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Cone-shaped figures can be recognized in many places in everyday life, such as when eating ice cream. This lesson will focus on the formulas for the volume and the surface area of a cone.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Explore

Heichi knows how to calculate the volume of a cylinder. He thinks that the volume of a cone can be found using cylinders. To do so, he makes a cone-shaped mold with a height and radius of $r.$ Then, he fills it with sand and pours it into a cylinder with the same radius and height.

How can the relationship between the volumes of a cone and a cylinder be expressed? What can Heichi assume at the end of this experiment?

Discussion

Before proceeding to the volume of a cone, the definition of a cone and its characteristics will be examined.

A cone is a three-dimensional solid with a circular base and a point, called the *vertex* or *apex*, that is not in the same plane as the base. The *altitude* of a cone is the segment that runs perpendicularly from the vertex to the base.

If a cone is not a right cone, it is called an oblique cone. Oblique cones do not have a uniform slant height.

Considering Heichi's experiment, the formula for the volume of a cone will be one third of the volume of a cylinder with the same radius and height.

Rule

The volume of a cone is one third the product of its base area and its height.

The base area $B$ is the area of the circle and the height $h$ is measured perpendicular to the base.

$V=31 Bh$

Since the base is a circle, its area depends on its radius. Therefore, the base area can also be expressed in terms of the radius $r.$

$V=31 πr_{2}h$

Consider a cone and a cylinder that have the same base area and height.

Therefore, the volume of a cone is one third the volume of the cylinder with the same base area and height.

A cone can be modeled as a stack of cylinders. The sum of the volumes of the small cylinders will be greater than the cone's volume. However, the higher the number of cylinders, the more the sum will approximate the volume of the cone.

Furthermore, the ratio of the sum of the volumes of each small cylinder to the volume of the big cylinder nears $31 $ as the number of stacked cylinders increases.

Number of Cylinders | $Volume of Large CylinderSum of Volumes of Stacked Cylinders $ |
---|---|

$4$ | $≈0.469$ |

$16$ | $≈0.365$ |

$64$ | $≈0.341$ |

$256$ | $≈0.335$ |

$1024$ | $≈0.334$ |

$4096$ | $≈0.333$ |

$∞$ | $31 $ |

$V_{cone}=31 V_{cylinder}⇓V_{cone}=31 Bh $

Since the base area is the area of a circle, that formula can be substituted for $B$ to find a more detailed formula for the volume of the cone.
$V_{cone}=31 πr_{2}h $

Example

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

The cathedral reaches $114$ meters in height, excluding the cross. Furthermore, its circular base has a radius of $50$ meters. Calculate its volume. Round the answer to the nearest cubic meter.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">m<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["298451"]}}

The volume of a cone is one third of the product of its base area and height.

To find the volume of the cathedral, the formula for the volume of a cone will be used.
The volume of the cone is approximately $298451$ cubic meters.

$V=31 πr_{2}h $

In this formula, $r$ is the radius of the base and $h$ is the height of the cone. The height of the conical cathedral is $114$ meters and its radius is $50$ meters. By substituting these values into the formula, the volume can be found.
$V=31 πr_{2}h$

SubstituteII

$r=50$, $h=114$

$V=31 π(50)_{2}(114)$

▼

Simplify right-hand side

CalcPow

Calculate power

$V=31 π(2500)(114)$

Multiply

Multiply

$V=31 285000π$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V=3285000π $

UseCalc

Use a calculator

$V=298451.302091…$

RoundInt

Round to nearest integer

$V≈298451$

Example

Tadeo is learning how to make a traditional Chinese conical hat. He notices that the craftsman uses $20-$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.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8238736249999999em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">cm<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["3217"]}}

Use the Pythagorean Theorem to find the height of the cone.

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.

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.$AC_{2}=AB_{2}+BC_{2}$

SubstituteII

$AC=20$, $BC=16$

$20_{2}=AB_{2}+16_{2}$

▼

Solve for $AB$

CalcPow

Calculate power

$400=AB_{2}+256$

SubEqn

$LHS−256=RHS−256$

$144=AB_{2}$

RearrangeEqn

Rearrange equation

$AB_{2}=144$

SqrtEqn

$LHS =RHS $

$AB_{2} =144 $

SqrtPowToNumber

$a_{2} =a$

$AB=12$

$V=31 πr_{2}h $

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.
$V=31 πr_{2}h$

SubstituteII

$r=16$, $h=12$

$V=31 π(16_{2})(12)$

▼

Simplify right-hand side

CalcPow

Calculate power

$V=31 π(256)(12)$

Multiply

Multiply

$V=31 3072π$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V=33072π $

UseCalc

Use a calculator

$V=3216.990877…$

RoundInt

Round to nearest integer

$V≈3217$

Example

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

The height of the cone part is $28$ inches. The prism below it is a square prism with side lengths of $14$ inches and height of $1$ inch. Find the radius of the cone. Write the answer to the nearest inch.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"in.","answer":{"text":["10"]}}

The volume occupied by the traffic cone is the sum of the volume of the prism base and the volume of the cone part.

