Question

the radius of two circles are 8cm and 6cm respectively find the radius of the circle having its area equal to the the sum of the areas of the two circles. with diagram​

Answer 1

Step-by-step explanation:

area of first circle = pi*(8^2)

= 64pi

are of second circle=pi*(6^2)

= 36pi

the area of resulting circle = 36pi+64pi

= 100pi

the radius of new circle = sqrt(100pi/pi)

= sqrt(100)

= 10

sqrt(x) represent square root of x

the radius of new circle = 10 cm

Answer 2

Given : Radius of first circle (r1) = 8cm & Radius of second circle (r2) = 6cm.

To Find : Find the radius of circle ?

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Solution : Let the radius of circle third circle be r3.

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• ${\sf\leadsto{Area ~of~ first ~circle~ =~ π ~× ~8^2}}$
• ${\sf\leadsto{Area ~of ~second ~circle ~= ~π ~×~ 6^2}}$
• ${\sf\leadsto{Area ~of ~third ~circle~ = ~π~×~r^2}}$

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◗Here, sum of the areas of first to circles is equal to the area of third circle.

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$\pmb{\sf{\underline{According~ to ~the~ Given~ Question~:}}}$

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$\qquad{\sf:\implies{π(r_{1})(r_{1})~+~π(r_{2})(r_{2})~=~π(r_{3})(r_{3})}}$

$\qquad{\sf:\implies{\dfrac{22}{7}~×~64~+~\dfrac{22}{7}~×~36~=~\dfrac{22}{7}~×~(r_{3})(r_{3})}}$

$\qquad{\sf:\implies{\dfrac{1408}{7}~+~\dfrac{792}{7}~=~\dfrac{22}{7}~×~(r_{3})(r_{3})}}$

$\qquad{\sf:\implies{\dfrac{2200}{7}~=~\dfrac{22}{7}~×~(r_{3})(r_{3})}}$

$\qquad{\sf:\implies{(r_{3})(r_{3})~=~\dfrac{\cancel{7}~×~2200}{\cancel{7}~×~22}}}$

$\qquad{\sf:\implies{(r_{3})(r_{3})~=~100}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{(r_{3})~=~10~cm}}}}}$★

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Hence,

$\therefore\underline{\sf{The~radius~of~the~circle~is~\bf{\underline{10~cm}}}}$