Areas Related To Circles Class 10th Mathematics CBSE Solution

Class 10^th Mathematics CBSE Solution

Exercise 12.1

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of…
2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the…
3. Fig. 12.3 depicts an archery target marked with its five scoring areas from the…
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions…
5. Tick the correct answer in the following and justify your choice: If the…

Exercise 12.2

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector…
2. Find the area of a quadrant of a circle whose circumference is 22 cm.…
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the…
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find…
5. In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre.…
6. A chord of a circle of radius 15 cm subtends an angle of 60 at the centre. Find…
7. A chord of a circle of radius 12 cm subtends an angle of 120 at the centre.…
8. A horse is tied to a peg at one corner of a square shaped grass field of side…
9. A brooch is made with silver wire in the form of a circle with diameter 35 mm.…
10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming…
11. A car has two wipers which do not overlap. Each wiper has a blade of length 25…
12. To warn ships for underwater rocks, a lighthouse spreads a red colored light…
13. A round table cover has six equal designs as shown in Fig. 12.14. If the…
14. Tick the correct answer in the following: Area of a sector of angle p (in…

Exercise 12.3

1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and…
2. Find the area of the shaded region in Fig. 12.20, if radii of the two…
3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side…
4. Find the area of the shaded region in Fig. 12.22, where a circular arc of…
5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm…
6. In a circular table cover of radius 32 cm, a design is formed leaving an…
7. In Fig. 12.25, ABCD is a square of side 14 cm. With centers A, B, C and D, four…
8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.…
9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O)…
10. The area of an equilateral triangle ABC is 17320.5cm2. With each vertex of the…
11. On a square handkerchief, nine circular designs each of radius 7 cm are made…
12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm.…
13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm,…
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7…
15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle…
16. Calculate the area of the designed region in Fig. 12.34 common between the two…

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Exercise 12.1

Question 1.

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.

Answer:

Given: Radius (r₁) of 1st circle = 19 cm

Radius (r₂) or 2nd circle = 9 cm

Now,

Let the radius of 3rd circle be r

Circumference of 1st circle = 2πr₁ = 2π (19) = 38π cm

Circumference of 2nd circle = 2πr₂ = 2π (9) = 18π cm

Circumference of 3rd circle = 2πr

Given:

Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle

2πr = 38π + 18π = 56π cm

r = 

= 28

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm

Question 2.

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.

Answer:

Radius (r1) of 1st circle = 8 cm

Radius (r2) of 2nd circle = 6 cm

Let the radius of 3rd circle be r

Area of 1^stcircle = r1² = π (8)² = 64π

Area of 2^nd circle = πr2² = π (6)² = 36 π

Given that,

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

πr² = πr1² + πr2²

πr² = 64 π + 36 π

πr² = 100 π

r² = 100

r = 

But we know that the radius cannot be negative. Hence, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm

Question 3.

Fig. 12.3 depicts an archery target marked with its five scoring areas 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.

Answer:

Radius (r₁) of gold region (i.e., 1st circle) =  = 10.5 cm

Given:

Each circle is 10.5 cm wider than the previous circle.

Hence,

Radius (r₂) of 2nd circle = 10.5 + 10.5

21 cm

Radius (r₃) of 3rd circle = 21 + 10.5= 31.5 cm

Radius (r₄) of 4th circle = 31.5 + 10.5= 42 cm

Radius (r₅) of 5th circle = 42 + 10.5= 52.5 cm

Area of gold region = Area of 1^stcircle

= πr1²

= π (10.5)²

= 346.5 cm²

Area of red region = Area of 2nd circle - Area of 1^stcircle

= πr₂² – πr₁²

= π (21)² – π (10.5)²

= 441π – 110.25π

= 330.75π

= 1039.5 cm²

Area of blue region = Area of 3^rdcircle - Area of 2^ndcircle

= πr₃² – πr2²

= π (31.5)² – π (21)²

= 992.25 π - 441 π

= 551.25 π

= 1732.5 cm²

Area of black region = Area of 4^thcircle - Area of 3rd circle

= πr₄² – πr₃²

= π (42)² – π (31.5)²

= 1764 π – 992.25 π

= 771.75 π

= 2425.5 cm²

Area of white region = Area of 5^thcircle - Area of 4^thcircle

= πr₅² – πr₄²

= π (52.5)² – π (42)²

= 2756.25 π - 1764 π

= 992.25 π

= 3118.5 cm²

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm², and 3118.5 cm² respectively

Question 4.

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?

