Exercise 27 Page 851

We are asked to find the diameter of the given cylinder. To do so, we will find the radius and then double the value obtained. Let's recall the formula for the surface area of a cylinder. S=2π rh+2π r^2 Here, r is the radius of the base and h is the height of the cylinder. Since we are given that the surface area is 256 π mm^2 and the height is 8 mm, we can substitute them into the formula and calculate the length of the radius.

S=2π rh+2π r^2

SubstituteII

S= 256π, h= 8

256 π=2π r( 8)+2π r^2

▼

.LHS /2π.=.RHS /2π.

FactorOut

Factor out 2π

256π = 2π (r(8)+r^2)

DivEqn

.LHS /2π.=.RHS /2π.

256π/2π=r(8)+r^2

CancelCommonFac

Cancel out common factors

256/2=r(8)+r^2

CalcQuot

Calculate quotient

128 = r(8) + r^2

Multiply

128 = 8r + r^2

SubEqn

LHS-128=RHS-128

0 = 8 r + r^2 - 128

CommutativePropAdd

Commutative Property of Addition

0 = r^2 + 8r - 128

RearrangeEqn

Rearrange equation

r^2 + 8r - 128=0

We obtained a quadratic equation. Let's identify the values of a, b, and c. r^2 + 8r - 128 = 0 ⇕ 1r^2+( 8)r+( - 128)=0 We see that a = 1, b = 8, and c = - 128. Next, we will substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

▼

Solve using the quadratic formula

SubstituteValues

Substitute values

x=- 8±sqrt(8^2-4( 1)( - 128))/2( 1)

CalcPow

Calculate power

x=-8±sqrt(64 - 4(1)(-128))/2(1)

Multiply

x=-8±sqrt(64 +512)/2

AddTerms