## Key Concepts

- What is a circle?
- Parts of the circle.
- What is circumference?
- Describe parts of a circle and find circumference to solve problems.
- Find the diameter using the circumference.
- Use circumference to solve a problem.

**8.5 Solve problems involving the circumference of a circle**

## Circle:

Collection of points that are equidistant from a fixed point called “center” is defined as a circle.

*(OR)*

A circle is a round shaped figure that has no corners or edges. In geometry, a circle can be defined as a closed, two-dimensional curved shape.

**Parts of a circle:**

**Radius: **The radius of a circle is the length of the line segment joining the center of the circle to any point on the circumference of the circle. A circle can have many radii, and they measure the same. Usually, the radius of the circle is denoted by ‘r’.

**Diameter: **The length of a line segment joining any two points on the circle and passing through the center is defined as a diameter. A circle can have many diameters, and they measure the same. Usually, the diameter of the circle is denoted by ‘d’.

**Circumference: **The distance around the circle is defined as the circumference of the circle. In other words, it is the perimeter of the circle.

**Chord: **The length of a line segment joining any two points on the circle is called a chord. Diameter is the biggest chord that is possible in the circle.

**Arc: **The portion of a circumference of the boundary of the circle is called its arc. The smaller portion of the circle’s boundary is called its minor arc, and the larger portion is called the major arc.

**Segment: **The region occupied by an arc and chord of the circle is defined as the segment of the circle.

**Sector: ** The region enclosed by two radii and arc of the circle is called the sector of the circle. Any two radii divide the circle into two sectors.

**8.5.1 Describe parts of a circle and find circumference to solve problems**

**Example 1: **A circle has a diameter of length 70 inches. Determine the length of the boundary of the circle.

**Method – I**

**Solution: **We know that the radius of the circle is half of the length of the diameter.

**Step 1: **Find the length of the radius to determine the length of the boundary.

Radius =Diameter / 2

=70 / 2

=35 inches

**Step 2: **We know that circumference, C = 2πr

C = 2 ×π× r

C = 2 × 3.14× 35 (Use π as 3.14)

C = 219.8 inches.

**Method – 2**

**Step 1: **Use the length of the diameter to find the circumference of the circle.

Circumference, C =πd

C =π× d

C =3.14× 70 (Useπ as 3.14)

C = 219.8 inches.

Therefore, the circumference of the circle is 219.8 inches.

**Example 2:** A circular pond has a diameter of 8 m and needs to be fenced for the protection of the children. What length of fencing is required?

**Method – I**

**Solution: **We know that the radius of the circle is half of the length of the diameter.

**Step 1: **Find the length of the radius to determine the length of the fencing.

Radius = Diameter / 2

=8/ 2

=4 m.

**Step 2: **We know that circumference, C = 2πr

C = 2 ×π× r

C = 2 ×3.14× 4

C = 25.12 m.

**Method – 2**

**Step 1: **Use the length of the diameter to find the length of the boundary of the pool.

Circumference, C =πd

C =π× d

C =3.14× 8

C =25.12m.

Therefore, the circumference of the pool is 25.12 m.

Hence, the length of fencing must be 25.12 m.

**8.5.2 Solve problems involving complementary and supplementary angles**

**Example 1: **Lawrence and Mayo are running in a circular park having 308 m as its circumference. Determine the largest possible distance between the two of them in the park.

**Solution: **

We know that the largest possible distance between the two of them in the park would be equal to the diameter of the park.

**Method – I****:** Use 3.14 as an approximation for

π𝜋

**Step 1: **We know that circumference C =π𝜋d

C =π𝜋× d

308 =3.143.14× d

d = 308 / 3.14

d ≈ 98 m

**Method – II: **Use22/7as an approximation forπ𝜋

**Step 1: **We know that circumference C =π𝜋d

C =π× d

308 =227× d

d =308 /22 /7

d = 308 ×7 / 22

d = 14 × 7

d = 98 m.

Therefore, the greatest possible distance between Lawrence and mayo is 98 m.