Answer:

The diameter of the wheel of the car = 80 cm

Radius (r) of the wheel of the car = 40 cm

Circumference of wheel = 2πr

⇒ 2π (40) = 80π cm

Speed of car = 66 km/hour

1 km = 1000 m and 1m = 100 cm 
⇒ 1 km = 100000 cm

= 110000 cm/min
As distance= speed × time

Distance traveled by car in 10 minutes

= 110000 × 10 
= 1100000 cm

Let the number of revolutions of the wheel of the car be "n".

n × Distance travelled in 1 revolution (i.e., circumference)
(In one revolution of the wheel, a wheel covers the distance equal to its circumference)

= Distance travelled in 10 minutes

n × 80 π = 1100000

n = 

= 

= 4375

Therefore, each wheel of the car will make 4375 revolutions.

Question 5.

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

Answer:

Let the radius of the circle be r

Circumference of circle = 2πr

Area of circle = πr²

Given that, the circumference of the circle and the area of the circle are equal.

This implies 2πr = πr²

2 = r

Therefore, the radius of the circle is 2 units

Hence, the correct answer is A

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Exercise 12.2

Question 1.

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

Answer:

Let OACB be a sector of the circle making 60° angle at centre O of the circle

Area of sector of angle θ =  ×πr²

Area of sector OACB = × (6)²

= 

=  cm²

Hence,

The area of the sector of the circle making 60° at the centre of the circle is cm²

Question 2.

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

Answer:

To find: Area of the quadrant of the circle
Given: Circumference of the circle 
Let the radius of the circle be r

Circumference = 22 cm

Circumference = 2πr

Therefore, 2πr = 22

Quadrant of circle means 1/4 th of the circle.

Area of such quadrant of the circle =

= 

= 

=  cm²
Hence, The area of the quadrant of the circle is cm²

Question 3.

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

Answer:

So 15 minutes = 90º

5 min = 5 × 6°
= 30°
So 5 minutes subtend an angle of 30º.

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area swept by min hand = Area of sector

Question 4.

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)

Answer:

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.



= 3.14 × 25
= 78.5 cm²

OAB is a right angles triangle, 
Area of triangle OAB = 1/2 (base x height)
Area of ΔOAB = x OA x OB

=  x 10 x 10

= 50 cm²

Area of minor segment ACB = Area of minor sector OACB -Area of ΔOAB

= 78.5 - 50

= 28.5 cm²
ii) Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

Area of major sector OACB =

=3/4 x 3.14 x 10 x 10

=235.5cm²

Question 5.

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.

Answer:

Radius (r) of circle = 21 cm

The angle subtended by the given arc = 60°

Length of an arc of a sector of angle θ =

(i) Length of arc ACB = 

=  × 2 × 22 × 3

= 22 cm

(ii) Area of sector OACB =


=

=  ×  × 21 × 21

= 231 cm²

(iii) In ΔOAB,

As OA = OB

⇒ ∠OAB = ∠OBA (Angles opposite to equal sides are equal)

⇒ ∠OAB + ∠AOB + ∠OBA = 180°

⇒ 2∠OAB + 60° = 180°

⇒ 2∠OAB = 180° - 60° 

⇒ ∠OAB = 60°

Hence,

ΔOAB is an equilateral triangle

Area of equilateral triangle = (Side)²

⇒ Area of ΔOAB = (Side)²

=  × (21)²

=  cm²

Area of segment ACB = Area of sector OACB - Area of ΔOAB

= (231 - ) cm²

Question 6.

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 = 1.73)

Answer:

Radius (r) of circle = 15 cm

Area of sector OPRQ = (60/360)×Πr²

= 1/6 × 3.14 × (15)²

= 117.75 cm²

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = ×(Side)²

=  × (15)²

= 

= 56.25 × 

= 97.3125 cm²

Area of segment PRQ = Area of sector OPRQ - Area of ΔOPQ

= 117.75 - 97.3125

= 20.4375 cm²

Area of major segment PSQ = Area of circle - Area of segment PRQ

= Π(15)² – 20.4375

= 3.14 ×225 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm²

Question 7.

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 √3 = 1.73)

Answer:

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST and the angle O.

SV = VT

In ΔOVS,
As, 

 = Cos 60^o

 = 

OV = 6 cm

 = Sin 60^o

 = 

SV = 6√3 cm

ST = 2 × SV

= 2 × 6√3

= 12√3 cm

Area of ΔOST = × 12√3 × 6

= 36√3

= 36 × 1.73

= 62.28 cm²

Area of sector OSUT = × π × (12)²

=  × 3.14 × 144

= 150.72 cm²

Area of segment SUT = Area of sector OSUT - Area of ΔOST

= 150.72 - 62.28

= 88.44 cm²

Question 8.

