Areas Related to Circles - NCERT Questions

The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

SOLUTION:
**Soln. :** We have, *r*_{1} = 19 cm and *r*_{2} = 9 cm

∴ Circumference of circle-I = 2π*r*_{1} = 2π(19) cm

and circumference of circle-II = 2π*r*_{2} = 2π (9) cm

Sum of the circumference of circle-I

and circle-II = 2π(19) + 2π(9) = 2π(19 + 9) cm = 2π(28) cm

Let *R* be the radius of the circle-III.

∴ Circumference of circle-III = 2πR

According to the condition, 2π*R* = 2π(28)

Thus, the radius of the new circle = 28 cm.

The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

SOLUTION:
**Soln. :** We have,

Radius of circle-I, *r*_{1} = 8 cm

Radius of circle-II, *r*_{2} = 6 cm

∴ Area of circle-I = π*r*_{1}^{2} = π(8)^{2} cm^{2}

Area of circle-II = π*r*_{2}^{2} = π(6)^{2} cm^{2}

Let the radius of the circle-III be *R*

∴ Area of circle-III = π*R*^{2}

Now, according to the condition,

π*r*_{1}^{2} + π*r*_{2}^{2} = π*R*^{2}

⇒ π(8)^{2} + π(6)^{2} = π*R*^{2}

⇒ π(8^{2} + 6^{2}) = π*R*^{2}

⇒ 8^{2} + 6^{2} = *R*^{2}

⇒ 64 + 36 = *R*^{2}

⇒ 100 = *R*^{2}

⇒ 10^{2} = *R*^{2} ⇒ *R* = 10

Thus, the radius of the new circle = 10 cm.

The given figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

**Soln. :** Diameter of the innermost region = 21 cm

Radius of the innermost (Gold Scoring) region

∴ Area of Gold region = π(10.5)^{2} cm^{2}

Area of the Red region = π(10.5 + 10.5)^{2} - π(10.5)^{2} = π(21)^{2} - π(10.5)^{2} = π[(21)^{2} - (10.5)^{2}]

= [(21 + 10.5) (21 - 10.5)]cm^{2}

Since each band is 10.5 cm wide

∴ Radius of Gold and Red region

= (10.5 + 10.5) = 21 cm.

Area of Blue region

= π[(21 + 10.5)^{2} -(21)^{2}]cm^{2}

[(31.5)^{2} - (21)^{2}] cm^{2}

= [(31.5 + 21) (31.5 - 21)]cm^{2}

Similarly,

Area of Black region

= π[(31.5 + 10.5)^{2} - (31.5)^{2}] cm^{2}

= [(42)^{2} - (31.5)^{2}] cm^{2}

= [(42 - 31.5) (42 + 31.5)] cm^{2}

Area of White region

= π[(42 + 10.5)^{2} - (42)^{2}] cm^{2}

= π[(52.5)^{2} - (42)^{2}] cm^{2}

= π[(52.5 + 42)(52.5 - 42)] cm^{2}

= 3118.5 cm^{2}

The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

SOLUTION:
**Soln. :** Diameter of a wheel = 80 cm

∴ Radius of the wheel =

So, circumference of the wheel

⇒ Distance covered by a wheel in one revolution

Distance travelled by the car in 1 hour

= 66 km = 66 × 1000 × 100 cm

∴ Distance travelled in 10 minutes

Now, number of revolutions

Thus, the required number of revolutions

= 4375.

Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

**Soln. :** (A) We have

[Numerical area of the circle]

= [Numerical circumference of the circle]

⇒ π*r*^{2} = 2πr

⇒ π*r*^{2} - 2π*r* = 0

⇒ *r*^{2} - 2*r* = 0

⇒ *r*(*r* - 2) = 0

⇒ *r* = 0 or *r* = 2

But *r* cannot be zero

∴ *r* = 2 units.

Thus, the radius of circle is 2 units.

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

SOLUTION:
**Soln. :** Here *r* = 6 cm and θ = 60°

∴ The Area of a sector

Find the area of a quadrant of a circle whose circumference is 22 cm.

