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The Cathedral of Maringá, one of the tallest churches in the world, was designed in the form of a cone by José Augusto Bellucci.

Tadeo is learning how to make a traditional Chinese conical hat. He notices that the craftsman uses $20\text{-}$centimeter bamboo sticks to make the framework.

If the radius of the base is $16$ centimeters, find the volume of the hat. Round the answer to the nearest cubic centimeter.

Solution

To find the volume of the conical hat, its height will be calculated first. Then, the formula for the volume of a cone will be used.

Finding Height of the Hat

The distance from the vertex of the cone to its base is the height of the cone. Each stick if the frame represents the slant height of the cone. Therefore, the cone has a slant height of $20$ centimeters and a radius of $16$ centimeters.

These three segments form a right triangle, $△ABC.$ From here, the height $AB$ can be determined by using the Pythagorean Theorem.

$A{C}^2=A{B}^2+B{C}^2$

SubstituteII

$AC=20$, $BC=16$

${20}^2=A{B}^2+{16}^2$

Solve for $AB$

CalcPow

Calculate power

$400=A{B}^2+256$

SubEqn

$\text{LHS}-256=\text{RHS}-256$

$144=A{B}^2$

RearrangeEqn

Rearrange equation

$A{B}^2=144$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$\sqrt{A{B}^2}=\sqrt{144}$

SqrtPowToNumber

$\sqrt{a^2}=a$

$AB=12$

The principal root was taken because negative values do not make sense for these measurements.

Finding the Volume of the Hat

Recall the formula for the volume of a cone.

$$\begin{array}{c}V=\frac{1}{3}\pi r^{2}h\end{array}$$

Here, $r$ is the radius and $h$ is the height of the cone. The height was found to be $12$ centimeters, and the radius of the base is $16$ centimeters. The volume of the hat can be found by substituting these values into the formula.

$V=\frac{1}{3}\pi r^{2}h$

SubstituteII

$r=16$, $h=12$

$V=\frac{1}{3}\pi ({16}^2)(12)$

Simplify right-hand side

CalcPow

Calculate power

$V=\frac{1}{3}\pi (256)(12)$

Multiply

Multiply

$V=\frac{1}{3}3072\pi$

MoveRightFacToNumOne

$\frac{1}{b}\cdot a=\frac{a}{b}$

$V=\frac{3072\pi }{3}$

UseCalc

Use a calculator

$V=3216.990877\dots$

RoundInt

Round to nearest integer

$V\approx 3217$

The volume of the cone is about $3217$ cubic centimeters.

The diagram shows a traffic cone, which has a volume of $3128$ cubic inches.

For the Jefferson High Science Fair, Ali is thinking about a chemistry experiment in which he will need a cylinder with a radius of $20$ centimeters and a height $81$ centimeters, with a cone inside. The cylinder must be open on both ends, and the cone must have an open bottom.

To conduct the experiment, Ali needs to answer some questions first. Help him find the answers in order to win the first prize in the fair!

Solution

a The volumes will be calculated one at a time.

Volume of Water

Ali will fill the cone with water. Therefore, the volume of the cone is needed. The height and radius of the cone are the same as the height and radius of the cylinder. Therefore, the height of the cone is $81$ centimeters and its radius is $20$ centimeters.

The volume of a cone is one third the product of $\pi ,$ the square of the radius, and the height.

$$\begin{array}{c}{V}_{\text{cone}}=\frac{1}{3}\pi r^{2}h\end{array}$$

In the this formula, $81$ and $20$ can be substituted for $h$ and $r,$ respectively.

$V_{\text{cone}}=\frac{1}{3}\pi r^{2}h$

SubstituteII

$h=81$, $r=20$

$V_{\text{cone}}=\frac{1}{3}\pi \left({20}^2\right)(81)$

Evaluate right-hand side

CalcPow

Calculate power

$V_{\text{cone}}=\frac{1}{3}\pi (400)(81)$

Multiply

Multiply

$V_{\text{cone}}=\frac{1}{3}\pi (32400)$

CommutativePropMult

Commutative Property of Multiplication

$V_{\text{cone}}=\frac{1}{3}(32400)\pi$

MoveRightFacToNumOne

$\frac{1}{b}\cdot a=\frac{a}{b}$

$V_{\text{cone}}=\frac{32000}{3}\pi$

CalcQuot

Calculate quotient

$V_{\text{cone}}=10800\pi$

Therefore, the volume of water that Ali needs is $10800\pi$ cubic centimeters. Finally, this number will be approximated to the nearest integer.

