Contents

**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.****Solution:Given: **radius of 1

^{st}circle (R

_{1}) = 19 cm

∴ Circumference of 1

^{st}circle = 2πR

_{1}= 2π(19) cm

Radius of 2

^{nd}circle (R

_{2}) = 9 cm

∴ Circumference of 2

^{nd}circle = 2πR

_{2}= 2π(9) cm

Let radius of 3

^{rd}circle be R

_{3}Circumference of 3

^{rd}circle = 2πR

_{3}According to question,

2πR

_{1}+ 2πR

_{2}= 2πR

_{3}

⇒ 2π(R

_{1}+ R

_{2}) = 2πR

_{3}

⇒ R

_{1}+ R

_{2}= R

_{3}⇒ 19 + 9 = R

_{3}

⇒ R

_{3}= 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.****Solution:Given:** radius of 1

^{st}circle (R

_{1}) = 8 cm

Area of 1

^{st}circle = πR

_{1}

^{2}= π(8)

^{2}cm

^{2}

Radius of 2

^{nd}circle (R

_{2}) = 6 cm

Area of 2

^{nd}circle = πR

_{2}

^{2}= π(6)

^{2}cm

^{2}Let radius of 3

^{rd}circle be R

_{3}Area of 3

^{rd}circle = πR

_{3}

^{2}According to question,

πR,

^{2}+ πR

_{2}

^{2}– πR

_{3}

^{2}⇒ R

_{1}

^{2}+ R

_{2}

^{2}= R

_{3}

^{2 }⇒ (8)

^{2}+ i6)

^{2}– R

_{3}

^{2}⇒ 64 + 36 = R

_{3}

^{2}⇒ R

_{3}= = 10 cm

**Question 3.****The 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.****Solution:**Diameter of the region representing gold score is 21 cm

⇒ Radius of the region representing gold region = cm

**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?****Solution:Given:** diameter of the wheels of the car = 80 cm

⇒ Radius of the wheel of the car = = 40 cm

Circumference of the wheel = 2πr = 2 x x 40 cm

Speed of the car = 66 km/h

Distance covered in 10 minutes = = 11 km

= 11 x 1000 x 100 cm = 11,00,000 cm

**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) n units

(c) 4 units

(d) 7 units**Solution:**

Let radius of the circle = r units

Perimeter of the circle = 2πr

Area of the circle = πr^{2}According to question,

Perimeter of the circle = Area of the circle

⇒ 2πr = πr^{2}⇒ r = 2 units

Hence, option (a) is correct.

**Exercise 12.2**

**Question 1.****Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.****Solution:**Radius of the sector (r) = 6 cm

**Central angle of the sector = 60°**

**Question 2.****Find the area of a quadrant of a circle whose circumference is 22 cm.****Solution:**Let radius of the circle = r

∴ Circumference of the circle = 2πr

According to question,

**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.****Solution:**Length of minute hand of the clock = 14 cm

**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 segment (Use π = 3.14)**Solution:Given:** radius of the circle = 10 cm

Angle subtended by chord at centre = 90°

(i) Area of the minor segment

= Area of the sector OAPB – Area of ΔAOB formed with radius and chord

**Question 5.****In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:**

(i) length of the arc.

(ii) area of the sector formed by the arc.

(iii) area of the segment formed by the corresponding chord.**Solution:**Radius of the circle = 21 cm

Angle at the centre = 60°

**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 √3 = 1.73)****Solution:**

**Question 7.****A chord of a circle of the 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).****Solution:**Radius of the circle = 15 cm

Angle subtended by chord at centre = 60°

Area of the corresponding segment = Area of the sector – Area of Δ formed by radii and chord

= 150 – 62.28 cm² = 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 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)**Solution:(i)** Length of the rope = Radius of the sector grazed by horse = 5 m

Here, angle of the sector = 90°

**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 figure.****Find:**

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.**Solution:**Length of one diameter = 35 mm

Total length of 5 diameters = 5 x 35 mm = 175 mm

**Question 10.****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.****Solution:**Radius of the circle = 45 cm Number of ribs = 8

central angle of the circle

**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.Solution:Given:** length of blade of wiper = radius of sector sweep by blade = 25 cm

