Exercise 12 Page 636

We are asked to draw a cone with a surface area that is between $50$ and $100$ square units.

Before we do that, let's remember the formula for the surface area of a cone. The surface area S.A. of a cone with slant height

ℓ

and radius

r

is given as follows.

S . A . = π r ℓ + π r^{2}

We should choose such values of

r

and

ℓ

that the expression on the right-hand side has a value between

50

and

100 .

50 < π r ℓ + π r^{2} < 100

There are infinitely many pairs of numbers that satisfy the inequality. We should choose just one. Since it is easier to focus on just one variable, let's say that the slant height will always be twice the radius.

ℓ = 2 r

We can now substitute

2 r

for

ℓ

in the expression and simplify.

50 < π r ℓ + π r^{2} < 100

Substitute

$ℓ = 2 r$

50 < π r (2 r) + π r^{2} < 100

CommutativePropMult

Commutative Property of Multiplication

50 < 2 \cdot π \cdot r \cdot r + π r^{2} < 100

ProdToPowTwoFac

$a \cdot a = a^{2}$

50 < 2 \cdot π \cdot r^{2} + π r^{2} < 100

Multiply

50 < 2 π r^{2} + π r^{2} < 100

AddTerms

Add terms

50 < 3 π r^{2} < 100

Now there is just one variable for us to consider. See that we need to choose a value of

r

that satisfies the last inequality.

50 < 3 π r^{2} < 100

Notice that

r = 3

satisfies the condition. We can check that using a calculator. Since the radius of our cone is

3

units, this means that the slant height is

2 (3) = 6 .

Let's now find the surface area of the cone.

S . A . = π r ℓ + π r^{2}

SubstituteII

$r = 3$ , $ℓ = 6$

S . A . = π (3) (6) + π (3)^{2}

CalcPow

Calculate power

S . A . = π (3) (6) + π (9)

Multiply

S . A . = 18 π + 9 π

AddTerms

Add terms

S . A . = 27 π

UseCalc

Use a calculator

S . A . = 84.823001 \dots

RoundDec

Round to $2$ decimal place(s)

S . A . \approx 84.82

The volume of the cone is about

84.82 units^{3} .

Since this more than

50

and less than

100,

the dimensions are correct. Last, let's draw the cone.

Note that this is just a sample solution. The slant height does not need to be twice the radius. Also, the radius does not have to be a natural number. We could choose any two numbers $r$ and $ℓ$ that satisfy the inequalities.