Question

the surface area of a cylinder, ( a ), in terms of the radius of the base, ( r ), and height, ( h ), is given by the equation ( a = 2pi r(r + h) ).
what is the radius of the base in terms of the surface area and height of the cylinder?

( \bigcirc ) a. ( r = sqrt{\frac{a}{2pi}} - \frac{h}{2} )

( \bigcirc ) b. ( r = sqrt{\frac{a}{2pi} - h} )

( \bigcirc ) c. ( r = sqrt{\frac{a}{2pi} + \frac{h^2}{2}} - \frac{h}{2} )

( \bigcirc ) d. ( r = sqrt{\frac{a}{2pi} + \frac{h^2}{4}} - \frac{h}{2} )

Explanation:

Step1: Start with the given formula

We have the surface area formula for a cylinder: \( A = 2\pi r(r + h) \)

Step2: Expand the right - hand side

Expand \( 2\pi r(r + h) \): \( A=2\pi r^{2}+2\pi rh \)

Step3: Rearrange into a quadratic equation in terms of r

Let's rewrite the equation as a quadratic equation \( 2\pi r^{2}+2\pi rh - A=0 \). For a quadratic equation of the form \( ax^{2}+bx + c = 0 \) (here \( x = r \), \( a = 2\pi \), \( b = 2\pi h \), \( c=-A \)), we can use the quadratic formula \( r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)
First, calculate the discriminant \( D=b^{2}-4ac=(2\pi h)^{2}-4\times(2\pi)\times(-A)=4\pi^{2}h^{2}+8\pi A \)
But we can also solve the equation \( A = 2\pi r^{2}+2\pi rh \) by completing the square.
Divide the entire equation by \( 2\pi \): \( \frac{A}{2\pi}=r^{2}+rh \)
Complete the square for the terms involving \( r \). The coefficient of \( r \) is \( h \), so we add and subtract \( (\frac{h}{2})^{2}=\frac{h^{2}}{4} \) on the right - hand side:
\( \frac{A}{2\pi}=r^{2}+rh+\frac{h^{2}}{4}-\frac{h^{2}}{4}=(r + \frac{h}{2})^{2}-\frac{h^{2}}{4} \)
Then, \( (r+\frac{h}{2})^{2}=\frac{A}{2\pi}+\frac{h^{2}}{4} \)
Take the square root of both sides: \( r+\frac{h}{2}=\pm\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}} \)
Since the radius \( r>0 \), we take the positive square root: \( r=\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}-\frac{h}{2} \)

Answer:

D. \( r=\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}-\frac{h}{2} \)