8. Circles
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Chapter 9
8.
Circles
Circles are fundamental in geometry and have diverse real-world applications, from architecture to engineering. This lesson provides insights into understanding key properties of circles, including how to measure their circumference and area. It explains how the number pi relates to circles and explores its significance in defining their boundaries and enclosed regions. Whether for academics or practical use, it helps build an intuitive grasp of circular shapes and their dimensions.
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| | 16 Theory slides |
| | 17 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Circles
Slide of 16
The world is full of circles. They are everywhere. Try to look around right now — it is very likely to find at least one circle. For this reason, it might be useful to learn about them and their dimensions. Grab some paper and a pencil and buckle up, because it is about to begin!
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Comparing Circle Dimensions
For his birthday, Tadeo's parents are taking him to a Detroit Pistons game, his favorite NBA team. Tadeo plans to bring some money to buy souvenirs at the stadium. He asks his mother to change the coins in his piggy bank to bills to make it easier to take his money with him.
His mother tells him that the coins all have something in common. She works with Tadeo to measure the distance around the coins and the width of their faces. Tadeo then calculates the ratio of the distance around each coin to its width.
Discussion
Circle
A circle is the set of all the points in a plane that are equidistant from a given point. There are a few particularly notable features of a circle.
- Center - The given point from which all points of the circle are equidistant. Circles are often named by their center point.
- Radius - A segment that connects the center and any point on the circle. Its length is usually represented algebraically by r.
- Diameter - A segment whose endpoints are on the circle and that passes through the center. Its length is usually represented algebraically by d.
- Circumference - The perimeter of a circle, usually represented algebraically by C.
circle O,since it is centered at O.
Pop Quiz
Identifying the Parts of a Circle
Click on the indicated part of the given circle.
Discussion
The Number Pi
When Tadeo opened his piggy bank and compared the measures of the coins, he found that dividing the circumference by the diameter always resulted in the same number. This fact is true for all circles and is so important that mathematicians gave a unique name to this number.
Concept
Pi
The number pi, usually written with the Greek letter 
Since π is an irrational number, its decimal part never repeats or terminates. However, its value is often rounded to 3.14 to make calculations easier. Alternatively, π can be approximated by 722.
π,is a constant defined as the ratio between the circumference and the diameter of a circle. This ratio is the same for all circles.
π=3.1415926…
Graphically, π is the number of times that the diameter of the circle fits on top of the circle.
Discussion
Calculating the Circumference of a Circle
Back in his room, Tadeo wants to measure the circumference of one of his coins. He does not have a tape measure and wonders whether there is another way to find the circumference of a circle. Maybe he could use a ruler? Good news! He can find this information if he knows either the circle's diameter or its radius. Remember, the diameter of a circle is twice the radius.
Rule
Circumference of a Circle
The circumference of a circle is calculated by multiplying its diameter by π.
C=πd
Since the diameter is twice the radius, the circumference of a circle can also be calculated by multiplying 2r by π.
C=2πr
Example
Basketball Court
Tadeo is finally at the Detroit Pistons stadium with his parents. During the halftime show, he was able to go onto the court for a contest. There, he realized how big the logo in the central circle is.
a The diameter of the blue outer circle is 12 feet. What is the circumference of this circle? Round the answer to one decimal place.
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b The radius of the red inner circle is 5 feet. What is the circumference of this circle? Round the answer to one decimal place.
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Hint
a The circumference of a circle is calculated by multiplying its diameter by π. Approximate π as 3.14.
b The diameter of a circle is twice its radius. The circumference of a circle equals twice the radius times π.
Solution
a The circumference of a circle is calculated by multiplying its diameter by π.
C=πd
The diameter of the outer circle is 12 feet. The value of π can be approximated as 3.14. Substitute these two values into the formula to find the circumference, then round to one decimal place.
C=πd
π≈3.14
C≈3.14d
Substitute
d=12
C≈3.14(12)
Multiply
Multiply
C≈37.68
RoundDec
Round to 1 decimal place(s)
C≈37.7
b This time the diameter of the inner circle is not given but its radius is — 5 feet. Remember, the diameter of a circle is twice its radius.
d=2r
Substitute this expression into the formula used in Part A.
