A circle with area A cm
is contained in the interior of a larger circle with area
(A-B)cm- and the radius of the larger circle is 4 cm If A, B. A+B are in AP
then the diameter in cm of the smaller circle is
Answers
Answer:
,Given :
Given :Area of smaller circle = A cm2
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− B
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π =316π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π =316πBut A=Area of smaller circle =π×radius2=316π
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π =316πBut A=Area of smaller circle =π×radius2=316π⟹ radius=34
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π =316πBut A=Area of smaller circle =π×radius2=316π⟹ radius=34∴ Diameter =38=383.
Given :Area of smaller circle = A cm2Area of larger circle = A+B cm2=πr2=16π∴ A =16π− BA, B, A+B i.e 16π− B, B, 16π are in arithmetic progression∴ (16π−B)−B=B−16π⟹ 32π=3B⟹ B=332π∴ A=16π−332π⟹ A=348π−32π =316πBut A=Area of smaller circle =π×radius2=316π⟹ radius=34∴ Diameter =38=383.Hence, option C is correct.
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