SOLUTION:
**Soln. :** Let radius of the circle = *r*

∴ 2π*r* = 22

Here θ = 90°

∴ Area of the quadrant of the circle,

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

SOLUTION:
**Soln. :** Length of minute hand = radius of the circle ⇒ *r* = 14 cm

Q Angle swept by the minute hand in

60 minutes = 360°

∴ Angle swept by the minute hand in 5 minutes

Now, area of the sector with *r* = 14 cm

and θ = 30°

Thus, the required area swept by the minute hand in 5 minutes

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major sector. (Use π = 3.14)

SOLUTION:
**Soln. :** Length of the radius (*r*) = 10 cm Sector angle θ = 90°

Area of the sector

Now, (i) Area of the minor segment

= [Area of the minor sector] -

[Area of right ∠*AOB*]

= 78.5 cm^{2} - 50 cm^{2} = 28.5 cm^{2}.

(ii) Area of the major segment

= [Area of the circle] -

[Area of the minor segment]

= π*r*^{2} - 78.5 cm^{2}

= (314 - 78.5) cm^{2} = 235.5 cm^{2}.

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

**Soln. :** Here, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2π*r*

∴ Length of arc *APB*

(ii) Area of the sector with sector angle 60°

= 11 × 21 cm^{2} = 231 cm^{2}

(iii) Area of the segment *APB* = [Area of the sector *AOB*] - [Area of ∠*AOB*] ...(1)

In ∠*AOB*, *OA* = *OB* = 21 cm

∴ ∠*A* = ∠*B* = 60° [Q ∠*O* = 60°]

⇒ *AOB* is an equilateral ∠.

∴ *AB* = 21 cm

...(2)

From (1) and (2), we have

Area of segment

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and)

SOLUTION:
**Soln. :** Here, radius (*r*) = 15 cm and

Sector angle (θ) = 60°

∴ Area of the sector

Since ∠*O* = 60° and *OA* = *OB* = 15 cm

∴ *AOB* is an equilateral triangle.

⇒ *AB* = 15 cm and ∠*A* = 60°

Draw *OM* ⊥ *AB*, in ∠*AMO*

Now, *ar*(∠*AOB*)

Now area of the minor segment

= (Area of minor sector) - (*ar* ∠*AOB*)

= (117.75 97.3125) cm^{2} = 20.4375 cm^{2}

Area of the major segment

= [Area of the circle] -

[Area of the minor segment]

= π*r*^{2} - 20.4375 cm^{2}

= 706.5 - 20.4375 cm^{2} = 686.0625 cm^{2}.

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

(Use π = 3.14 and )

**Soln. :** Here θ = 120° and *r* = 12 cm

∴ Area of the sector

= 150.72 cm^{2} ...(1)

Now, area of ∠*AOB*

[Q *OM* ⊥*AB*] ...(2)

In ∠*OAB*, ∠*O* = 120°

⇒ ∠*A* + ∠*B* = 180° - 120° = 60° Q *OB* = *OA* = 12 cm

⇒ ∠*A* = ∠*B* = 30°

So,

⇒

Now, from (2),

Area of ∠*AOB*

= 36 × 1.73 cm^{2} = 62.28 cm^{2} ...(3)

From (1) and (3)

Area of the minor segment

= [Area of sector] - [Area of ∠*AOB*]

= [150.72 cm^{2}] - [62.28 cm^{2}] = 88.44 cm^{2}.

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find :

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m.

(Use π = 3.14)

**Soln. :** Here, Length of the rope = 5 m

∴ Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed

(ii) When length of the rope is increased to 10 m, ∴ *r* = 10 m

⇒ Area of the new circular portion grazed

∴ Increase in the grazing area

= (78.5 - 19.625) m^{2} = 58.875 m^{2}.

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find:

(i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

**Soln. :** Diameter of the circle = 35 mm

∴ Radius (*r*) =

(i) Circumference = 2π*r*

Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm

∴ Length of 5 pieces = 5 × 35 = 175 mm

∴ Total length of the silver wire

= 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

∴ Sector angle θ

⇒ Area of each sector

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

**Soln. :** Here, radius (*r*) = 45 cm

Since circle is divided in 8 equal parts,

∴ Sector angle corresponding to each part

⇒ Area of a sector (part)

∴ The required area between the two ribs

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

SOLUTION:
**Soln. :** Here, radius (*r*) = 25 cm.