$10800\pi$

UseCalc

Use a calculator

$33929.200658\dots$

RoundInt

Round to nearest integer

$33929$

Ali needs about $33929$ cubic centimeters of water.

Volume of Foam

Ali will fill the part of the cylinder not occupied by the cone with foam. Therefore, the volume of this portion is needed. It is given that the height of the cylinder is $81$ centimeters and that its radius is $20$ centimeters.

The volume of a cylinder is the product of $\pi ,$ the square of the radius, and the height.

$$\begin{array}{c}{V}_{\text{cylinder}}=\pi r^{2}h\end{array}$$

One more time, $81$ and $20$ can be substituted for $h$ and $r,$ respectively.

$V_{\text{cylinder}}=\pi r^{2}h$

SubstituteII

$h=81$, $r=20$

$V_{\text{cylinder}}=\pi \left({20}^2\right)(81)$

Evaluate right-hand side

CalcPow

Calculate power

$V_{\text{cylinder}}=\pi (400)(81)$

Multiply

Multiply

$V_{\text{cylinder}}=32400\pi$

The volume of the cylinder is $32400\pi$ cubic centimeters. The volume of the space inside the cylinder that is not occupied by the cone is the difference between the volume of the cylinder and the volume of the cone, which is $10800\pi$ cubic centimeters.

$$\begin{array}{c}32400\pi -10800\pi =21600\pi \end{array}$$

The volume of the cylinder that is not occupied by the cone is $26600\pi$ cubic centimeters and is the volume of foam that Ali will need. This number will be approximated to the nearest integer.

$21600\pi$

UseCalc

Use a calculator

$67858.401317\dots$

RoundInt

Round to nearest integer

$67858$

Ali needs about $67858$ cubic centimeters of foam.

b In Part $A$ it was found that the volume of the cylinder is $32400\pi$ cubic centimeters. It was also found that the volume of the portion of the cylinder not occupied by the cone is $21600\pi$ cubic centimeters. The ratio of the second value to the first value will result in the desired percentage.

$\frac{21600\pi }{32400\pi }$

Evaluate

CancelCommonFac

Cancel out common factors

$\frac{21600\cancel{\pi }}{32400\cancel{\pi }}$

SimpQuot

Simplify quotient

$\frac{21600}{32400}$

ReduceFrac

$\frac{a}{b}=\frac{a/10800}{b/10800}$

$\frac{2}{3}$

Convert to percent

FracToDiv

$\frac{a}{b}=a÷b$

$0.666666\dots$

WritePercent

Convert to percent

$66.666666\dots \%$

RoundDec

Round to $1$ decimal place(s)

$66.7\%$

It can be concluded that the volume of the cylinder not occupied by the cone represents $66.7\%$ of the interior of the cylinder.

For his experiment for the science fair, Ali plans to make the figure by himself.

Solution

Since the cost is given per square meter, all the measures will be converted from centimeters to meters. To do this, the measures need to be multiplied by a conversion factor, $\frac{1\text{m}}{100\text{cm}}.$

$$\begin{array}{rl}\text{Height:}81\cancel{\text{cm}}\cdot \frac{1\text{m}}{100\cancel{\text{cm}}}& =0.81\text{ m}\\ & \\ \text{Radius:}20\cancel{\text{cm}}\cdot \frac{1\text{m}}{100\cancel{\text{cm}}}& =0.2\text{ m}\end{array}$$

With this information, update the measures on the diagram.

The lateral areas of the cylinder and the cone will be calculated one at a time. Then, their sum will be multiplied by the cost per square meter.

Lateral Area of the Cylinder

Recall that the cylinder is open in both ends. The lateral area of a cylinder is twice the product of $\pi ,$ the radius, and the height.

$$\begin{array}{c}L{A}_{\text{cylinder}}=2\pi rh\end{array}$$

In the above formula, $0.81$ and $0.2$ can be substituted for $h$ and $r,$ respectively.