The traffic cone is basically composed of two solids — a cone and a square prism. Therefore, the volume occupied by the traffic cone is the sum of the volume of the prism $V_{p}$ and the volume of the cone $V_{c}.$
### Finding the Volume of the Cone

The base of the prism is a square with side lengths of $14$ inches, so its area is the square of $14.$
### Finding the Radius of the Cone

Finally, using the formula for the volume of a cone, the radius of the cone can be found.
The radius of the cone is about $10$ inches.

$V=V_{p}+V_{c} $

First the volume of the cone will be found. Then, the volume formula of a cone will be used to calculate the radius of the base of the cone. $B=14_{2}⇔B=196 $

Since the volume of a prism is its base area times its height, the volume of the square prism can be found as follows.
Given that the total volume of the traffic cone is $3128$ cubic inches, the volume of the cone part can be found now. $V_{c}=31 πr_{2}h $

To do so, substitute $2932$ for $V_{c}$ and $28$ for $h$ into the formula and solve for $r.$
$V_{c}=31 πr_{2}h$

SubstituteII

$V_{c}=2932$, $h=28$

$2932=31 πr_{2}(28)$

▼

Solve for $r$

MultEqn

$LHS⋅3=RHS⋅3$

$8796=πr_{2}(28)$

CommutativePropMult

Commutative Property of Multiplication

$8796=28πr_{2}$

DivEqn

$LHS/28π=RHS/28π$

$28π8796 =r_{2}$

RearrangeEqn

Rearrange equation

$r_{2}=28π8796 $

SqrtEqn

$LHS =RHS $

$r_{2} =28π8796 $

SqrtPowToNumber

$a_{2} =a$

$r=28π8796 $

UseCalc

Use a calculator

$r=9.999738…$

RoundInt

Round to nearest integer

$r≈10$

Example

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!

a Ali will fill the cone with water and the interior portion of the cylinder not occupied by the cone with foam. How many cubic centimeters of water and foam does Ali need? Write the answers to the nearest integer.

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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Foam:","formTextAfter":"cm<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["67858","67 858","67,858"]}}

b What percent of the interior of the cylinder is *not* occupied by the cone? Round the answer to one decimal place.

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

b Use the exact values for the volumes obtained in Part A.

a The volumes will be calculated one at a time.
### Volume of Water

Therefore, the volume of water that Ali needs is $10800π$ cubic centimeters. Finally, this number will be approximated to the nearest integer.
Ali needs about $33929$ cubic centimeters of water. ### Volume of Foam

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 $π,$ the square of the radius, and the height.$V_{cone}=31 πr_{2}h $

In the this formula, $81$ and $20$ can be substituted for $h$ and $r,$ respectively.
$V_{cone}=31 πr_{2}h$

SubstituteII

$h=81$, $r=20$

$V_{cone}=31 π(20_{2})(81)$

▼

Evaluate right-hand side

CalcPow

Calculate power

$V_{cone}=31 π(400)(81)$

Multiply

Multiply

$V_{cone}=31 π(32400)$

CommutativePropMult

Commutative Property of Multiplication

$V_{cone}=31 (32400)π$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$V_{cone}=332000 π$

CalcQuot

Calculate quotient

$V_{cone}=10800π$

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 $π,$ the square of the radius, and the height.$V_{cylinder}=πr_{2}h $

One more time, $81$ and $20$ can be substituted for $h$ and $r,$ respectively.
The volume of the cylinder is $32400π$ 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 $10800π$ cubic centimeters. $32400π−10800π=21600π $

The volume of the cylinder that is not occupied by the cone is $26600π$ cubic centimeters and is the volume of foam that Ali will need. This number will be approximated to the nearest integer.
Ali needs about $67858$ cubic centimeters of foam. b In Part $A$ it was found that the volume of the cylinder is $32400π$ cubic centimeters. It was also found that the volume of the portion of the cylinder not occupied by the cone is $21600π$ cubic centimeters. The ratio of the second value to the first value will result in the desired percentage.
It can be concluded that the volume of the cylinder not occupied by the cone represents $66.7%$ of the interior of the cylinder.

$32400π21600π $

▼

Evaluate

CancelCommonFac

Cancel out common factors

$32400π 21600π $

SimpQuot

Simplify quotient

$3240021600 $

ReduceFrac

$ba =b/10800a/10800 $

$32 $

▼

Convert to percent

FracToDiv

$ba =a÷b$

$0.666666…$

WritePercent

Convert to percent

$66.666666…%$

RoundDec

Round to $1$ decimal place(s)

$66.7%$

Discussion

Consider a right cone with radius $r$ and slant height $ℓ.$

The surface area of a right cone is the sum of the base area and the lateral area. The area of the base is given by $πr_{2}$ and the lateral area is $πrℓ.$

$SA=πr_{2}+πrℓ$

The surface area of a cone is made of two main components, the area of the circular base and the lateral area.
Finally, the expressions for the area of the circular base and the lateral area can be substituted to find the expression for the surface area of a cone.

