﻿ Class 10 NCERT Math Solution
﻿
TOPICS
Exercise - 12.3

Question-1 :-  Find the area of the shaded region in Fig., if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. (Take π = 22/7).

Solution :-
```  Here, RQ is the diameter of the circle.
Therefore, ∠RPQ will be 90°.
PQ = 24 cm, PR = 7 cm
By applying Pythagoras theorem in ΔPQR,
RP² + PQ² = RQ²
(7)² + (24)² = RQ²
49 + 576 = RQ²
RQ² = 625
RQ = ±25 cm
Radius of circle, OR = RQ/2 = 25/2 cm
Since RQ is the diameter of the circle, it divides the circle in two equal parts.

Area of semi-circle RPQOR = 1/2 x πr²
= 1/2 x 22/7 x (25/2)²
= 11/7 x 625/4
= 6875/28 cm²

Area of ΔPQR = 1/2 x PQ x PR
= 1/2 x 24 x 7
= 84 cm²
Area of shaded region = Area of semi-circle RPQOR − Area of ΔPQR
= 6875/28 - 84
= (6875 - 2352)/28
= 4523/28 cm²
```

Question-2 :-  Find the area of the shaded region in Fig. , if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°. (Take π = 22/7).

Solution :-
```  Radius of inner circle = 7 cm
Radius of outer circle = 14 cm
∠ AOC = 40°

Area of shaded region = Area of sector OAFC − Area of sector OBED
= 40/360 x 22/7 x 14 x 14 - 40/360 x 22/7 x 7 x 7
= 1/9 x 22/7 x 196 - 1/9 x 22/7 x 49
= 1/9 x 22/7(196 - 49)
= 1/9 x 22/7 x 147
= 1/9 x 22 x 21
= 154/3 cm²
```

Question-3 :-  Find the area of the shaded region in Fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles. (Take π = 22/7).

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

Area of each semi-circle = 1/2 x πr²
= 1/2 x 22/7 x 7 x 7
= 11 x 7
= 77 cm²

Area of square ABCD = (Side)² = (14)² = 196 cm²
= 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., 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. (Take π = 22/7).

Solution :-
```  We know that each interior angle of an equilateral triangle is of measure 60°.
Radius of circle = 6 cm

Area of sector OCDE = θ/360 x πr²
= 60/360 x 22/7 x 6 x 6
= 1/6 x 22/7 x 6 x 6
= 22/7 x 6
= 132/7 cm²

Area of ΔOAB = √3/4 x (side)²
= √3/4 x 12 x 12
= √3 x 36
= 36√3 cm²

Area of circle = πr²
= 22/7 x 6 x 6
= 792/7 cm²

Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE
= 36√3 + 792/7 - 132/7
= (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. Find the area of the remaining portion of the square. (Take π = 22/7).

Solution :-
```  Each quadrant is a sector of 90° in a circle of 1 cm radius.

Area of each quadrant = θ/360 x πr²
= 90/360 x 22/7 x 1 x 1
= 1/4 x 22/7
= 11/14 cm²

Area of square = (Side)² = (4)² = 16 cm²
Area of circle = πr² = π (1)² = 22/7 cm²

Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant
= 16 - 22/7 - 4 x 11/14
= 16 - 22/7 - 22/7
= 16 - 44/7
= (112 - 44)/7
= 68/7 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. Find the area of the design. (Take π = 22/7).

Solution :-
```  Radius (r) of circle = 32 cm
AD is the median of ΔABC.

In ΔABD,
AB² = (48)² + (AB/2)²
AB² = 2304 + AB²/4
AB² - AB²/4 = 2304
3AB²/4 = 2304
AB² = (2304 x 4)/3
AB² = 9216/3
AB = 96/√3
By rationalization
AB = 96/√3 x √3/√3
AB = 96√3/3
AB = 32√3 cm

Area of equilateral triangle, ΔABC = √3/4 x (side)²
= √3/4 x 32√3 x 32√3
= 96 x 8 x √3
= 768√3 cm²

Area of circle = πr²
= 22/7 x 32 x 32
= 22/7 x 1024
= 22528/7 cm²

Area of design = Area of circle − Area of ΔABC
= (22528/7 - 768√3) cm²
```

Question-7 :-  In Fig., 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. (Take π = 22/7).

Solution :-
```  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.

Area of each sector = θ/360 x πr²
= 90/360 x 22/7 x 7 x 7
= 1/4 x 22 x 7
= 77/2 cm²

Area of square ABCD = (Side)² = (14)² = 196 cm²
Area of shaded portion = Area of square ABCD − 4 × Area of each sector
= 196 - 4 x 77/2
= 196 - 154
= 42 cm²
```

Question-8 :-  In Fig. depicts a racing track whose left and right ends are semicircular. (Take π = 22/7). 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 :-
```  Distance between the two inner parallel Line segments = 60 m
So, radius is = 60/2 = 30 m
length of segment lines = 106 m

Distance around the track along its inner edge
= AB + arc BEC + CD + arc DFA
= 106 + 1/2 x 2πr + 106 + 1/2 x 2πr
= 212 + 2πr
= 212 + 2 x 22/7 x 30
= 212 + 1320/7
= (1484 + 1320)/7
= 2804/7 m

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)
= (106 x 80 - 106 x 60)
+ (1/2 x 22/7 x 40 x 40 - 1/2 x 22/7 x 30 x 30)
+ (1/2 x 22/7 x 40 x 40 - 1/2 x 22/7 x 30 x 30)
= 106(80 - 60) + 22/7 x 40 x 40 - 22/7 x 30 x 30
= 106 x 20 + 22/7(1600 - 900)
= 2120 + 22/7 x 700
= 2120 + 2200
= 4320 m²
```

Question-9 :-  In Fig., 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. (Take π = 22/7).