Sector angle (θ) = 115°

∴ Area cleaned by each sweep of the blades

[Q there are 2 blades]

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)

**Soln. :** Here, Radius (*r*) = 16.5 km and Sector angle (θ) = 80°

∴ Area of the sea surface over which the ships are warned

A round table cover has six

equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm^{2}. (Use )

**Soln. :** Here, *r* = 28 cm

Since, the circle is divided into six equal sectors.

∴ Sector angle

∴ Area of each sector

...(1)

Now, area of 1 design

= Area of segment *APB*

= Area of sector *ABO* - Area of ∠*AOB* ...(2)

In ∠*AOB*, ∠*AOB* = 60°, *OA* = *OB* = 28 cm

∴ ∠*OAB* = 60° and ∠*OBA* = 60°

⇒ ∠*AOB* is an equilateral triangle.

⇒ *AB* = *AO* = *BO* ⇒ *AB* = 28 cm

Draw *OM* ⊥ *AB*

∴ In right ∠A*OM*, we have

∴ Area of ∠*AOB*

= 14 × 14 × 1.7 cm^{2} = 333.3 cm^{2} ...(3)

Now, from (1), (2) and (3), we have:

Area of segment *APB* = 410.67 cm^{2} - 333.2 cm^{2}

= 77.47 cm^{2}

⇒ Area of 1 design = 77.47 cm^{2}

∴ Area of the 6 equal designs

= 6 × (77.47) cm^{2} = 464.82 cm^{2}

Cost of making the design at the rate of

₹0.35 per cm^{2}

= ₹ 0.35 × 464.82 = ₹ 162.68.

Tick the correct answer in the following :

Area of a sector of angle π (in degrees) of a circle with radius *R* is

(A) (B)

(C) (D)

**Soln. :** (D) Here, radius (*r*) = *R*

Angle of sector (θ) = *p*°

∴ Area of the sector

Find the area of the shaded region in the given figure. *PQ* = 24 cm, *PR* = 7cm and *O* is the centre of the circle.

**Soln. :** Since *O* is the centre of the circle,

∴ *QOR* is a diameter.

⇒ ∠*RPQ* = 90° [Angle in a semi-circle]

Now, in right ∠*RPQ*, *RQ*^{2} = *PQ*^{2} + *PR*^{2}

[Pythagoras theorem]

⇒ *RQ*^{2} = 24^{2} + 7^{2} = 576 + 49 = 625

∴ radius of circle

∴ Area of ∠*RPQ*

= 12 × 7 cm^{2} = 84 cm^{2}

Now, area of semi-circle

∴ Area of the shaded portion

= 245.54 cm^{2} - 84 cm^{2} = 161.54 cm^{2}.

Find the area of the shaded region in figure, if radii of the two concentric circles with centre *O* are 7 cm and 14 cm respectively and

∠*AOC* = 40°.

**Soln. :** Radius of the outer circle = 14 cm

and θ = 40°

∴ Area of the sector *AOC*

Radius of the inner circle = 7 cm and θ = 40°

∴ Area of the sector *BOD*

Now, area of the shaded region

= Area of sector *AOC* - Area of sector *BOD*

Find the area of the shaded region in the figure, if *ABCD* is a square of side 14 cm and *APD* and *BPC* are semi-circles.

**Soln. :** Side of the square = 14 cm ∴ Area of the square *ABCD* = 14 × 14 cm^{2}

= 196 cm^{2}

Now, diameter of the circle

= (Side of the square) = 14 cm

⇒ Radius of each of the circles

∴ Area of the semi-circle *APD*

Area of the semi-circle *BPC*

∴ Area of the shaded region

= Area of the square - [Area of semi-circle *APD*

+ Area of semi-circle *BPC*]

= 196 - [77 + 77] cm^{2} = (196 - 154) cm^{2}

= 42 cm^{2}.

Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex *O* of an equilateral triangle *OAB* of side 12 cm as centre.

