CBSE Class 10 Mathematics Areas Related to Circles Notes Set 07

Perimeter and Area of a Circle

The collection of all points in a plane which are at a fixed distance from a fixed point is called a circle. The fixed point is the centre and the fixed distance is the radius. The diameter of a circle is twice its radius. In the figure, O is the centre and OP is the radius.

The length of the boundary of a circle is called its circumference or perimeter. The ratio of the circumference of a circle to its diameter is always a constant, denoted by the Greek letter π. \[ \frac{\text{Circumference}}{\text{Diameter}} = \pi, \qquad \text{Circumference} = \pi \times 2r = 2\pi r. \] The exact value of π is not known, because π is an irrational number. For practical purposes the value of π is taken as \( \dfrac{22}{7} \) or 3.14.

If r is the radius of a circle, then:
1. Circumference = 2πr or πd, where d = 2r is the diameter of the circle.
2. Area = πr².
3. Area of a semicircle = \( \dfrac{\pi r^2}{2} \).
4. Perimeter of a semicircle = πr + 2r.
5. Area enclosed by two concentric circles of radii R and r = πR² − πr² = π(R² − r²) = π(R + r)(R − r).

Some important points:
1. If two circles touch each other externally, the distance between their centres is equal to the sum of their radii.
2. If two circles touch each other internally, the distance between their centres is equal to the difference of their radii.
3. The distance moved by a rotating wheel in one revolution is equal to the circumference of the wheel.

Solved Examples on Circumference and Area

Example 1. Find the area of a circle whose circumference is 22 cm.
Answer:
Let r be the radius. Then 2πr = 22, so \( 2 \times \dfrac{22}{7} \times r = 22 \) and \( r = \dfrac{7}{2} \) cm.
Area \( = \pi r^2 = \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} = \) 38.5 cm².

Example 2. If the perimeter of a semi-circular protractor is 66 cm, find the diameter of the protractor (take \( \pi = \dfrac{22}{7} \)).
Answer:
Let the radius be r cm. Perimeter = πr + 2r = 66, so \( r\left(\dfrac{22}{7} + 2\right) = 66 \), that is \( r \times \dfrac{36}{7} = 66 \).
\( r = \dfrac{66 \times 7}{36} = \dfrac{77}{6} \) cm, so the diameter = 2r = \( \dfrac{77}{3} \) cm.

Example 4. A race track is in the form of a ring whose inner circumference is 352 m and the outer circumference is 396 m. Find the width of the track.
Answer:
Let the outer and inner radii be R and r metres. Then 2πR = 396 and 2πr = 352.
\( R = 396 \times \dfrac{7}{22} \times \dfrac{1}{2} = 63 \) m and \( r = 352 \times \dfrac{7}{22} \times \dfrac{1}{2} = 56 \) m.
Width of the track = R − r = 63 − 56 = 7 m.

Example 5. The diameter of a cycle wheel is 28 cm. How many revolutions will it make in moving 13.2 km?
Answer:
Distance travelled in one revolution = \( 2\pi r = 2 \times \dfrac{22}{7} \times 14 = 88 \) cm.
Total distance = 13.2 × 1000 × 100 = 1320000 cm.
Number of revolutions = \( \dfrac{1320000}{88} = \) 15000.

Example 6. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.
Answer:
Length of the wire = circumference of the circle = \( 2 \times \dfrac{22}{7} \times 28 = 176 \) cm.
Let the side of the square be x cm. Perimeter of the square = length of the wire: 4x = 176, so x = 44.
The side of the square is 44 cm.

Check Your Progress: Circumference and Area

Check Your Progress 1. The difference between the circumference and the radius of a circle is 37 cm. Find the area of the circle.
Answer:
2πr − r = 37, so \( r\left(\dfrac{44}{7} - 1\right) = r \times \dfrac{37}{7} = 37 \) and r = 7 cm.
Area = \( \dfrac{22}{7} \times 7 \times 7 = \) 154 cm².
In simple words: The two circumference terms combine into one fraction that cancels with 37.

Teacher's Note:
1. Write the difference as r(2π − 1).
2. Solve for r first.

Check Your Progress 2. A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.
Answer:
Distance in one revolution = 88000 m ÷ 1000 = 88 m = 2πr.
\( r = \dfrac{88 \times 7}{2 \times 22} = \) 14 m.
In simple words: The distance in one turn equals the circumference.

Teacher's Note:
1. Divide the total distance by the number of revolutions.
2. Equate it to 2πr.

Check Your Progress 3. A circular road runs round a circle. If the difference between the circumference of the outer circle and the inner circle is 66 m, find the width of the road.
Answer:
Let the radii be R and r. Then 2πR − 2πr = 66, so \( 2\pi(R - r) = 66 \).
\( R - r = \dfrac{66 \times 7}{2 \times 22} = \) 10.5 m (= 21/2 m).
In simple words: The width of the road is the difference of the two radii.

Teacher's Note:
1. Take 2π as common.
2. R − r is the width.

Check Your Progress 4. If the diameter of a circle is increased by 100%, find the percentage increase in its area.
Answer:
Let the old diameter be d. The new diameter is 2d, so the radius becomes double.
Area is proportional to r², so the new area is 4 times the old area. The increase is 3 times the old area, that is 300%.
In simple words: Doubling the radius makes the area four times.

Teacher's Note:
1. Area depends on the square of the radius.
2. Increase = new − old.