C=πd⇒C=π(2r)
The circumference of a circle equals twice the radius times π. Substitute 5 for r and 3.14 for π and simplify. C=π(2r)
π≈3.14
C=3.14(2r)
Substitute
r=5
C≈3.14(2(5))
Multiply
Multiply
C≈3.14(10)
Multiply
Multiply
C≈31.4
Example
A Decisive Free Throw
It is the last seconds of the fourth quarter. The Detroit Pistons are about to lose by two points. Alec Burks manages to score a 2-point shot and is fouled. The clock stops at 0.2 seconds. He prepares to take a free throw. If he makes it, the Pistons will win.
a The circumference of the free throw circle is about 37 feet long. How long is the free throw line inside the circle? Round the answer to one decimal place.
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b The basketball hoop has a circumference of about 56.5 inches. What is the radius of the hoop? Round the answer to one decimal place.
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Hint
a The free throw line is a diameter of the circle. The circumference of a circle equals its diameter times π. Substitute 3.14 for π and solve the resulting equation for the diameter.
b The circumference is equal to twice the radius times π.
Solution
a Notice that the free throw line is the diameter of the circle. This circle has a circumference of 37 feet, according to the given information.
C=37
Remember that the circumference of a circle is equal to the diameter multiplied by π.
C=πd
Substitute 37 for C and 3.14 for π to estimate the diameter of the free throw circle.
C=πd
π≈3.14
C≈3.14d
Substitute
C=37
37≈3.14d
DivEqn
LHS/3.14=RHS/3.14
3.1437≈d
RearrangeEqn
Rearrange equation
d≈3.1437
CalcQuot
Calculate quotient
d≈11.783439…
RoundDec
Round to 1 decimal place(s)
d≈11.8
b Start by recalling that the circumference of a circle is twice its radius times π.
C=π(2r)
It is given that the circumference of the basketball hoop is about 56.5 inches. Substitute 56.5 for C and 3.14 for π into the formula. Then, solve it for r to determine the radius of the basketball hoop.
C=π(2r)
π≈3.14
C≈3.14(2r)
Substitute
C=56.5
56.5≈3.14(2r)
Multiply
Multiply
56.5≈6.28r
DivEqn
LHS/6.28=RHS/6.28
6.2856.5≈r
RearrangeEqn
Rearrange equation
r≈6.2856.5
CalcQuot
Calculate quotient
r≈8.996815…
RoundDec
Round to 1 decimal place(s)
r≈9.0
Pop Quiz
Determining the Dimensions of the Circle
Discussion
The Space Inside a Circle
The amount of space inside a two-dimensional figure is known as the area of the figure. The area can usually be calculated if some dimensions of the figure are known. In the particular case of a circle, only its radius is needed.
Rule
Area of a Circle
Example
Area of a Circular Rug
When they got home from the game, Tadeo's parents gave him another birthday present. They had bought him a circular rug with the Detroit Pistons logo on it for his bedroom. Tadeo was so happy that he ran to his room and placed it on the floor next to his bed.
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Hint
b The area of a circle is π times the radius squared.
Solution
a The space occupied by a plane figure refers to its area. Since the rug is circular, its area can be calculated by multiplying π by the radius of the rug squared.
A=πr2
The radius of the rug is not given, but its diameter is. The radius of the rug can be found by using the fact that the diameter is twice the radius. The diameter of the rug is 4 feet.
The radius of the rug is 2 feet. Next, substitute 2 for r into the formula for the area of a circle.
A=πr2
Substitute
r=2
A=π(2)2
CalcPow
Calculate power
A=π(4)
UseCalc
Use a calculator
A=12.566370…
RoundDec
Round to 2 decimal place(s)
A≈12.57
b The area of the circular mirror can be found by multiplying π by its radius squared.
A=πr2
The radius of the mirror is 10 inches. Substitute 10 for r into the formula and simplify.
A=πr2
Substitute
r=10
A=π(10)2
CalcPow
Calculate power
A=π(100)
UseCalc
Use a calculator
A=314.159265…
RoundDec
Round to 2 decimal place(s)
A≈314.16
Example
Length of the Clock Hands
Tadeo has a cool basketball hoop-shaped clock on his bedroom wall.
The area of this clock is 144π square inches.
a The length of the hour hand is half the radius of the clock. How long is the hour hand?
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b The length of the minute hand is one-third the diameter of the clock. How long is the minute hand?
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Hint
a The area of a circle is equal to π times the radius squared.
b The diameter of a circle is twice the radius.
Solution
a According to the given information, the length of the hour hand is half the radius of the clock. Let h be the length of the minute hand and r be the radius of the clock. The following equation represents the relation between h and r.
h=2r
The radius of the clock is not given but its area. The radius can be calculated by using the fact that the area of a circle is π times the radius squared.