**Soln. :** Area of the circle with radius 6 cm

Area of equilateral triangle, having side
*a* = 12 cm, is given by

Q Each angle of an equilateral triangle = 60°

∴ ∠*AOB* = 60°

∴ Area of sector *COD*

Now, area of the shaded region,

= [Area of the circle] + [Area of the equilateral triangle] - [Area of the sector *COD*]

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

SOLUTION:
**Soln. :** Side of the square = 4 cm ∴ Area of the square *ABCD* = 4 × 4 cm^{2}

= 16 cm^{2}

Q Each corner has a quadrant circle of radius 1 cm.

∴ Area of all the 4 quadrant squares

Diameter of the middle circle = 2 cm

⇒ Radius of the middle circle = 1 cm

∴ Area of the middle circle

Now, area of the shaded region

= [Area of the square *ABCD*] -

[(Area of the 4 quadrant circles)

+ (Area of the middle circle)]

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle *ABC* in the middle as shown in figure. Find the area of the design.

**Soln. :** Area of the circle = π*r*^{2}

'*O*' is the centre of the circle,

∴ *AO* = *OB* = *OC* = 32 cm

⇒ ∠*AOB* = ∠*BOC* = ∠*AOC* = 120°

Now, in ∠*AOB*, ∠1 + ∠2 = 60°

and *OA* = *OB* ⇒ ∠1 = ∠2

∴ ∠1 = 30°

If *OM* ⊥*AB*, then

Also,

Now, area of ∠*AOB*,

Since area of ∠*ABC* = 3 ×[area of ∠*AOB*]

Now, area of the design = [Area of the circle] - [Area of the equilateral triangle]

In figure, *ABCD* is a square of side 14 cm. With centres *A*, *B*, *C* and *D* four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

**Soln. :** Side of the square *ABCD* = 14 cm

∴ Area of the square *ABCD* = 14 × 14 cm^{2}

= 196 cm^{2}.

Q Circles touch each other

Radius of the circle

Now, area of a sector of radius 7 cm and sector angle θ as 90°

Area of 4 sectors

Area of the shaded region =

[Area of the square *ABCD*]

- [Area of the 4 sectors]

= 196 cm^{2} - 154 cm^{2} = 42 cm^{2}.

The figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segment is 60 m and they are each 106 m long.

If the track is 10 m wide, find:

(i) the distance around the track along its inner edge

(ii) the area of the track.

**Soln. :** (i) Distance around the track along its inner edge

= *BC* + *EH* +

(ii) Now, area of the track = Area of the shaded region = (Area of rectangle *ABCD*) + (Area of rectangle *EFGH*) + 2 [(Area of semi-circle of radius 40 m) (Area of semi-circle of radius 30 cm)] [Q The track is 10 m wide]

⇒ Area of the track

= (106 × 10 m^{2}) + (106 × 10 m^{2})

= 1060 m^{2} + 1060 m^{2}

[(40 + 30) × (40 - 30)] m^{2}

= 2120 m^{2} + 2200 m^{2} = 4320 m^{2}.

In the figure, *AB* and *CD* are two diameters of a circle (with centre *O*) perpendicular to each other and *OD* is the diameter of the smaller circle. If *OA* = 7 cm, find the area of the shaded region.

**Soln. :** *O* is the centre of the circle, *OA* = 7 cm

⇒ *AB* = 2*OA* = 2 × 7 = 14 cm
* OC* = *OA* = 7 cm

Q *AB* and *CD* are perpendicular to each other

⇒ *OC* ⊥*AB*

∴ Area of ∠*ABC*

Again *OD* = *OA* = 7 cm

∴ Radius of the small circle

∴ Area of the small circle

Radius of the big circle

Area of the semi-circle *OACB*

= 11 × 7 cm^{2} = 77 cm^{2}

Now, Area of the shaded region

= [Area of the small circle] + [Area of the big semi-circle *OABC*] - [Area of ∠*ABC*]

The area of an equilateral triangle *ABC* is 17320.5 cm^{2}. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.

(Use π = 3.14 and = 1.73205).

**Soln. :** Area of ∠*ABC* = 17320.5 cm^{2}

∴ ∠*ABC* is an equilateral triangle and area of an equilateral

∴

⇒ (side)^{2} = 40000

⇒ (side)^{2} = (200)^{2} ⇒ side = 200 cm

∴ Radius of each circle

Since each angle of an equilateral triangle is 60°,

∴ ∠*A* = ∠*B* = ∠*C* = 60°

∴ Area of a sector having angle of sector as 60° and radius 100 cm.

