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Class 10 Maths Chapter 12 Area Related To Circles

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Exercise 12.1 (Page 230)

Question 1:
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Answer 1:
The radius of the 1^st circle = 19 cm (given)

∴ Circumference of the 1^st circle = 2π×19 = 38π cm
The radius of the 2^nd circle = 9 cm (given)

∴ Circumference of the 2^nd circle = 2π×9 = 18π cm
So,

The sum of the circumference of two circles = 38π+18π = 56π cm
Now, let the radius of the 3^rd circle = R

∴ The circumference of the 3^rd circle = 2πR
It is given that sum of the circumference of two circles = circumference of the 3^rd circle

Hence, 56π = 2πR
Or, R = 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 2:
Radius of 1st circle = 8 cm (given)

∴ Area of 1st circle = π(8)2 = 64π
Radius of 2nd circle = 6 cm (given)
∴ Area of 2nd circle = π(6)2 = 36π

So,

The sum of 1st and 2nd circle will be = 64π+36π = 100π
Now, assume that the radius of 3rd circle = R

∴ Area of the circle 3rd circle = πR2
It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle

Or, πR2 = 100πcm2
R2 = 100cm2
So, R = 10cm

Question 3:
Fig. 12.3 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.

Answer 3:
The radius of 1^st circle, r₁ = 21/2 cm (as diameter D is given as 21 cm)
So, area of gold region = π r₁² = π(10.5)² = 346.5 cm²

Now, it is given that each of the other bands is 10.5 cm wide,
So, the radius of 2^nd circle, r₂ = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2^nd circle − Area of gold region = (πr₂²−346.5) cm²
= (π(21)² − 346.5) cm²
= 1386 − 346.5
= 1039.5 cm²

Similarly,

The radius of 3^rd circle, r₃ = 21 cm+10.5 cm = 31.5 cm
The radius of 4^th circle, r₄ = 31.5 cm+10.5 cm = 42 cm
The Radius of 5^th circle, r₅ = 42 cm+10.5 cm = 52.5 cm

For the area of n^th region,
A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle
= π(31.5)² – 1386 cm²
= 3118.5 – 1386 cm²
= 1732.5 cm²

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle
= π(42)² – 1386 cm²
= 5544 – 3118.5 cm²
= 2425.5 cm²

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

= π(52.5)² – 5544 cm²
= 8662.5 – 5544 cm²
= 3118.5 cm2

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 4:
The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)
So, the circumference of wheels = 2πr = 80 π cm
Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km
Converting km into cm we get,

Distance covered by the car in 1hr = (66×10⁵) cm
In 10 minutes, the distance covered will be = (66×10⁵×10)/60 = 1100000 cm/s

∴ Distance covered by car = 11×10⁵ cm
Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)
=( 11×10⁵)/80 π = 4375.

Question 5:
Tick the correct Solution: 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 5:
Since the perimeter of the circle = area of the circle,
2πr = πr²
Or, r = 2
So, option (A) is correct i.e. the radius of the circle is 2 units.

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Exercise 12.2 (Page 230)

Question 1:
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer 1:
It is given that the angle of the sector is 60°
We know that the area of sector = (θ/360°)×πr²

∴ Area of the sector with angle 60° = (60°/360°)×πr² cm²
= (36/6)π cm²
= 6×22/7 cm² = 132/7 cm²

Question 2:
Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer 2:
Circumference of the circle, C = 22 cm (given)

It should be noted that a quadrant of a circle is a sector which is making an angle of 90°.
Let the radius of the circle = r

As C = 2πr = 22,
R = 22/2π cm = 7/2 cm

∴ Area of the quadrant = (θ/360°) × πr²
Here, θ = 90°
So, A = (90°/360°) × π r² cm²

= (49/16) π cm²
= 77/8 cm² = 9.6 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 3:
Length of minute hand = radius of the clock (circle)
∴ Radius (r) of the circle = 14 cm (given)

Angle swept by minute hand in 60 minutes = 360°
So, the angle swept by the minute hand in 5 minutes = 360° × 5/60 = 30°

We know,

Area of a sector = (θ/360°) × πr2
Now, area of the sector making an angle of 30° = (30°/360°) × πr2 cm2

= (1/12) × π142
= (49/3)×(22/7) cm2
= 154/3 cm2

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 4:

Here AB be the chord which is subtending an angle 90° at the center O.
It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360°)×πr²
= (¼)×(22/7)×10²

