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# The difference between the circumference and radius of a circle is $30\pi$. Find the diameter of the circle.A. $\dfrac{{30\pi }}{{2\pi - 1}}$B. $\dfrac{{60\pi }}{{2\pi - 1}}$C. $\dfrac{{2\pi - 1}}{{60\pi }}$D. None

Last updated date: 20th Jun 2024
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Hint: We use the formula of circumference of the circle and make an equation subtracting radius from the circumference. Calculate the length of the diameter after calculating the radius.
* If radius of a circle is ‘r’, then circumference is given by $2\pi r$
* Diameter of circle having radius ‘r’ is ‘2r’

Complete step-by-step solution:
We have a circle with the radius ‘r’.
Since we know circumference of circle is given by $2\pi r$
We form an equation subtracting radius from the circumference and equating it to the given values
$\Rightarrow 2\pi r - r = 30\pi$
Take ‘r’ common in LHS of the equation
$\Rightarrow r(2\pi - 1) = 30\pi$
Divide both sides of the equation by $(2\pi - 1)$
$\Rightarrow \dfrac{{r(2\pi - 1)}}{{(2\pi - 1)}} = \dfrac{{30\pi }}{{(2\pi - 1)}}$
Cancel same terms from numerator and denominator on both sides of the equation
$\Rightarrow r = \dfrac{{30\pi }}{{(2\pi - 1)}}$
Now we know that the length of diameter of the circle is twice the length of the radius of the circle.
$\Rightarrow$Diameter of the circle with radius ‘r’ $= 2 \times \dfrac{{30\pi }}{{(2\pi - 1)}}$
$\Rightarrow$Diameter of the circle with radius ‘r’ $= \dfrac{{60\pi }}{{(2\pi - 1)}}$
$\therefore$Diameter of the circle with radius ‘r’ is $\dfrac{{60\pi }}{{(2\pi - 1)}}$

$\therefore$Correct option is B.

Note: Students are likely to make mistakes while shifting the values from one side of the equation to another side of the equation as they forget to change the sign of the value shifted. Keep in mind we always change the sign of the value from positive to negative and vice versa when shifting values from one side of the equation to another side of the equation. Also, many students substitute the value of $\pi$ in the end, we need not substitute the value as the options have the value $\pi$ directly in it.