$L{A}_{\text{cylinder}}=2\pi rh$

SubstituteII

$h=0.81$, $r=0.2$

$L{A}_{\text{cylinder}}=2\pi (0.2)(0.81)$

Evaluate right-hand side

CommutativePropMult

Commutative Property of Multiplication

$L{A}_{\text{cylinder}}=2(0.2)(0.81)\pi$

Multiply

Multiply

$L{A}_{\text{cylinder}}=0.324\pi$

The lateral area of the cylinder is about $0.324\pi$ square meters.

Lateral Area of the Cone

The lateral area of a cone is the product of $\pi ,$ the radius, and the slant height.

$$\begin{array}{c}L{A}_{\text{cone}}=\pi r\ell \end{array}$$

The slant height $\ell$ is the hypotenuse of the right triangle formed by the radius, the height, and the segment that connects the center of the base of the cylinder with a point on the circumference of the opposite base.

The missing value can be found by using the Pythagorean Theorem.

$a^2+b^2=c^2$

SubstituteValues

Substitute values

${0.81}^2+{0.2}^2={\ell }^2$

Solve for $\ell$

CalcPow

Calculate power

$0.6561+0.04={\ell }^2$

AddTerms

Add terms

$0.6961={\ell }^2$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$\sqrt{0.6961}=\ell$

RearrangeEqn

Rearrange equation

$\ell =\sqrt{0.6961}$

The slant height of the cone is about $\sqrt{0.6961}$ meters. Now, the formula for the lateral area of a cone can be used. Substitute $0.83$ and $0.2$ for $\ell$ and $r,$ respectively.

$L{A}_{\text{cone}}=\pi r\ell$

SubstituteII

$\ell =\sqrt{0.6961}$, $r=0.2$

$L{A}_{\text{cone}}=\pi (0.2)\left(\sqrt{0.6961}\right)$

CommutativePropMult

Commutative Property of Multiplication

$L{A}_{\text{cone}}=0.2\sqrt{0.6961}\pi$

Total Cost

Finally, the sum of the lateral areas will be multiplied by the cost per square meter.

$(0.324\pi +0.2\sqrt{0.6961}\pi )125$

Evaluate

UseCalc

Use a calculator

$(1.017876\dots +0.524222\dots )125$

AddTerms

Add terms

$(1.542098\dots )125$

Multiply

Multiply

$192.76$

To make the shape that Ali needs, he has to spend about $\$192.76.$

Tiffaniqua's teacher gives her a piece of paper on which a circle and a sector are drawn. The paper is a square of side length $10$ centimeters.

The teacher also gives the following set of information.

• The circle is tangent to the top and right sides of the paper as well as to the sector.
• The centers of the circle and sector are on the diagonal of the paper.
• The circle and the sector can form a cone.

Help Tiffaniqua answer the following questions.

Solution

a Let $\ell$ be the radius of the sector and $r$ be the radius of the circle. To find these values, a system of two equations should be written first. The equations can be written using the trigonometric ratios and the properties of cones. Then, the system of equations will be solved using the Substitution Method.

Writing the System of Equations

Start by drawing a right triangle whose hypotenuse is $\ell +r$ centimeters.

Since the diagonal of the square bisects the angle, the right triangle is also an isosceles triangle. Using the sine ratio of $45^{\circ },$ an equation describing the relationship between $r$ and $\ell$ can be written.

$$\begin{array}{c}\sin 45^{\circ }=\frac{10-r}{\ell +r}\end{array}$$

Since $\sin 45^{\circ }=\frac{1}{\sqrt{2}}$, the above equation can be solved for $\ell .$

$\sin 45^{\circ }=\frac{10-r}{\ell +r}$

Substitute

$\sin 45^{\circ }=\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}=\frac{10-r}{\ell +r}$

Solve for $\ell$

MultEqn

$\text{LHS}\cdot (\ell +r)=\text{RHS}\cdot (\ell +r)$

$\frac{1}{\sqrt{2}}\cdot (\ell +r)=10-r$

MultEqn

$\text{LHS}\cdot \sqrt{2}=\text{RHS}\cdot \sqrt{2}$

$\ell +r=10-r(\sqrt{2})$

Distr

Distribute $\sqrt{2}$

$\ell +r=10\sqrt{2}-r\sqrt{2}$

SubEqn

$\text{LHS}-r=\text{RHS}-r$

$\ell =10\sqrt{2}-r\sqrt{2}-r$

The first equation is obtained. It is also known that the sector of circle with radius $\ell$ and the circle with radius $r$ form a cone.