Or, Area of minor sector = 78.5 cm²
Also, area of ΔAOB = ½×OB×OA
Here, OB and OA are the radii of the circle i.e. = 10 cm
So, area of ΔAOB = ½×10×10
= 50 cm²

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50
= 28.5 cm²

(ii) Area of major sector = Area of circle – Area of minor sector
= (3.14×10²)-78.5
= 235.5 cm²

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 5:

Given,
Radius = 21 cm
θ = 60°

(i) Length of an arc = θ/360°×Circumference(2πr)
∴ Length of an arc AB = (60°/360°)×2×(22/7)×21
= (1/6)×2×(22/7)×21
Or Arc AB Length = 22cm

(ii) It is given that the angle subtend by the arc = 60°
So, area of the sector making an angle of 60° = (60°/360°)×π r² cm²
= 441/6×22/7 cm²
Or, the area of the sector formed by the arc APB is 231 cm²

(iii) Area of segment APB = Area of sector OAPB – Area of ΔOAB
Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔOAB is an equilateral triangle. So, its area will be √3/4×a² sq. Units.
Area of segment APB = 231-(√3/4)×(OA)²
= 231-(√3/4)×21²
Or, Area of segment APB = [231-(441×√3)/4] 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 √3 = 1.73)

Answer 6:

Given,
Radius = 15 cm
θ = 60°

So,
Area of sector OAPB = (60°/360°)×πr² cm²
= 225/6 πcm²

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°
So, Area of ΔAOB = (√3/4) ×a²
Or, (√3/4) ×15²

∴ Area of ΔAOB = 97.31 cm²
Now, area of minor segment APB = Area of OAPB – Area of ΔAOB
Or, area of minor segment APB = ((225/6)π – 97.31) cm² = 20.43 cm²

And,

Area of major segment = Area of circle – Area of segment APB
Or, area of major segment = (π×15²) – 20.4 = 686.06 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 7:

Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

So, AD = DB

Now, the area of the minor sector = (θ/360°)×πr²
= (120/360)×(22/7)×12²
= 150.72 cm²

Consider the ΔAOB,
∠ OAB = 180°-(90°+60°) = 30°
Now, cos 30° = AD/OA
√3/2 = AD/12
Or, AD = 6√3 cm

We know OD bisects AB. So,
AB = 2×AD = 12√3 cm
Now, sin 30° = OD/OA
Or, ½ = OD/12
∴ OD = 6 cm

So, the area of ΔAOB = ½ × base × height
Here, base = AB = 12√3 and
Height = OD = 6

So, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm²
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
= 150.72 cm²– 62.28 cm² = 88.44 cm²

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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 8:
As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with θ = 90°) of the field with radius 5 m.

Here, the length of rope will be the radius of the circle i.e. r = 5 m
It is also known that the side of square field = 15 m

(i) Area of circle = πr² = 22/7 × 5² = 78.5 m²
Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle) = 78.5/4 = 19.625 m²

(ii) If the rope is increased to 10 m,
Area of circle will be = πr² =22/7×10² = 314 m²
Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle)
= 314/4 = 78.5 m²

∴ Increase in the grazing area = 78.5 m² – 19.625 m² = 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 9:
Diameter (D) = 35 mm
Total number of diameters to be considered= 5

Now, the total length of 5 diameters that would be required = 35×5 = 175
Circumference of the circle = 2πr
Or, C = πD = 22/7×35 = 110

Area of the circle = πr²
Or, A = (22/7)×(35/2)² = 1925/2 mm²

(i) Total length of silver wire required = Circumference of the circle + Length of 5 diameter
= 110+175 = 185 mm

(ii) Total Number of sectors in the brooch = 10

So, the area of each sector = total area of the circle/number of sectors
∴ Area of each sector = (1925/2)×1/10 = 385/4 mm²

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 10:
The radius (r) of the umbrella when flat = 45 cm
So, the area of the circle (A) = πr² = (22/7)×(45)² =6364.29 cm²

Total number of ribs (n) = 8
∴ The area between the two consecutive ribs of the umbrella = A/n
6364.29/8 cm²

Or, The area between the two consecutive ribs of the umbrella = 795.5 cm²

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 11:
Given,
Radius (r) = 25 cm
Sector angle (θ) = 115°
Since there are 2 blades,
The total area of the sector made by wiper = 2×(θ/360°)×π r²

= 2×(115/360)×(22/7)×25²
= 2×158125/252 cm²
= 158125/126 = 1254.96 cm²

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 12:
Let O bet the position of Lighthouse.