A=πr2
Tadeo's clock has an area of 144π square inches. Substitute this value for A and solve the equation for r.
A=πr2
Substitute
A=144π
144π=πr2
▼
Solve for r
DivEqn
LHS/π=RHS/π
π144π=r2
CrossCommonFac
Cross out common factors
π144π=r2
SimpQuot
Simplify quotient
144=r2
RearrangeEqn
Rearrange equation
r2=144
SqrtEqn
LHS=RHS
r2=144
SqrtPowToNumber
a2=a
r=144
CalcRoot
Calculate root
r=12
h=212⇒h=6
The minute hand is 6 inches long.
b Let m be the length of the minute hand and d be the diameter of the clock. It is given that the length of the minute hand is one-third of the diameter of the clock.
m=3d
The diameter of the clock was not given but remember that the diameter of a circle is twice its radius.
d=2r
The radius of the clock is 12 inches, according to Part A. Therefore, substitute 12 for r to find the diameter.
d=2(12)⇒d=24
The diameter of Tadeo's clock is 24 inches. Lastly, substitute 24 for d into the first equation to calculate the length of the minute hand.
m=324⇒m=8
The minute hand is 8 inches long.
Pop Quiz
Determining the Area and Other Dimensions of a Circle
Discussion
Half of a Circle
What is the shape of the space that a car's rear wiper cleans? It looks like half of a circle, right?
Next, the definition of this figure and the formulas to calculate its area and perimeter are introduced. 
Concept
Semicircle
A semicircle is half of a circle. It is a two-dimensional figure obtained when a circle is cut into two halves. Its shape consists of an arc and a segment.
The radius of a semicircle is defined as the distance from the midpoint of the segment to any point of the arc.
The perimeter of a semicircle with radius r is the length of the segment plus half the circumference of a circle with radius r. The area of a semicircle with radius r is half the area of a circle with radius r.
| Perimeter | Area |
|---|---|
| P=2r+πr | A=21πr2 |
Example
Half of a Pizza
Tadeo's parents bought pizza to celebrate both Tadeo's birthday and the Pistons' victory. It took Tadeo a while to get to the table because he was playing a video game. When he arrived, his parents had already eaten half of the supreme pizza.
External credits: @macrovector
The radius of the pizza is 6 inches.
a What is the perimeter of the remaining pizza? Write the answer in terms of π.
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b How much space does the remaining pizza take up on the plate? Write the answer in terms of π.
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Hint
a Half of the pizza is a semicircle. The perimeter of a semicircle is twice its radius plus πr.
b The part of the plate occupied by the pizza is the area of a semicircle. The area of a semicircle is 21πr2.
Solution
a Start by identifying the shape of half of the pizza. It looks like a semicircle with a radius of 6 inches.
P=2r+πr
Substitute 6 for r into the formula and simplify. Remember to keep the result in terms of π.
P=2r+πr
Substitute
r=6
P=2(6)+π(6)
Multiply
Multiply
P=12+π(6)
CommutativePropMult
Commutative Property of Multiplication
P=12+6π
b The space occupied by the pizza is its area, which is the area of the semicircle.
A=21πr2
The radius of the pizza is 6 inches. Substitute 6 into the area's formula and simplify. Keep the answer in terms of π.
A=21πr2
Substitute
r=6
A=21π(6)2
CalcPow
Calculate power
A=21π(36)
MoveRightFacToNum
ca⋅b=ca⋅b
A=2π(36)
ReduceFrac
ba=b/2a/2
A=π(18)
CommutativePropMult
Commutative Property of Multiplication
A=18π
Closure
Measuring Wheel
While watching the sports news with his dad, Tadeo saw an interesting object referees use to measure how far an athlete throws a javelin. Tadeo said that it looks like a unicycle.
Calculate the circumference of the following objects. Use 3.14 for π and round each answer to one decimal place.
External credits: Wikipedia
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External credits: Davod
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The circumference of a circle is equal to its diameter multiplied by π. C = π d We can see that the diameter of the quarter is 0.955 inches. Let's substitute this value for d and 3.14 for π to determine the circumference of the coin.
The circumference of the quarter is about 3 inches.
We are given a CD with a radius of 6 centimeters. We can calculate its circumference by multiplying π by twice its radius. C = 2π r Let's substitute 6 for r and 3.14 for π into the formula and evaluate the right-hand side.
The circumference of the CD is about 37.7 centimeters.
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Consider the following tractor.