Area of 3 equal sectors

Now, area of the shaded region

= [Area of the equilateral triangle *ABC*]

- [Area of 3 equal sectors]

= 17320.5 cm^{2} - 15700 cm^{2} = 1620.5 cm^{2}.

On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

SOLUTION:
**Soln. :** Q The circles touch each other. ∴ The side of the square *ABCD*

= 3 × diameter of a circle

= 3 × (2 × radius of a circle) = 3 × (2 × 7cm)

= 42 cm

⇒ Area of the square *ABCD* = 42 × 42 cm^{2}

= 1764 cm^{2}.

Now, area of one circle

Q There are 9 squares

∴ Total area of 9 circles = 154 × 9 = 1386 cm^{2}

∴ Area of the remaining portion of the handkerchief = (1764 - 1386) cm^{2} = 378 cm^{2}.

In the figure, *OACB* is a quadrant of a circle with centre *O* and radius 3.5 cm. If *OD* = 2 cm, find the area of the

(i) quadrant *OACB*,

(ii) shaded region.

**Soln. :** (i) Here, centre of the circle is *O* and radius = 3.5 cm.

∴ Area of the quadrant *OACB*

(ii)

∴ Area of the shaded region

= (Area of the quadrant *OACB*)

- (Area of ∠*BOD*)

In the figure, a square *OABC* is inscribed in a quadrant *OPBQ*. If *OA* = 20 cm, find the area of the shaded region (Use π = 3.14)

**Soln. :** *OABC* is a square such that its side

*OA* = 20 cm

∴ *OB*^{2} = *OA*^{2} + *AB*^{2}

⇒ *OB*^{2} = 20^{2} + 20^{2}

= 400 + 400 = 800

⇒ Radius of the circle

Now, area of the quadrant

Area of the square *OABC* = 20 × 20 cm^{2}

= 400 cm^{2}

∴ Area of the shaded region

= 628 cm^{2} - 400 cm^{2} = 228 cm^{2}.

*AB* and *CD* are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre *O* (see figure). If ∠*AOB* = 30°, find the area of the shaded region.

**Soln. :** Q Radius of bigger circle *R* = 21 cm and sector angle θ = 30°

∴ Area of the sector *OAB*

Again, radius of the smaller circle, *r* = 7 cm

Also, the sector angle is 30°

∴ Area of the sector *OCD*

∴ Area of the shaded region

In the figure, *ABC* is a quadrant of a circle of radius 14 cm and a semi-circle is drawn with *BC* as diameter. Find the area of the shaded region.

**Soln. :** Radius of the quadrant = 14 cm

Therefore, area of the quadrant *ABPC*

= 22 × 7 cm^{2} = 154 cm^{2}

Area of right ∠*ABC*

⇒ Area of segment *BPC* = 154 cm^{2} - 98 cm^{2}

= 56 cm^{2}

Now, in right ∠*ABC*,
*AC*^{2} + *AB*^{2} = *BC*^{2}

⇒ 14^{2} + 14^{2} = *BC*^{2}

⇒ 196 + 196 = *BC*^{2}

⇒ *BC*^{2} = 392

∴ Radius of the semi-circle *BQC*

∴ Area of the semi-circle *BQC*

= 11 × 7 × 2 cm^{2} = 154 cm^{2}

Now, area of the shaded region

= [Area of semi-circle *BQC*]

- [Area of segment *BPC*]

= 154 cm^{2} - 56 cm^{2} = 98 cm^{2}.

Calculate the area of the designed region in the figure, common between the two quadrants of circles of radius 8 cm each.

**Soln. :** Q Side of the square = 8 cm

∴ Area of the square (*ABCD*) = 8 × 8 cm^{2}

= 64 cm^{2}

Now, radius of the quadrant *ADQB* = 8 cm

∴ Area of the quadrant *ADQB*

Similarly, area of the quadrant

Sum of the two quadrant

Now, area of design

= [Sum of the area of the two quadrant] -

[Area of the square *ABCD*]