Check Your Progress 5. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.
Answer:
In 2 days the hour hand (short, 4 cm) makes 4 revolutions and the minute hand (long, 6 cm) makes 48 revolutions.
Distance = \( 4 \times 2\pi(4) + 48 \times 2\pi(6) = 32\pi + 576\pi = 608\pi \) cm.
With \( \pi = \dfrac{22}{7} \): \( 608 \times \dfrac{22}{7} = \) 1910.86 cm (approximately).
In simple words: Count how many full turns each hand makes in 48 hours.

Teacher's Note:
1. The hour hand makes 2 turns a day and the minute hand 24 turns an hour.
2. Add the two circumferences times their turns.

Sector and Segment of a Circle

The region bounded by an arc of a circle and its two bounding radii is called a sector of the circle. If the arc is a minor arc, the corresponding sector is called the minor sector, and the remaining part bounded by the major arc is the major sector.

Length of an arc and area of a sector. Let r be the radius of the circle with centre O and let AOB be a sector with ∠AOB = θ.

An arc that subtends 180° at the centre has length πr (a semicircle). So an arc subtending θ at the centre has length \[ \ell = \frac{\pi r}{180^\circ} \times \theta = \frac{\pi r \theta}{180^\circ}. \] Similarly an arc subtending 180° bounds a semicircular region of area \( \dfrac{1}{2}\pi r^2 \), so the area of the sector is \[ A = \frac{\pi r^2}{2} \times \frac{1}{180^\circ} \times \theta = \frac{\pi r^2 \theta}{360^\circ}. \] Dividing the two results gives \( \dfrac{A}{\ell} = \dfrac{r}{2} \), so \( A = \dfrac{\ell r}{2} \).

Some important points:
1. The minute hand describes 360° in 60 minutes, so it rotates through 6° in one minute.
2. The hour hand describes 360° in 12 hours, so it rotates through 30° in one hour, that is \( \left(\dfrac{1}{2}\right)^\circ \) in one minute.

Segment of a Circle

The region enclosed by an arc and a chord is called a segment of the circle. The segment containing the minor arc is the minor segment and the remaining segment containing the major arc is the major segment.

Area of a segment. Let r be the radius of the circle with centre O and let AB be an arc subtending θ at the centre. Draw AP ⊥ OB.

Area of the segment AMB = area of sector OAMB − area of △OAB \[ = \frac{\pi r^2 \theta}{360^\circ} - \frac{1}{2} \times OB \times AP. \] In △AOP, \( AP = OA \sin\theta = r\sin\theta \), so \[ \text{Area of the segment} = \frac{\pi r^2 \theta}{360^\circ} - \frac{1}{2} r^2 \sin\theta. \]

Solved Examples on Sector and Segment

Example 12. A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of its arc and area.
Answer:
\( \ell = \dfrac{\theta}{360^\circ} \times 2\pi r = \dfrac{150}{360} \times 2 \times \dfrac{22}{7} \times 21 = \) 55 cm.
\( A = \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{150}{360} \times \dfrac{22}{7} \times 21^2 = \dfrac{1155}{2} = \) 577.5 cm².

Example 14. The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 9 A.M. and 9.35 A.M.
Answer:
The minute hand turns through 6° in one minute, so in 35 minutes it turns through 6 × 35 = 210°.
Area swept = area of a sector of angle 210° and radius 10 cm = \( \dfrac{210}{360} \times \dfrac{22}{7} \times 10^2 = \) 183.3 cm² (approximately).

Check Your Progress: Sector and Segment

Check Your Progress 6. Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
2πr = 22, so \( r = \dfrac{22 \times 7}{2 \times 22} = \dfrac{7}{2} \) cm.
Area of the quadrant = \( \dfrac{1}{4}\pi r^2 = \dfrac{1}{4} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} = \dfrac{77}{8} = \) 9.625 cm².
In simple words: A quadrant is one quarter of the whole circle.

Teacher's Note:
1. Find the radius from the circumference.
2. Take one quarter of πr².

Check Your Progress 7. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.
Answer:
Perimeter = arc + 2r, so the arc length ℓ = 27.2 − 11.2 = 16 cm.
Area = \( \dfrac{1}{2}\ell r = \dfrac{1}{2} \times 16 \times 5.6 = \) 44.8 cm².
In simple words: The perimeter of a sector includes the two radii.

Teacher's Note:
1. Subtract 2r to get the arc.
2. Use A = ℓr/2.

Check Your Progress 8. A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of the pendulum.
Answer:
The length of the pendulum is the radius. \( \ell = \dfrac{\theta}{360} \times 2\pi r \), so \( 8.8 = \dfrac{30}{360} \times 2 \times \dfrac{22}{7} \times r \).
\( r = \dfrac{8.8 \times 12 \times 7}{44} = \) 16.8 cm.
In simple words: The arc is one twelfth of the whole circumference.

Teacher's Note:
1. Put θ = 30° in the arc formula.
2. Solve for r.

Check Your Progress 9. The perimeter of a sector of a circle of radius 7 cm is 44 cm. Find the area of the sector.
Answer:
Arc length = 44 − 14 = 30 cm.
Area = \( \dfrac{1}{2} \times 30 \times 7 = \) 105 cm².
In simple words: The area of a sector is half of arc length times radius.

Teacher's Note:
1. Remove the two radii from the perimeter.
2. Apply A = ℓr/2.

Check Your Progress 10. An arc of a circle is of length 5π cm and the sector it bounds has an area of 25π cm². Find the radius of the circle.
Answer:
\( A = \dfrac{1}{2}\ell r \), so \( 25\pi = \dfrac{1}{2} \times 5\pi \times r \).
\( r = \) 10 cm.
In simple words: The two given quantities connect through A = ℓr/2.

Teacher's Note:
1. Substitute ℓ and A.
2. The π cancels.