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The circumference of the front tire is 188.5 inches. We can find its diameter by using the fact that the circumference of a circle equals its diameter times π. C = π d Let's substitute 188.5 for C and 3.14 for π into this formula to find the diameter of the tire.
The diameter of the tractor's front tire is about 60 inches.
The circumference of the rear tire is 254.5 inches. We can calculate its radius by using the fact that the circumference of a circle is twice its radius multiplied by π.
C = 2π r
First, let's substitute 254.5 for C and 3.14 for π. Then, we solve the resulting equation for r.
The rear tractor tire has a radius of about 40.5 inches.
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Consider the following pair of circles.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">r<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["2.5"]}}
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Let's begin by recalling the definitions of a radius and a diameter of a circle.
| Radius | Diameter |
|---|---|
| A segment that connects the center and any point on the circle. The length of this segment is the radius of the circle. | A segment whose endpoints are on the circle and passes through the center. The length of this segment is the diameter of the circle. |
From the definition, we can see that a diameter can be seen as two radii. This means that the diameter of a circle is twice its radius.
The radius of ⊙ P is 5. Therefore, the diameter of ⊙ P is 10.
Even though we are not given any explicit information about ⊙ Q, can can use the information we found in Part A to help us. Let's start by drawing a diameter of ⊙ Q that has point P as one endpoint.
Notice that the diameter we drew is also a radius of ⊙ P, so we know that PR= 5. We also know that the diameter of a circle is twice its radius. This means that the radius is half the diameter. d = 2r ⇒ r = d/2 Let's substitute 5 for d into the equation to determine the radius of ⊙ Q. r = 5/2 ⇒ r = 2.5 The radius of ⊙ Q is 2.5.
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Consider the following three circles.
Use 3.14 for π in the computations.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.891661em;vertical-align:-0.208331em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">A<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.328331em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2299<\/span><span class=\"mord mathdefault mtight\" style=\"margin-right:0.02778em;\">O<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.208331em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["50.24"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.891661em;vertical-align:-0.208331em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">A<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.328331em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2299<\/span><span class=\"mord mathdefault mtight\" style=\"margin-right:0.00773em;\">R<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.208331em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["12.56"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.891661em;vertical-align:-0.208331em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">A<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.328331em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mtight\">\u2299<\/span><span class=\"mord mathdefault mtight\" style=\"margin-right:0.13889em;\">T<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.208331em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["7.065"]}}
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The area of a circle is π times the radius squared. A = π r^2 In the given diagram, we can see that the radius of ⊙ O is 4.
Let's substitute 4 for r and 3.14 for π into the formula for the area.
The area of the largest circle ⊙ O is 50.24.
The area of a circle is π multiplied by the radius squared. For ⊙ R we are not given its radius, but we do know that it has a diameter of 4.
The diameter of a circle is twice its radius, so the radius is half the diameter. Therefore, the radius of ⊙ R is 2. r = d/2 ⇒ r = 4/2= 2 Now we are ready to find the area of ⊙ R. Let's substitute 2 for r and 3.14 for π into the area formula.
The area of ⊙ R is 12.56.
Let's begin by noticing that we are not given either the radius or the diameter of ⊙ T. However, we are given its circumference.
We can find the radius of ⊙ T by remembering that the circumference of a circle is twice the radius multiplied by π. C= 2π r Let's substitute 9.42 for C and 3.14 for π into the formula. Then, we will solve the resulting equation for r.
The radius of ⊙ T is 1.5. Finally, we can find its area the same way we did before. Let's do it!
The area of the smallest circle, ⊙ T, is 7.065.
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The following circle has an area of 78.5 square inches.
{"type":"text","form":{"type":"list","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"ordermatters":true,"numinput":3,"listEditable":false,"hideNoSolution":true},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Dimensions <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.36687em;vertical-align:0em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:1.80002em;vertical-align:-0.65002em;\"><\/span><span class=\"mord\"><span class=\"delimsizing size2\">{<\/span><\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1.80002em;vertical-align:-0.65002em;\"><\/span><span class=\"mord\"><span class=\"delimsizing size2\">}<\/span><\/span><\/span><\/span><\/span>","answer":{"text":["5","10","31.4"]}}
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Let's begin by recalling that the area of a circle is equal to π times the radius squared. A = π r^2 We know that the area of the given circle is 78.5 square inches. We can substitute this value for A into the formula and solve it for r. Here, we will use 3.14 for π.
We found that the radius of the circle is 5 inches. We can use this to find the diameter of the circle because the diameter of a circle is twice its radius. d = 2r & ⇒ d = 2( 5) & ⇒ d = 10 The circle has a diameter of 10 inches. Finally, we can find the circumference of the circle by multiplying π by the diameter. Remember to use 3.14 for π.