Here the radius will be the distance over which light spreads.
Given, radius (r) = 16.5 km
Sector angle (θ) = 80°

Now, the total area of the sea over which the ships are warned = Area made by the sector
Or, Area of sector = (θ/360°)×πr²

= (80°/360°)×πr² km²
= 189.97 km²

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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 ₹ 0.35 per cm² . (Use √3 = 1.7)

Answer 13:

Total number of equal designs = 6
AOB= 360°/6 = 60°
Radius of the cover = 28 cm
Cost of making design = ₹ 0.35 per cm²

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be (√3/4)×a² sq. units

Here, a = OA
∴ Area of equilateral ΔAOB = (√3/4)×28² = 333.2 cm²

Area of sector ACB = (60°/360°)×πr² cm²
= 410.66 cm²

So, area of a single design = area of sector ACB – area of ΔAOB
= 410.66 cm² – 333.2 cm² = 77.46 cm²

∴ Area of 6 designs = 6×77.46 cm² = 464.76 cm²
So, total cost of making design = 464.76 cm² ×Rs.0.35 per cm²
= Rs. 162.66

Question 14:
Tick the correct solution in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR
(B) p/180 × π R²
(C) p/360 × 2πR
(D) p/720 × 2πR²

Answer 14:
The area of a sector = (θ/360°)×πr²
Given, θ = p
So, area of sector = p/360×πR²

Multiplying and dividing by 2 simultaneously,
= (p/360)×2/2×πR²
= (2p/720)×2πR²

So, option (D) is correct.

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Exercise 12.3 (Page 234)

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 1:
Here, P is in the semi-circle and so,
P = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.
∴ QR = D

Using Pythagorean theorem,
QR² = PR²+PQ²
Or, QR² = 7²+24²
QR= 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm
Now, the area of the semicircle = (πR²)/2

= (22/7)×(25/2)×(25/2)/2 cm²
= 13750/56 cm² = 245.54 cm²

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 cm²
= 84 cm²
Hence, the area of the shaded region = 245.54 cm²-84 cm²

= 161.54 cm²

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 2:

Given,
Angle made by sector = 40°,
Radius the inner circle = r = 7 cm, and
Radius of the outer circle = R = 14 cm

We know,
Area of the sector = (θ/360°)×πr²
So, Area of OAC = (40°/360°)×πr² cm²
= 68.44 cm²

Area of the sector OBD = (40°/360°)×πr² cm²
= (1/9)×(22/7)×7² = 17.11 cm²

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD
= 68.44 cm² – 17.11 cm² = 51.33 cm²

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 3:
Side of the square ABCD (as given) = 14 cm
So, Area of ABCD = a²
= 14×14 cm² = 196 cm²

We know that the side of the square = diameter of the circle = 14 cm
So, side of the square = diameter of the semicircle = 14 cm
∴ Radius of the semicircle = 7 cm

Now, area of the semicircle = (πR²)/2
= (22/7×7×7)/2 cm² =
= 77 cm²

∴ Area of two semicircles = 2×77 cm² = 154 cm²
Hence, area of the shaded region = Area of the Square – Area of two semicircles

= 196 cm² -154 cm²
= 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 4:
It is given that OAB is an equilateral triangle having each angle as 60°
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = (√3/4) (OA)²= (√3/40×12² = 36√3 cm²
Area of the circle = πR² = (22/7)×6² = 792/7 cm²
Area of the sector making angle 60° = (60°/360°) ×πr² cm²

= (1/6)×(22/7)× 6² cm² = 132/7 cm²

Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
= 36√3 cm² +792/7 cm²-132/7 cm²
= (36√3+660/7) 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 5:
Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)²= 4² = 16 cm²
Area of the quadrant = (πR²)/4 cm² = (22/7)×(1²)/4 = 11/14 cm²

∴ Total area of the 4 quadrants = 4 ×(11/14) cm² = 22/7 cm²

Area of the circle = πR² cm² = (22/7×1²) = 22/7 cm²
Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)

= 16 cm²-(22/7) cm²+(22/7) cm²
= 68/7 cm²

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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.

Answer 6:
Radius of the circle = 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

⇒ (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 ΔADB = √3/4 ×(32√3)² cm² = 768√3 cm²
Area of circle = πR² = (22/7)×32×32 = 22528/7 cm²
Area of the design = Area of circle – Area of ΔADB

= (22528/7 – 768√3) cm²

Question 7:
In Fig. 12.25, 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.