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 Fig. 12.11). 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)

Answer:

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius

(i) Area that can be grazed by horse = Area of sector

=  × r²

=  × 3.14 × 5 × 5

= 19.625 m²

The area that can be grazed by the horse when the length of rope is 10 m long =  ×  (10)²

=  × 3.14 × 100

= 78.5 m²

(ii) Increase in grazing area = Area grazed by a horse now - Area grazed previously
(78.5 - 19.625)

= 58.875 m²

Question 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 Fig. 12.12. Find:
(i) The total length of the silver wire required

(ii) The area of each sector of the brooch

Answer:

(i) To Calculate total length of wire required to make brooch, we need to find Circumference of Brooch and the length of 5 diameters of Brooch.

Diameter of Circle = 35 mm



Radius of Circle = 35/2 mm

Circumference of Circle = 2 π r



Total Length of wire required = Circumference of Circle + 5 x Diameter of Circle 

Total Length of wire required = 110 + 5 x 35

Total Length of wire required = 285 mm

(ii)

A complete Circle subtends an angle of 360º

10 sectors = 360º

1 sector will subtend = 36º

Question 10.

An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer:

There are 8 ribs in the given umbrella. The area between two consecutive ribs is subtending = 45^o at the centre of the assumed flat circle

Now we know that area of a sector of a circle subtending an angle x is given by

Question 11.

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.

Answer:

It can be seen from the figure that each blade of wiper would sweep an area of a sector of 115° in a circle with 25 cm radius

Area of sector = 



Area swept by 2 blades :


= 1254.96 cm²

=1255 cm² (approx)

Question 12.

To warn ships for underwater rocks, a lighthouse spreads a red colored 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)

Answer:

It can be seen from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius

Area of sector OACB = × r²

=  × 3.14 × 16.5 × 16.5

= 189.97 km²

Question 13.

A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm². (Use  = 1.7)

Answer:

It can be concluded that these designs are segments of the circle.

Let us take segment APB.

Chord AB is a side of the regular hexagon.

And,

Each chord will substitute = 60^o at the centre of the circle

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB = 180° - 60°
= 120°

∠OAB = 60°

Hence,

ΔOAB is an equilateral triangle.

Area of ΔOAB = (Side)²

=  × (28)²

= 196 

= 333.2 cm²

Area of sector OAPB = × r²

=  ×  × 28 × 28

=  cm²

Now,

Area of segment APB = Area of sector OAPB - Area of ΔOAB

= ( - 333.2) cm²

Area of design = 6 × ( - 333.2)

= 2464 – 1999.2

= 464.8 cm²

Cost of making 1 cm² designs = Rs 0.35

Cost of making 464.76 cm² designs = 464.8 × 0.35

= Rs 162.68

Hence, the cost of making such designs would be Rs 162.68.

Question 14.

Tick the correct answer in the following:

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

A. 

B. 

C. 

D. 

Answer:

Area of sector of angle θ = ×R²

Where θ = angle , r = radius of circle

Here θ = p and radius = R

Area of sector of angle p = (R²)

Multiply and divide by 2,

Area of sector of angle p = ()(2R²)

Hence, (D) is the correct answer

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Exercise 12.3

Question 1.

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer:

As, ΔPQR is in the semicircle and we know angle in the semicircle is a right angle, therefore PQR is a right-angled triangle.
[ If an angle is inscribed in a semi-circle, that angle measures 90 degrees.]
Here, QR is the hypotenuse of ΔPQR as lines from two ends of diameter always make a right angle when they meet at the circumference of that circle.

Hence by Pythagoras theorem we get,
Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

QR² = PQ² + PR²

QR² = 24² + 7²

QR² = 576 + 49

QR = 25 cm

Diameter = 25 cm

Or, radius = 12.5cm

Area of right angled triangle triangle = 

= 12 × 7
= 84 cm²

= 245.3125 sq cm

Hence,

Area of shaded region = Area of semicircle - area of ΔPQR
= 245.3125 – 84

Area of shaded region = 161.3125 sq cm

Question 2.

Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°.

Answer:

Area of the shaded region = Area of Bigger sector – Area of Smaller Sector

If r₂ and r₁ are the two radii of Bigger and Smaller circles respectively, then:

Area of bigger sector = 40/360 πr₂²

Area of smaller sector = 40/360 πr₁²

Question 3.

Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer:

It can be observed from the figure that the radius of each semi-circle is 7 cm

Area of each semi-circle =r²

=  *  * (7)²

= 77 cm²

Area of square ABCD = (Side)²

= (14)²

= 196 cm²

Area of the shaded region= Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC

= 196 - 77 - 77

= 196 - 154

= 42 cm²

Question 4.

Find the area of the shaded region in Fig. 12.22, 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.

Answer:

We know that each interior angle of an equilateral triangle is of measure 60°

Area of sector = r²

=  x  x 6 x 6

=  cm²

Area of triangle OAB =  x (12)²

= 36  cm²

Area of circle = πr²

=  × 6 × 6

=  cm²

Area of shaded region = Area of ΔOAB + Area of circle - Area of sector

=  - 

= (36 + ) cm²

Question 5.

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 Fig. 12.23. Find the area of the remaining portion of the square.

Answer:

Each quadrant is a sector of 90° in a circle of 1 cm radius

Area of each quadrant =  * r²

=  *  * (1)²

=  cm²

Area of square = (Side)²

= (4)²

= 16 cm²

Area of circle = πr² = π (1)²

=  cm²

Area of the shaded region = Area of square - Area of circle - 4 × Area of quadrant

= 16 -  - 4 * 

= 16 -  - 

= 16 - 

= 

=  cm²

Question 6.

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design (shaded region).

Answer:

Radius of the circle "R" = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD [ By the property of equilateral triangle inscribed in a circle]

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm
In ΔADB,

By Pythagoras theorem,

AB² = AD² + BD²

⇒ AB² = 48² + (AB/2)²

⇒ AB² = 2304 + AB²/4
⇒ 3/4 (AB²)= 2304

⇒ AB² = 3072

⇒ AB= 32√3 cm
Area of ΔABC = √3/4 × (32√3)² cm²
= 768√3 cm²

Area of circle = π R²
= 22/7 × 32 × 32 
= 22528/7 cm²

Area of the design = Area of circle - Area of ΔABC

= (22528/7 - 768√3) cm²

Question 7.

In Fig. 12.25, ABCD is a square of side 14 cm. With centers 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.

Answer:

To Find: Area of shaded region

Given: Side of square ABCD = 14 cm

Radius of circles with centers A, B, C and D = 14/2 = 7 cm


Area of shaded region = Area of square - Area of four sectors subtending right angle


Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius. So, Area of four sectors will be equal to Area of one complete circle

Area of 4 sectors = Πr²

Area of square ABCD = (Side)²

Area of square ABCD = (14)²

Area of square ABCD = 196 cm²

Area of shaded portion = Area of square ABCD - 4 × Area of each sector

= 196 – 154

= 42 cm²

Therefore, the area of shaded portion is 42 cm²

Question 8.

Fig. 12.26 depicts a racing track whose left and right ends are semicircular.



The distance between the two inner parallel line segments 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.

Answer:

(i) Distance around the track along its inner edge = AB + semicircle BEC + CD + semicircle DFA

= 106 + ( × 2πr )+ 106 +(  × 2πr) [Perimeter of semicircle is half of circle and perimeter of circle is 2πr)

= 212 +(  × 2 ×  × 30) +(  × 2 ×  × 30)

= 212 + (2 ×  ×30)

= 

=  m

(ii) Area of the track = (Area of GHIJ - Area of ABCD) + (Area of semi-circle HKI - Area of semi-circle BEC) + (Area of semi-circle GLJ - Area of semi-circle AFD)
Area of semicircle is half of circle and area of circle is πr² 
Also area of rectangle = length x breadth

= (106 × 80 – 106 × 60 )+(  ×  × (40)² -  ×  × (30)²) +(  ×  × (40)² -  ×  × (30)² )

=[ 106 (80 – 60) ]+[  × (40)² -  × (30)²]

= [106 (20)] +[  (40)² – (30)²]

= 2120 +[  (40 – 30) (40 + 30)]

= 2120 + (22/7) (10) (70)

= 2120 + 2200

= 4320 m²

Question 9.

In Fig. 12.27, 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.