**Example 2: **A wheel goes through 20 revolutions to cover 66 feet. Determine the diameter of the wheel.

**Solution:**

We know that the distance covered in one revolution would be equal to the circumference of the wheel.

Circumference of the wheel = 44 feet ÷ 20

C = 44 / 20 = 22 / 10 ft.

**Method – I:** Use 3.14 as approximation for π𝜋

**Step 1: **We know that circumference C =πd

C =π× d

22 / 10ft =

3.14 × d

d = 22 / 10÷ 3.14

d ≈ 0.700 ft.

**Method – II: **Use22/7as approximation for π𝜋

**Step 1: **We know that circumference C =π𝜋d

C =π× d

22 / 10ft =227 × d

d =22 / 10 /22 /7

d =22 / 10× 7 /22

d =7 / 10ft

d = 0.7 ft.

Therefore, the diameter of the wheel is 0.7 feet.

**8.5.3 Use circumference to solve a problem**

**Example 1: **How many full revolutions does a car tire with a diameter of 25 inches make when the car travels one mile?

**Solution: **We know that 1 revolution of the tire is equal to the circumference of the tire.

**Step 1: **Use the length of the diameter to find the length of the boundary of the pool.

Circumference, C =πd

C =π× d

C =3.14× 25

C =78.5inches.

Therefore, the circumference of the tire is 78.5 inches.

**Step 2: **To find the number of revolutions in a mile, find the number of inches in 1 mile.

We know that 1 mile = 63360 inches.

Number of revolutions made by the tire = Number of inches in a mile / Circumference of the car tire

Number of revolutions made by the tire = 63360 / 78.5

= 63360 ×10/785

=633600 / 78

≈** **807

Therefore, we conclude that a car tire with a diameter of 25 inches revolves 807 times to cover 1 mile.

**Example 2: **A unicycle wheel makes five rotations to travel 37.94 feet. Find the diameter of the wheel in inches. Use 3.14 for π. Round to the nearest tenth of an inch.

**Solution: **We know that 1 revolution of the wheel is equal to the circumference of the wheel.

**Step 1: **Determine the distance covered in one revolution.

Circumference, C =37.94 / 5

C =7.588 feet.

**Step 2: **We know that circumference C =πd

C =π× d

7.588ft =3.14 × d

d =7.588÷ 3.14

d ≈ 2.41 ft.

d ≈ 2. 5 feet

**Step 3: **Convert the diameter in feet to inches.

We know that 1 feet = 12 inches.

Let 2.5 feet be ‘x’ inches.

1 / 12 = 2.5 / x

Multiply by 12 on both sides of the equation.

1 / 12× 12 =

2.5 / x × 12

1 = 30 / x

Multiply by 12 on both sides of the equation.

1 × x =

30 / x × x

x = 30 inches.

Therefore, we conclude that the diameter of the unicycle wheel is 30 inches

**Exercise**

- Determine the circumference if the radius of the circle is 21 inches.
- Determine the circumference if the diameter of the circle is 98 feet.
- If the diameter of the circle is doubled, then what will happen to the circumference of the circle?
- If the circumference of the circle is 176 m, find its diameter.
- The distance around the circular field is noticed as 1540 feet after three rounds by Michael. What would be its diameter?
- Find the circumference of a wheel whose radius is 35 cm. Find the distance covered in 60 seconds if it revolves 5 times per second.
- A dog is tied to a wooden stake in a yard. His leash is 7 m long, and he runs around in circles pulling the leash as far as it can go. How much distance will he cover in 10 rounds?
- An asteroid hit the earth and created a huge round crater. Scientists measured the distance around the crater as 78.5 miles. What is the diameter of the crater?
- A well of diameter 160 cm has a stone parapet around it. If the length of the outer edge of the parapet is 516 cm, find the width of the parapet.
- Are there any circles for which the relation between the diameter and circumference cannot be explained by ? Explain

### What have we learned:

- What a circle is?
- Parts of the circle.
- What a circumference is?
- Finding circumference to solve problems.
- Finding the diameter using the circumference.
- Using circumference to solve a problem.

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