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 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., 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.
  AO = 2/3 x AD
  32 = 2/3 x AD
  AD = 48 cm

  In ΔABD,
  AB² = AD² + BD²
  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
(i) quadrant OACB,
(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 shaded region 
= 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