Answer 7:
Side of square = 14 cm
Four quadrants are included in the four sides of the square.

∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 14² = 196 cm²
Area of the quadrant = (πR²)/4 cm² = (22/7) ×7²/4 cm²

= 77/2 cm²

Total area of the quadrant = 4×77/2 cm² = 154cm²
Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm² – 154 cm²
= 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 8:
Width of the track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m

DE = CF = 60 m
Radius of inner semicircle, r = OD = O’C
= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B
= 30+10 m = 40 m

Also, AB = CD = EF = GH = 106 m
Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)

= 106+106+(2×πr) m = 212+(2×22/7×30) m
= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)
= (AB×CD)+(EF×GH)+2×(πr²/2) -2×(πR²/2) m²
= (106×10)+(106×10)+2×π/2(r²-R²) m²
= 2120+22/7×70×10 m²
= 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 9:
Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = (½)×7×14 = 49 cm²
Area of larger circle = πR² = (22/7)×7² = 154 cm²
Area of larger semicircle = 154/2 cm² = 77 cm²
Area of smaller circle = πr² = (22/7)×(7/2)×(7/2) = 77/2 cm²
Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area of smaller circle
Area of the shaded region = (154-49-77+77/2) cm²

= 133/2 cm² = 66.5 cm²

Question 10:
The area of an equilateral triangle ABC is 17320.5 cm². 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 Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Answer 10:
ABC is an equilateral triangle.
∴ ∠ A = ∠ B = ∠ C = 60°

There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm²

⇒ √3/4 ×(side)² = 17320.5
⇒ (side)² =17320.5×4/1.73205
⇒ (side)² = 4×10⁴
⇒ side = 200 cm

Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°)×π r² cm²

= 1/6×3.14×(100)² cm²
= 15700/3cm²

Area of 3 sectors = 3×15700/3 = 15700 cm²
Thus, area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors

= 17320.5-15700 cm² = 1620.5 cm²

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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 11:
Number of circular designs = 9
Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.
∴ Side of the square = 3×diameter of circle = 3×14 = 42 cm

Area of the square = 42×42 cm² = 1764 cm²
Area of the circle = π r² = (22/7)×7×7 = 154 cm²
Total area of the design = 9×154 = 1386 cm²

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design = 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 12:
Radius of the quadrant = 3.5 cm = 7/2 cm

(i) Area of quadrant OACB = (πR²)/4 cm²

= (22/7)×(7/2)×(7/2)/4 cm²
= 77/8 cm²

(ii) Area of triangle BOD = (½)×(7/2)×2 cm²
= 7/2 cm²

Area of shaded region = Area of quadrant – Area of triangle BOD

= (77/8)-(7/2) cm² = 49/8 cm²
= 6.125 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 13:
Side of square = OA = AB = 20 cm
Radius of the quadrant = OB

OAB is right angled triangle
By Pythagoras theorem in ΔOAB,

OB² = AB²+OA²

⇒ OB² = 20² +20²
⇒ OB² = 400+400
⇒ OB² = 800
⇒ OB= 20√2 cm

Area of the quadrant = (πR²)/4 cm² = (3.14/4)×(20√2)² cm² = 628cm²
Area of the square = 20×20 = 400 cm²
Area of the shaded region = Area of the quadrant – Area of the square

= 628-400 cm² = 228cm²

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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 14:
Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles = 30°

Area of the larger sector = (30°/360°)×πR² cm²

= (1/12)×(22/7)×21² cm²
= 231/2cm²

Area of the smaller circle = (30°/360°)×πr² cm²

= 1/12×22/7×7² cm²
=77/6 cm²

Area of the shaded region = (231/2) – (77/6) cm²
= 616/6 cm² = 308/3cm²

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 15:
Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.

By Pythagoras theorem in ΔABC,
BC² = AB² +AC²

⇒ BC² = 14² +14²
⇒ BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm
Area of ΔABC =( ½)×14×14 = 98 cm²
Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm²
Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm²
Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant
= 154 +98-154 cm² = 98cm²

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 16:
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = (½)×8×8 = 32 cm²
Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×8²

= 352/7 cm²

Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)
= (352/7 -32)+(352/7- 32) cm²
= 2×(352/7-32) cm²
= 256/7 cm²

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