Solution :-
```  Radius (r₁) of larger circle = 7 cm
Radius (r₂) of smaller circle = 7/2 cm

Area of smaller circle = πr₁²
= 22/7 x 7/2 x 7/2
= 77/2 cm²

Area of semi-circle AECFB of larger circle = 1/2 x πr₂²
= 1/2 x 22/7 x 7 x 7
= 77 cm²

Ara of ΔABC = 1/2 x AB x OC
= 1/2 x 14 x 7
= 49 cm²

Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC
= 77/2 + 77 - 49
= 28 + 77/2
= 28 + 38.5
= 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.). Find the area of the shaded region. (Use π = 3.14 and 3 = 1.73205).

Solution :-
```  Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm²
√3/4 x a² = 17320.5
1.73205/4 x a² = 17320.5
a² = (4 x 17320.5)/1.73205
a² = 4 x 10000
a = √40000
a = 200 cm

Each sector is of measure 60°.
Area of sector ADEF = θ/360 x πr²
= 60/360 x 3.14 x 100 x 100
= 1/6 x 3.14 x 10000
= 15700/6 cm²

Area of shaded region = Area of equilateral triangle − 3 × Area of each sector
= 17320.5 - 3 x 15700/3
= 17320.5 - 15700
= 1620.5 cm²
```

Question-11 :-  On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief. (Take π = 22/7).

Solution :-
```  Here, side of the square is = 42 cm
Radius of each circular design = 7 cm

Area of square = (Side)²
= (42)²
= 1764 cm²

Area of each circle = πr²
= 22/7 x 7 x 7
= 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., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(ii) shaded region (Take π = 22/7).

Solution :-
```(i) Since OACB is a quadrant, it will subtend 90° angle at O.
Radius of circle = 3.5 cm

Area of quadrant OACB = θ/360 x πr²
= 90/360 x 22/7 x 3.5 x 3.5
= 1/4 x 22 x 0.5 x 3.5
= 1/2 x 11 x 5/10 x 35/10
= 1/2 x 11 x 1/2 x 7/2
= 77/8 cm²

(ii) Area of ΔOBD = 1/2 x OB x OD
= 1/2 x 3.5 x 2
= 3.5
= 35/10
= 7/2 cm²

Area of the shaded region = Area of quadrant OACB − Area of ΔOBD
= 77/8 - 7/2
= (77 - 28)/2
= 49/2 cm²
```

Question-13 :-  In Fig., 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 :-
```  A square OABC is inscribed in a quadrant OPBQ.
OA = 20 cm

In ΔOAB,
OB² = OA² + AB² = (20)² + (20)²
OB² = 2 x (20)²
OB = √2 x 20
OB = 20√2 cm
So, Radius (r) of circle = 20√2 cm

Area of quadrant OPBQ = θ/360 x πr²
= 90/360 x 3.14 x 20√2 x 20√2
= 1/4 x 3.14 x 400 x 2
= 3.14 x 100 x 2
= 3.14 x 200
= 628 cm²

Area of OABC = (Side)² = (20)² = 400 cm²
Area of shaded region = Area of quadrant OPBQ − Area of OABC
= (628 − 400) cm²
= 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.). If ∠ AOB = 30°, find the area of the shaded region. (Take π = 22/7).

Solution :-
```  Two concentric circles of radii AB and CD are 21 cm and 7 cm respectively.
∠ AOB = 30°

Area of the shaded region = Area of sector OAEB − Area of sector OCFD
= 30/360 x 22/7 x 21 x 21 - 30/360 x 22/7 x 7 x 7
= 1/12 x 22/7(441 - 49)
= 1/6 x 11/7 x 392
= 1/6 x 11 x 56
= 308/3 cm²
```

Question-15 :-  In Fig., 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. (Take π = 22/7).

Solution :-
```  As ABC is a quadrant of the circle, ∠BAC will be of measure 90º.

In ΔABC,
BC² = AC² + AB² = (14)² + (14)²
BC² = 2 x (14)²
BC = √2 x 14
BC = 14√2 cm
So, Radius (r1) of semi-circle drawn on BC = 14√2/2 = 7√2 cm

Area of ΔABC = 1/2 x AB x AC
= 1/2 x 14 x 14
= 98 cm²

Area of sector ABDC = θ/360 x πr²
= 90/360 x 22/7 x 14 x 14
= 1/4 x 22 x 2 x 14
= 11 x 14
= 154 cm²

Area of semi circle drawn on BC = 1/2 x πr²
= 1/2 x 22/7 x 7√2 x 7√2
= 1/2 x 22/7 x 98
= 11 x 14
= 154 cm²

= Area of semi-circle - (Area of sector ABDC - Area of ΔABC)
= 154 - (154 - 98)
= 154 - 56
= 98 cm²
```

Question-16 :-  Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each. (Take π = 22/7)

Solution :-
```  The designed area is the common region between two sectors BAEC and DAFC.
Radius of circle = 8 cm

Area of sector BAEC = θ/360 x πr²
= 90/360 x 22/7 x 8 x 8
= 1/4 x 22/7 x 64
= 22/7 x 16
= 352/7 cm²

Area of ΔBAC = 1/2 x BA x BC
= 1/2 x 8 x 8
= 32 cm²

Area of the designed portion = 2 × (Area of segment AEC)
= 2 × (Area of sector BAEC − Area of ΔBAC)
= 2 x (352/7 - 32)
= (2 x 128)/7
= 256/7 cm²
```
CLASSES