Question

The circumference of two circles are in the ratio 8 : 6. Find the ratio of their areas.​

Answer 1

$\sf \large \underbrace{ \underline{Understanding \: the \:Concept }}$

Here's the concept of circumference as well as area of Circle are used, where ratio of circumference is given and we have to find ratio of area. So first we have to find the radius of circle by apply formula of circumference of circle and then we will be able to find the ratio of areas.

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Let's start!

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Let us suppose radius of circle 1 be R and radius of circle 2 be r.

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$\color{green} \sf\dfrac{Circumference\: of\: circle_1}{Circumference \:of\: circle_2} = \dfrac{8}{6}$

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$\underline{ \boxed{ \blue{ \bold{\sf★ \: Circumference \: of \: circle=2πr}}}}$

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$\sf \to\dfrac{2πR}{2πr}= \dfrac{8}{6}$

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$\sf\to \cancel\dfrac{2πR}{2πr}= \dfrac{8}{6}$

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$\large \underline{ \boxed{ \color{grey} \sf \dfrac{R}{r}= \dfrac{8}{6}}}$

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So from here we can conclude that:

⊙ Radius of circle 1 is R=8cm

⊙ Radius of circle 2 is r=6cm

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Now we have to find the ratio of area of circle

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$\color{blue} \large \underline{ \boxed{ \sf&#10029 \: Area \: of \: circle=πr²}}$

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$\sf \dfrac{Area \: of \: circle_1}{Area \: of \: circle_2} = \dfrac{πR²}{πr²}$

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$\sf \to\dfrac{Area \: of \: circle_1}{Area \: of \: circle_2} = \dfrac{π \times 8 \times 8}{π \times 6 \times 6}$

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$\sf\to \dfrac{Area \: of \: circle_1}{Area \: of \: circle_2} = \cancel \dfrac{π \times 8 \times 8}{π \times 6 \times 6}$

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$\large \underline{ \boxed{ \color{red} \sf \dfrac{Area \: of \: circle_1}{Area \: of \: circle_2} = \dfrac{ 16}{ 9}}}$

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Hence the ratio of the Area of circle 1 with area of circle 2 is 16:9.

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$\sf \large \underbrace{ \underline{More \: Formulae \: to \: know}}$

$\sf \: &#10029 \: Area \: of \: parallelogram=Base×height$

$\sf &#10029Area \: of \: triangle=\frac{1}{2}×Base×Height$

$\sf &#10029 \: Area \: of \: square=side²$

$\sf &#10029 \: Area \: of \: rectangle=Length×breadth$

Answer 2

Answer ::

The ratio of their areas is 16 : 9

Step-by-step explanation ::

To Find :-

• The ratio of their areas

Solution :-

Given that,

• The circumference of two circles are in the ratio 8 : 6

Therefore,

Let us assume the radii as $\sf{{r}_{1}}$ and $\sf{{r}_{2}}$

As we know that,

Circumference of circle = 2πr, therefore

Given that,

$\mapsto \: \sf{ \dfrac{2\pi \: {r}_{1}}{2\pi \: {r}_{2}} = \dfrac{8}{6}} \\ \\ \\ \mapsto \: \sf{ \dfrac{2\pi \: {r}_{1}}{2\pi \: {r}_{2}} = \cancel{\dfrac{8}{6}}} \\ \\ \\ \mapsto \: \sf{ \dfrac{ \cancel{2}\pi \: {r}_{1}}{ \cancel{2}\pi \: {r}_{2}} = \dfrac{4}{3}} \\ \\ \\ \mapsto \: \sf{ \dfrac{ \cancel{\pi} \: {r}_{1}}{ \cancel{\pi} \: {r}_{2}} = \dfrac{4}{3}} \\ \\ \\ \mapsto \: \sf{ \dfrac{{r}_{1}}{{r}_{2}} = \dfrac{4}{3}}$

Therefore, the ratio of radius of circle are 4 : 3.

ACCORDING THE QUESTION,

Now, area of the circles, as we know that,

$\dag \: \boxed{ \sf{Area \: of \: circle = \pi \: {r}^{2}}}$

Therefore, the area is,

As area of circle = πr² [ radius = square (r²) ] so we need to also square the given ratio of radius,

$\mapsto \: \sf{ \dfrac{{\pi \: r ^{2} }_{1}}{{\pi \: r^{2}}_{2}} = \dfrac{4^{2} }{3^{2}}} \\ \\ \\ \mapsto \: \sf{ \dfrac{{\pi \: r}_{1}}{{\pi \: r}_{2}} = \dfrac{16}{9}}$

Therefore, the areas of the circle is 16 : 9.

Formula used :-

• Circumference of circle = 2πr
• Area of circle = πr²