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  • If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then (A) R1 + R2 =R (B) R1^{2} + R2^{2}=R^{2} (C) R1 + R2 <R (D)R1^{2} + R2^{2}<R^{2}

If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then

(A) R1 + R2 =R   (B) R12 + R22=R2  (C) R1 + R2 <R (D)R12 + R22<R2  

Answers (1)

(B) R12 + R22=R2

Solution

Radius of first circle= R1

Area of first circle =πR12

Radius of second circle =R2

Area of second circle =πR22

Radius of third circle = R

Area of third circle=πR2

According to question

πR12 + πR22=πR2

π(R12 + R22)=πR2

R12 + R22=R2

Hence option B is correct.

 

Posted by

infoexpert24

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