Area cleaned by each sweep of the blade = area of sector sweep by blade Angle of the sector formed by blade of wiper =115°

**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)**Solution:**Angle of the sector = 80°

Distance covered = 16.5 km

**Question 13.**

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 ₹0.35 per cm². (Use √3= 1.7)**Solution:** Radius of the cover = 28 cm

∵ There are six equal designs

**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**Solution:**Sector angle is p in degrees

Radius of the circle = R

**Exercise 12.3**

**Question 1.****Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.**

**Solution:**

In ∆PQR, ∠QPR = 90° (angIe in semicircle)

and QR² = PQ² + RP²

(as ∆ QPR is a right angled triangle)

∴ OR² = (24)² + (7)² = 576 + 49

⇒ QR² = 625 cm²

⇒ QR = cm²

⇒ QR=25

Area of the shaded region = Area of the semicircle – Area of the triangle

**Question 2.****Find the area of the shaded region in the given figure, ¡f radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ****∠****AOC = 40°.****Solution:**

∠AOC = 40° (given)

Radius of the sector AOC = 14 cm

**Question 3.Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.**

**Solution:**

ABCD is a square

**Given:**side of the square = 14 cm

Area of the square = (side)

^{2}= (14)

^{2}= 196 cm

^{2 }

Radius of the semicircle APD = (side of square) = x 14 = 7 cm

Similarly, area of the semicircle BPC = 77 cm

^{2}Total area of both the semicircles = 77 + 77 = 154 cm

^{2}Area of the shaded region = Area of square – area of both semicircles

= 196 – 154 = 42 cm

^{2}

**Question 4.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.**

**Solution:**

Area of the equilateral triangle OAB

**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 the figure. Find the area of the remaining portion of the square.**

**Solution:**

side of the square ABCD = 4 cm

Area of the square ABCD = 4 x 4 = 16 cm

^{2}Radius of the quadrant at corner = 1 cm

**Question 6.In a circular table cover of (he radius 32 cm, a design Is formed leaving an equilìtcral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).**

**Solution:**

Radius ofthe circle (r) = 32 m

Area of the circle =πr²

**Question 7.In the 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.**

**Solution:**

Side of the square ABCD = 14 cm

Area of the squat e = (side)

^{2}= 14 x 14 = 196 cm

^{2}

**Question 8.**The given figure 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.

**Solution:**

**Question 9.**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.

**Solution:**

OA = 7 cm

Radius of the semicircle ABC = OA = 7 cm

**Question 10.**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 √ 3 = 1.73205).

**Solution:**

Given:area of an equilateral triangle ABC = 17320.5 cm

Given:

^{2}Let side of the triangle ABC be ‘a’

**Question 11.On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.**

**Solution:**

Radius of the one circular design = 7 cm

Area of the one circular design = πr^{2} = x7x7 = 154 cm²

Now, area of the 9 circular designs = 9 x 154 = 1386 cm^{2}Diameter of the circular design = 7 x 2 = 14 cm

Side of the square = 3(diameter of one circle) = 3 x 14 = 42 cm

Area of the square = 42 x 42 = 1764 cm^{2}Area of the remaining portion of handkerchief

= Area of the square- (Area of the 9 circular designs)

= 1764-1386 = 378 cm^{2}

**Question 12.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.**

**Solution:**

(i) Radius of the quadrant OACB = 3.5 cm

**Question 13.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)Solution:Given:** side of the square OABC = OA = 20 cm

Area of the square — 20 x 20 = 400 cm

^{2}(Diagonal of the square)

^{2}= (side of the square)

^{2}+ (side of the square)

^{2 }(By pythagoras theorem)

Diagonal of the square = x (side of the square)

= x (20) = 20 cm

^{2}

**Question 14.AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ZAOB=30°, find the area of the shaded region.Solution:Given:** ∠AOB = 30°

Radius of the sector AOB = 21 cm

**Question 15.In the figure, 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.**

**Solution:**

Radius of the quadrant of the circle = 14 cm

**Question 16.Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.**

**Solution:**

Side of the square = 8 cm

Area of the square = 8 x 8 = 64 cm

^{2}Radius of the quadrant (formed at vertex) = 8 cm

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