Question

The areas of the two circles are in the ratio 81 : 100, then ratio of their circumferences are​

Answer 1

Answer :-

$\:\\\large{\boxed{\sf{What\;the\;Question\;Says\:?\;:-}}}$

Here the concept of Areas of Circle and Circumference of Circle has been used. We are given ratio of areas of circle. Now we can apply the values and simply them. We can take those radius as r and r'. Then we can apply the value of Circumference and find its ratio.

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★ Formula Used :-

$\:\\\large{\boxed{\sf{Area\;of\;Circle\;=\;\bf{\pi r^{2}}}}}$

$\:\\\large{\boxed{\sf{\dfrac{\pi r^{2}}{\pi (r')^{2}}\;=\;\bf{\dfrac{81}{100}}}}}$

$\:\\\large{\boxed{\sf{Circumference\;of\;Circle\;=\;\bf{2 \pi r}}}}$

$\:\\\large{\boxed{\sf{\dfrac{2 \pi r}{\pi (r')^{2}}\;=\;\bf{Required\;Ratio}}}}$

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★ Question ::

The areas of the two circles are in the ratio 81 : 100, then ratio of their circumferences are ?

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★ Solution :-

Given,

» Ratio of areas of two circles = 81 : 100

• Let the radius of the smaller circle be r

• Let the radius of the bigger circle be r'

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~ For area of both circles :-

$\:\\\sf{\rightarrow\;\;\:Area\;of\;Circle\;=\;\bf{\pi r^{2}}}$

$\:\\\sf{\rightarrow\;\;\:Area\;of\;Circle_{(r)}\;=\;\bf{\pi r^{2}}}$

$\:\\\sf{\rightarrow\;\;\:Area\;of\;Circle_{(r')}\;=\;\bf{\pi (r')^{2}}}$

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~ For the ratio of r : r'

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:\dfrac{\pi r^{2}}{\pi (r')^{2}}\;=\;\bf{\dfrac{81}{100}}}}$

Cancelling π from numerator and denominator, we get,

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:\dfrac{r^{2}}{(r')^{2}}\;=\;\bf{\dfrac{81}{100}}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:\dfrac{r}{(r')}\;=\;\bf{\sqrt{\dfrac{81}{100}}\;\:=\;\;\dfrac{9}{10}}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\:r\;:\;r'\;=\;\;\underline{\underline{\bf{9\;:\;10}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Ratio\;\;of\;\;their\;\;radii,\;\;r\;:\;r'\;=\;\bf{9\;:\;10}}}}}$

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~ For ratio of their Circumference :-

From above we got that,

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:\dfrac{r}{(r')}\;=\;\;\;\bf{\dfrac{9}{10}}}}$

Now multiplying both numerator and denominator on both by 2π, we get,

$\:\\\qquad\large{\sf{:\Longrightarrow\;\;\:\dfrac{2\pi r}{2\pi(r')}\;=\;\;\;\bf{\dfrac{2\pi 9}{2\pi 10}}}}$

Here we can cancel the common term 2π from numerator and denominator at RHS. Then,

$\:\\\large{\boxed{\boxed{\tt{Ratio\;\;of\;\;their\;\;circumferencs,\;\;2\pi r\;:\;2\pi r'\;=\;\bf{9\;:\;10}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus,ratio\;of\;circumference\;of\;given\;circles\;is\;\;\boxed{\bf{9\;:\;10}}}}}}$

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★ More to know :-

$\:\\\mathtt{\leadsto\;\;\; Area\;\;of\;\;Square\;\;=\;\;(Side)^{2}}$

$\:\\\mathtt{\leadsto\;\;\; Area\;\;of\;\;Rectangle\;\;=\;\;Length\:\times\:Breadth}$

$\:\\\mathtt{\leadsto\;\;\; Area\;\;of\;\;Parallelogram\;\;=\;\;Base\:\times\:Height}$

$\:\\\mathtt{\leadsto\;\;\; Area\;\;of\;\;Triangle\;\;=\;\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}$

$\:\\\mathtt{\leadsto\;\;\; Perimeter\;\;of\;\;Square\;\;=\;\;4\;\times\;Side}$

$\:\\\mathtt{\leadsto\;\;\; Perimeter\;\;of\;\;Rectangle\;\;=\;\;2\;\times\;(Length\:+\:Breadth)}$

Answer 2

Given,

The areas of the two circles are in the ratio 81 : 100

To find :

Ratio of their circumferences .

Solution :

Let the radius of 1st circle be r and radius of 2nd radius be R

We know, area of circle = πr²

So atq,

⇒ πr² : πR² = 81 : 100

⇒ πr²/πR² = 81/100

⇒ r²/R² = 81/100

⇒ (r/R)² = (9/10)²

⇒ r/R = 9/10

Now , circumference of circle = 2πr

⇒ Ratio of circumferences = 2πr : 2πR ​

⇒ Ratio of circumferences = 2πr/2πR

⇒ Ratio of circumferences = r/R

⇒ Ratio of circumferences = 9/10

⇒ Ratio of circumferences = 9 : 10

Therefore,

Ratio of their circumferences = 9 : 10.