The circumference of the circle is 31.4 inches. Let's summarize the results we found! Dimensions = { 5, 10, 31.4 }
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Consider the following window.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.5478515625em;vertical-align:-0.009765625em;\"><\/span><span class=\"mord text\"><span class=\"mord Roboto-Regular\">cm<\/span><\/span><\/span><\/span><\/span>","answer":{"text":["231.3"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","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\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["3179.25"]}}
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Start by noticing that the window has the shape of a semicircle with a radius of 45 centimeters.
The perimeter of a semicircle is twice its radius plus π r. P = 2r + π r Let's substitute 45 for r and 3.14 for π into the formula.
The window has a perimeter of 231.3 centimeters.
The area of the window is the area of a semicircle with radius 45 centimeters.
The area of a semicircle is equal to half the area of a circle, which is π times the radius squared. A = 1/2π r^2 Let's substitute 45 for r and 3.14 for π into the formula to find the area of the window.
The area of the window is 3179.25 square centimeters.
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The perimeter of the protractor is 35.98 centimeters.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"centimeters","answer":{"text":["14"]}}
A car's rear wiper cleans a region of 402 square inches.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"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:0.48312em;vertical-align:0em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":"inches","answer":{"text":["16"]}}
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The protractor has the shape of a semicircle. This means that its perimeter is equal to twice the radius plus π times its radius.
Let's find the radius by solving the perimeter formula for r. First, we will substitute 35.98 for P and 3.14 for π. Then we can factor out r from the right-hand side.
The radius of the protractor is 7 centimeters. The length of the straight side of the protractor is twice the radius, which means that the straight side is 14 centimeters long.
The area cleaned by the rear wiper is a semicircle with an area of 402 square inches. Let's begin by recalling that the area of a semicircle is half the area of a circle.
The length of the wiper is the radius of the semicircle. Let's substitute 402 for A and 3.14 for π into the area's formula. Then, we solve the resulting equation for r.
The wiper has a length of about 16 inches.
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The outline of the following figure is formed by four semicircles.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":["PI"]},"rendered":false},"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:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.13889em;\">P<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"centimeters","answer":{"text":["5\\pi"]}}
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We are told that four semicircles form the outline of the figure. This means that the perimeter of the figure can be found by finding the perimeters of the semicircles. First, let's identify the radius of each part.
We can see that the larger semicircles are 4 centimeters wide, so they both have a radius of 4÷2=2 centimeters. The two smaller semicircles have a radius of 1÷2=0.5 centimeters for the same reason.
The perimeter of a semicircle is twice the radius plus π times the radius, 2r+π r. Let's substitute the corresponding values into the formula to find the four perimeters.
| Semicircle | Radius | P=2r+π r | Simplify |
|---|---|---|---|
| Upper Part | r= 2 | P_1 = 2( 2)+π( 2) | P_1 = 4+2π |
| Right Part | r= 0.5 | P_2 = 2( 0.5)+π( 0.5) | P_2 = 1+0.5π |
| Bottom Part | r= 2 | P_3 = 2( 2)+π( 2) | P_3 = 4+2π |
| Left Part | r= 0.5 | P_4 = 2( 0.5)+π( 0.5) | P_4 = 1+0.5π |
Now let's add the four values in the right-most column. crcr & 4 & + & 2π & 1 & +& 0.5π & 4 & + & 2π + & 1 & + & 0.5π & 10 & + & 5π However, notice that this sum includes the straight segments of each semicircle.
We have to subtract the lengths of these four straight segments. In total, they added up 10 extra centimeters. Let's subtract 10 from the previous sum. P = 10+5π - 10 ⇒ P=5π In conclusion, the perimeter of the given figure is 5π centimeters.
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The area of the figure is 0.9 times the area of the same figure with its mouth closed.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":["PI"]},"rendered":false},"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:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">A<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","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\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["8.1\\pi"]}}
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Notice that the given figure looks like a circle. However, there is some missing part. We are told that its area is 0.9 times the area of the same figure with its mouth closed. Let's draw that figure.
We can find the area of this circle by multiplying π by the radius squared. Let's do it!
The closed-mouth figure has an area of 9π square centimeters. We can determine the area of the open-mouth figure by multiplying 9π by 0.9. Area of The Figure 9π * 0.9 = 8.1π The area of the open-mouth figure is 8.1π square centimeters.
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