Answer:

Radius (r₁) of larger circle = 7 cm
As OD = r₂ = 7 cm

Radius (r₂) of smaller circle =  cm

Area of smaller circle = π (r₂)²

=  x  x 

=  cm²

Area of semi-circle ACB = 1/2 Πr₁²

=  x  x (7)²

= 77 cm²

Area of triangle ABC =  x AB x OC
As AB = OB + OA
⇒ AB = 7 + 7
⇒ AB = 14 cm

Area of triangle ABC =  x 14 x 7

= 49 cm²

Area of the shaded region

= Area of smaller circle + Area of semi-circle ACB - Area of ΔABC

=  + 77 – 49

= 28 + 

= 28 + 38.5

= 66.5 cm²

Question 10.

The area of an equilateral triangle ABC is 17320.5cm2. With each vertex of the triangle as centre, acircle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and= 1.73205)

Answer:

Let the side of the equilateral triangle be a

Area of equilateral triangle = 17320.5 cm²

 (a)² = 17320.5

 (a)² = 17320.5

a² = 4 × 10000

a = 200 cm

Each sector is of measure 60° as each angle of an equilateral triangle is 60º.

Area of each sector =  × π × r²

=  × π × (100)²

= 

=  cm²

Area of shaded region = Area of equilateral triangle - 3 × Area of each sector

= 17320.5 – 3 * 

= 17320.5 – 15700

= 1620.5 cm²

Question 11.

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Answer:

Radius of circle is 7 cm.
Diameter of the circle is 14 cm.
As 3 circles are there on one side of the square.
The side of the square will be 14 × 3 = 42 cm

Area of square = (Side)²

= (42)²

= 1764 cm²

Area of each circle = πr²

=  × (7)²

= 154 cm²

Area of 9 circles = 9 × 154

= 1386 cm²

Area of the remaining portion of the handkerchief = 1764 - 1386

= 378 cm²

Question 12.

In Fig. 12.30, 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.

Answer:

(i) Since OACB is a quadrant, it will subtend 90° angle at O

Area of quadrant OACB =  * r²

=  *  * (3.5)²

=  *  * ()²

= 

=  cm²

(ii) Area of ΔOBD =  * OB * OD

=  * 3.5 * 2

=  *  * 2

=  cm²

Area of the shaded region = Area of quadrant OACB - Area of ΔOBD

=  - 

= 

=  cm²

Question 13.

In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Answer:

In ΔOAB,

OB² = OA² + AB²

= (20)² + (20)²

= 20

Radius (r) of circle = 20 cm

Area of quadrant OPBQ =  × 3.14 × (20)²

=  × 3.14 × 800

= 628 cm²

Area of OABC = (Side)²

= (20)²

= 400 cm²

Area of shaded region = Area of quadrant OPBQ - Area of OABC

= (628 - 400)

= 228 cm²

Question 14.

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠ AOB = 30°, find the area of the shaded region

Answer:

Area of the shaded region = Area of sector OAEB - Area of sector OCFD

=  *  * (21)² -  *  * (7)²

=  *  [(21)² – (7)²]

=  *  [(21 – 7) (21 + 7)]

= 

= 8624/84

=  cm²

Question 15.

In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Answer:

As quadrant is 1/4 th of a circle and area of circle is Πr²
Area of quadrant = 

=154 cm²

Area of triangle= 

= 7 × 14

= 98 cm²
In right angle triangle ABC,
BC² = AB² + AC²
BC² = (14)² + (14)²
BC² = 2(14)²
BC= 14 √2 cm
Hence diameter is 14 √2 cm.
Radius is (14 √2)/2 = 7 √2 cm
So, area of the semicircle = 1/2 πr²

= 154 cm²

Area of segment BC = Area of semi-circle - Area of Triangle = 154 - 98 = 56cm²

Area of the shaded region = Area of semi Circle - Area of segment BC = 154 - 56

= 98 cm²

Question 16.

Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each

Answer:

First, let us understand how we can acquire the given Area
From the figure first we need to subtract Area of quadrant from the square and then again subtract another quadrant from the square, by this, we will get the Area of the unshaded region
And finally to find Area of the shaded region we can subtract Area of the unshaded region from Area of square.
Radius of quadrant = length of square 
= r
= 8 cm
We can interpret from above that:
Area of unshaded region = 2(Area of the square - Area of the quadrant of the circle)
Area of shaded region = Area of square - 2(Area of square - Area of quadrant of circle)

Area of shaded region = 2 (Area of quadrant of circle) - Area of square

= 2 × (1/4)πr² - 8 × 8

= [(1/2) × π × 8²] - 64

= 32π - 64

= 32(π-2)

= 32 × 1.14

= 36.54 cm²

Therefore, area of shaded region is 36.54 cm²