As shown above, $\ell$ will be the slant height of the cone and $r$ will be the radius of its circular base. Recall that in a cone, the circumference of the circular base is equal to the arc length of the sector. Since the measure of the sector is $90^{\circ },$ its arc length can be found using the relationship between arc length and arc measure.

$$\begin{array}{rl}\text{Arc Length: }& \frac{90}{360}\cdot 2\pi \ell \\ \text{Circumference of Base:}& 2\pi r\end{array}$$

Because these two lengths are equal, another equation can be obtained.

$\frac{90}{360}\cdot 2\pi \ell =2\pi r$

Solve for $\ell$

DivEqn

$\text{LHS}/2\pi =\text{RHS}/2\pi$

$\frac{90}{360}\cdot \ell =r$

ReduceFrac

$\frac{a}{b}=\frac{a/90}{b/90}$

$\frac{1}{4}\cdot \ell =r$

MultEqn

$\text{LHS}\cdot 4=\text{RHS}\cdot 4$

$\ell =4r$

Now, a system of equations can be written using the two values of $\ell .$

$$\{\begin{array}{ll}\ell =10\sqrt{2}-r\sqrt{2}-r& \text{(I)}\\ \ell =4r& \text{(II)}\end{array}$$

Solving the System of Equations

The Substitution Method will be used to find the unknowns. First use the value of $\ell$ to find the value of $r.$

$\{\begin{array}{ll}\ell =10\sqrt{2}-r\sqrt{2}-r& \text{(I)}\\ \ell =4r& \text{(II)}\end{array}$

Substitute

$\text{(I): }$ $\ell =4r$

$\{\begin{array}{l}4r=10\sqrt{2}-r\sqrt{2}-r\\ \ell =4r\end{array}$

Solve for $r$

AddEqn

$\text{(I): }$ $\text{LHS}+r\sqrt{2}+r=\text{RHS}+r\sqrt{2}+r$

$\{\begin{array}{l}5r+r\sqrt{2}=10\sqrt{2}\\ \ell =4r\end{array}$

FactorOut

$\text{(I): }$ Factor out $r$

$\{\begin{array}{l}r(5+\sqrt{2})=10\sqrt{2}\\ \ell =4r\end{array}$

DivEqn

$\text{(I): }$ $\text{LHS}/(5+\sqrt{2})=\text{RHS}/(5+\sqrt{2})$

$\{\begin{array}{l}r=\frac{10\sqrt{2}}{5+\sqrt{2}}\\ \ell =4r\end{array}$

UseCalc

$\text{(I): }$ Use a calculator

$\{\begin{array}{l}r=2.204812\dots \\ \ell =4r\end{array}$

RoundDec

$\text{(I): }$ Round to $1$ decimal place(s)

$\{\begin{array}{l}r\approx 2.2\\ \ell =4r\end{array}$

Next, the value of $r$ can be substituted into the other equation to find the value of $\ell .$

$\{\begin{array}{l}r\approx 2.2\\ \ell =4r\end{array}$

$\text{(II): }$ $r\approx 2.2$

$\{\begin{array}{l}r\approx 2.2\\ \ell \approx 4(2.2)\end{array}$

Multiply

$\text{(II): }$ Multiply

$\{\begin{array}{l}r\approx 2.2\\ \ell \approx 8.8\end{array}$

The radii of the circle and the sector are about $2.2$ and $8.8$ centimeters, respectively.

b The surface area of a cone is the sum of the base area and the lateral area.

$$\begin{array}{c}S=\pi r^2+\pi r\ell \end{array}$$

By substituting $\ell =8.8$ and $r=2.2,$ the surface area of the cone will be determined.

$S=\pi r^2+\pi r\ell$

SubstituteII

$\ell =8.8$, $r=2.2$

$S=\pi (2.2{)}^2+\pi (2.2)(8.8)$

Evaluate right-hand side

CalcPow

Calculate power

$S=\pi (4.84)+\pi (2.2)(8.8)$

Multiply

Multiply

$S=\pi (4.84)+\pi (19.36)$

AddTerms

Add terms

$S=\pi (24.2)$

UseCalc

Use a calculator

$S=76.026542\dots$

RoundInt

Round to nearest integer

$S\approx 76$

The surface area of the cone is about $76$ square centimeters.