Let $LA$ be the lateral area of a cone and $BA$ the area of the circular base. The surface area $SA$ of a cone is made of the sum of the area of the circular base and the lateral area.

$SA=BA+LA $

The area of the circular base is found using the formula for the area of a circle.
$BA=πr_{2} $

The lateral area can be better visualized in two dimensions. Suppose that the cone is cut and the lateral area is expanded.
It should be noted that the figure obtained is a sector of a circle of radius $ℓ,$ the slant height of the cone. Then, the lateral area can be obtained using the formula for the area of a sector of a circle of radius $ℓ.$

$Area of a Sector=360_{∘}θ ⋅πℓ_{2} $

But from the image it should also be noted that the arc of the sector has the same length as the circumference of the circular base of radius $r.$ Using the formula for the length of an arc relative to its measure, it is possible to write an equation to equate these quantities.
$Circumference of Base=Arc Length⇓2πr=360_{∘}θ ⋅2πℓ $

Dividing both sides of the equation by $2π,$ this equation can be simplified.
$r=360_{∘}θ ⋅ℓ $

Now it is possible to write an expression for the lateral area. First, the expression is the same as the formula for the area of a sector.
$LA=360_{∘}θ ⋅πℓ_{2}$

CommutativePropMult

Commutative Property of Multiplication

$LA=(360_{∘}θ ⋅ℓ)⋅πℓ$

Substitute

$360_{∘}θ ⋅ℓ=r$

$LA=rπℓ$

$SA=BA+LA⇓SA=πr_{2}+πrℓ $

Example

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

The material to be used to create the figure costs $$125$ per{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">$<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["192.76"]}}

The area of the lateral surface of a cone is the product of $π,$ the radius, and the slant height. A cylinder's lateral surface area is twice the product of $π,$ the radius, and the height.

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, $100cm1m .$
### 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 $π,$ the radius, and the height.
The lateral area of the cylinder is about $0.324π$ square meters. ### Lateral Area of the Cone

The lateral area of a cone is the product of $π,$ the radius, and the slant height.
The slant height of the cone is about $0.6961 $ meters. Now, the formula for the lateral area of a cone can be used. Substitute $0.83$ and $0.2$ for $ℓ$ and $r,$ respectively.
### Total Cost

Finally, the sum of the lateral areas will be multiplied by the cost per square meter.
To make the shape that Ali needs, he has to spend about $$192.76.$

$Height:81cm ⋅100cm 1m Radius:20cm ⋅100cm 1m =0.81m=0.2m $

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.

$LA_{cylinder}=2πrh $

In the above formula, $0.81$ and $0.2$ can be substituted for $h$ and $r,$ respectively.
$LA_{cylinder}=2πrh$

SubstituteII

$h=0.81$, $r=0.2$

$LA_{cylinder}=2π(0.2)(0.81)$

▼

Evaluate right-hand side

CommutativePropMult

Commutative Property of Multiplication

$LA_{cylinder}=2(0.2)(0.81)π$

Multiply

Multiply

$LA_{cylinder}=0.324π$

$LA_{cone}=πrℓ $

The slant height $ℓ$ 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

$0.81_{2}+0.2_{2}=ℓ_{2}$

▼

Solve for $ℓ$

CalcPow

Calculate power

$0.6561+0.04=ℓ_{2}$

AddTerms

Add terms

$0.6961=ℓ_{2}$

SqrtEqn

$LHS =RHS $

$0.6961 =ℓ$

RearrangeEqn

Rearrange equation

$ℓ=0.6961 $

$LA_{cone}=πrℓ$

SubstituteII

$ℓ=0.6961 $, $r=0.2$

$LA_{cone}=π(0.2)(0.6961 )$

CommutativePropMult

Commutative Property of Multiplication

$LA_{cone}=0.20.6961 π$

$(0.324π+0.20.6961 π)125$

$192.76$

Example

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.

a Find the radii of the circle and the sector. Write the answers to one decimal place.

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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69444em;vertical-align:0em;\"><\/span><span class=\"mord\">\u2113<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"cm","answer":{"text":["8.8"]}}

b Find the surface area of the cone. Round the answer to the nearest square centimeter.

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a The circumference of the circle is equal to the arc length of the sector.

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

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

$sin45_{∘}=ℓ+r10−r $

Since $sin45_{∘}=2 1 $, the above equation can be solved for $ℓ.$
$sin45_{∘}=ℓ+r10−r $

Substitute

$sin45_{∘}=2 </$