Question

How do you solve for R in a = $2\pi {R^2} + 2\pi Rh$ ?

Answer 1

Hint: In the above question you were asked to solve a = $2\pi {R^2} + 2\pi Rh$ and find the value in terms of h. You can make this equation a quadratic one and then you can apply the quadratic equation’s formula to solve this. The given equation is the formula of the surface area of a cylinder. So let us see how we can solve this problem.

Complete Step by Step Solution:
We have to solve a = $2\pi {R^2} + 2\pi Rh$ for R.
First, we have to convert this equation into a quadratic equation.
 $\Rightarrow a = \;2\pi {R^2} + 2\pi Rh$
On subtracting both sides with we get,
 $\Rightarrow 0 = 2\pi {R^2} + 2\pi Rh - a$
This can also be written as
 $\Rightarrow 2\pi {R^2} + 2\pi Rh - a = 0$
Using the quadratic equation formula which is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we get,
 $\Rightarrow R = \dfrac{{ - 2\pi h \pm \sqrt {{{(2\pi h)}^2} - 2(2\pi )( - a)} }}{{4\pi }}$
 $\Rightarrow R = \dfrac{{ - 2\pi h \pm \sqrt {4{\pi ^2}{h^2} + 8\pi a} }}{{4\pi }}$
After separating $2\pi h$ from the denominator we get,
 $\Rightarrow R = - \dfrac{h}{2} \pm \dfrac{{\sqrt {4{\pi ^2}{h^2} + 8\pi a} }}{{4\pi }}$
We can write $4\pi$ as $\sqrt {16{\pi ^2}}$
 $\Rightarrow R = - \dfrac{h}{2} \pm \sqrt {\dfrac{{4{\pi ^2}{h^2} + 8\pi a}}{{16{\pi ^2}}}}$
 $\Rightarrow R = - \dfrac{h}{2} \pm \sqrt {\dfrac{{{h^2}}}{4} + \dfrac{a}{{2\pi }}}$

Therefore, R in a = $2\pi {R^2} + 2\pi Rh$ is $- \dfrac{h}{2} \pm \sqrt {\dfrac{{{h^2}}}{4} + \dfrac{a}{{2\pi }}}$.

Note:
In the above solution, we first make the equation a quadratic one and then applied the quadratic equation’s formula to find the value of R. Moreover, The above equation in the question is the formula of the surface area of a cylinder where r is the radius of the cylinder, h is the height of the cylinder, and the value of $\pi$ = 3.142 or $\dfrac{{22}}{7}$