around it​

Answer 1

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✺Answer:

♦️GiveN:

• Outer perimeter of total figure = 154 cm
• Inner perimeter of circular park = 132 cm
• The path is made around the park, that it looks like a circular ring.

♦️To FinD:

• Find the width of the path.

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✺Explanation of Q.

Before solving this question, let us understand the diagram of the required park and the path surrounding it.

↪ Refer to the attachment

The park is circular (yellow) and have a circumference of 132 cm. From here, we can find the radius by using peimeter(2πr) formula for circle. Then, it's given that a path runs around the circle externally (pink) and the circumference of total shape is now 154cm. So, here we can deduce Radius of total space.

Then, if you observe carefully, the width of the path is the difference of Radius of total shape and radius of park. In this way, we are going to solve the question.

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✺Concept to be used:

We don't need much formulas except one, The perimeter of circle formula, i.e

$\large{ \ddag \:{ \boxed{ \rm{ \red{ perimeter \: of \: circle = 2\pi \: r}}}}}$

Only using this, we can solve the question.

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✺Solution:

Let the radius of circular park = R1

Given, circumference of park = 132 cm

That means,

$\large{ \rm{ \purple{ \rightarrow \: 2\pi \: R1= 132cm}}}$

By solving this equation,

$\large{ \rm{ \rightarrow \: 2 \times \frac{22}{7} \times R1 = 132cm}} \\ \\ \large{ \rm{ \rightarrow \: R1 = \frac{ \cancel{132} \: \: 3 \: \times 7}{ \cancel{44}} cm}} \\ \\ \large{ \rm{ \rightarrow \: R1 = \boxed{ \rm{ \green{21 \: cm}}}}}$

Let the radius of total shape( park and path) be R2

Given, circumference of total shape = 154 cm

That means,

$\large{ \purple{ \rm{ \rightarrow \: 2\pi \: R2 = 154 cm}}}$

By solving this equation,

$\large{ \rm{ \rightarrow \: 2 \times \frac{22}{7} \times \: R2 = 154 m}} \\ \\ \large{ \rm{ \rightarrow \: R2 = \frac{ \cancel{154} \: 7 \times 7}{2 \times \cancel{22}} m}} \\ \\ \large{ \rm{ \rightarrow \:R2 = \frac{49}{2} m}} \\ \\\large{ \rm{ \rightarrow \: R2 = \boxed{ \rm{ \green{24.5 \: m}}}}}$

So, Width of the park = Difference between these radius of total and only park.

That means,

$\large{ \rm{ \rightarrow \: width \: of \: path = R2 - R1}} \\ \\ \large{ \rm{ \rightarrow \: width \: of \: park = 24.5 - 21\: m}} \\ \\\large{ \rm{ \rightarrow \: width \: of \: park \: = \boxed{ \green{ \rm{ 3.5 \: m}}}}}$

$\large{ \ddag\:{ \boxed{ \rm{ \pink{so \: width \: of \: park = 3.5 \: m}}}}}$

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Answer 2

Correct question : -

The inner circumference of circular park is 132 m and outer perimeter is 154 m. What is the length of a equal width path around it​.

Answer -

Given :

The outer perimeter i.e outer circumference of circular park = 154 m

The inner perimeter i.e inner circumference of circular park = 132 m

We can find the radius,

The formula for the circumference = 2πr unit.

So, the outer circumference,

⇒ 2πR = 154

$\bf\implies 2 \times (\frac{22}{7}) \times R = 154 \\ \\ \implies (\frac{44}{7}) \times R = 154 \\ \\ \implies R = 154 \times \frac{7}{44}\\ \\ \implies R = \frac{49}{2}m$

Now, the inner circumference,

⇒ 2πR = 132

$\bf \implies 2 \times \frac{22}{7} \times r = 132\\ \\ \implies \frac{44}{7}\times r = 132 \\ \\ \implies r = 132 \times \frac{7}{44}\\ \\ \implies r = 21 m$

Finding the width of the path

= radius R - radius r

⇒ R - r

= (49)/2 - 21

= (49 - 42)/2

∵ [Taking LCM 2]

= 7/2

= 3.5 m

Hence,

The width of the circular park = $\red{\boxed{\boxed{\tt{3.5 m}}}}$