Chapter 12 Areas Related to Circles

12 Areas Related to Circles Chapter In the Chapter In this chapter, you will be studying the following points: Circumference of a circle = 2 π r. O Area of a circle = π r2. O O Length of an arc of a sector of a circle with radius r and angle with degree measure θ is = O Area of a sector of a circle with radius r and angle with degrees measure θ is = O θ 360° θ 360° × π r2. × π r2. Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle. TEXT BOOK QUESTION (SOLVED) EXERCISE 12.1 Q.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. Ans. Let the radius of the bigger circle be R cm and radii of two smaller circles are r1 and r2, then according to question. 2 πR = 2πr1 + 2πr2 ⇒ 2 πR = 2π (19) + 2π (9) ⇒ 2 πR = 2π (19 + 9) ⇒ R = 28 Hence, the radius of the circle is 28 cm Q.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. Ans. Let the radius of the required circle be R and radii of two given circle be r1 and r 2, then according to question, 2 πR2 = 2πr12 + 2πr22 ⇒ 2 πR2 =2π (r12 + r22) ⇒ R2 = 64 + 36 ⇒ ⇒ R2 = 100 R = 10 cm Q.3. In Fig. 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. Ans. We have, r = Radius of the region representing Gold score = 10.5 cm r1 = Radius of the region representing Gold and Red scoring areas = (10.5 + 10.5) cm = 21cm = 2r cm r2 = Radius of the region representing Gold, Red and Blue scoring areas = (21+10.5) cm = 31.5 cm = 3r cm (217) 218 r3 = r4 = = = Now, A1 = Radius of the region representing Gold, Red, Blue and Black scoring areas (31.5 + 10.5) cm = 42 cm = 4r cm Radius of the region representing Gold, Red, Blue, Black and white scoring areas (42 + 10.5) cm = 52.5 cm = 5r cm Area of the region representing Gold scoring area 22 22 × (10.5)2 = × 10.5 × 10.5 7 7 = 22 × 1.5 × 10.5 = 346.5 cm2 A 2 = Area of the region representing Red scoring area = π (2r)2 – πr2 = 3πr2 = 3A1 = 3 × 346.5 cm2 = 1039.5 cm2 A 3 = Area of the region representing Blue scoring area = π(3r)2 – π(2r)2 = 9πr2 – 4πr2 = 5πr2 = 5A1 = 5 × 346.5 cm2 = 1732.5 cm2 A 4 = Area of the region representing Black scoring area = π(4πr)2 – π(3r)2 = 7πr2 = 7A1 = 7 × 346.5 cm2 = 2425.5 cm2 A 5 = Area of the region representing White scoring area = π(5r)2 – π(4r)2 = 9πr2 = 9A1 = 9 × 346.5 cm2 = 3118.5 cm2 Q.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 = πr2 = at a speed of 66 km per hour? Ans. Given, diameter of wheel d1 = 80 cm d1 80 = = 40 cm 2 2 ∴ Circumference of wheel of a car = ∴ Radius, r = 22 1760 × 40 = cm 7 7 Circumference of circle = 2πr 2πr = 2 × 66× 1000 m min–1 60 = 1100m min–1 = 110000 cm min–1 ∴ Wheel of car moves in 1 min = 110000 cm ∴ Wheel of car moves in 10 min = 1100000 cm ∴ Number of complete revolution Since speed of car = 66 kmh–1 = 1100000 7700000 = 1760 / 7 = 1760 = 4375 Hence, the wheel make complete revolution in 10 min is 4375. Q.5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numeri cally equal, then the radius of the circle is (A) 2 units (B) p units (C) 4 units (D) 7 units Ans. (a) Area of circle = πr2 and area of perimeter = 2πr According to the given condition, Area of circle = Area of perimeter ∴ πr2 = 2πr ⇒ r = 2 units EXERCISE 12.2 Q.1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. Ans. Here, we have r = 6 cm and θ = 60. We know that, Area of sector = π r × 2 θ 360° 60° ⎞ ⎛ 22 ⎟ cm2 = ⎜ × 6× 6× 360° ⎠ ⎝ 7 = 132 cm2 = 18.86 cm2 7 Q.2. Find the area of a quadrant of a circle whose circumference is 22 cm. Ans. Since, circumference of circle = 22 cm ⇒ 2π r = 22 ⇒ 2× 22 r = 22 7 7 cm 2 ∴ Area of a quadrant of a circle = ∴ Radius of circle, r = 2 ⎛7⎞ ⎜ ⎟ 2 π r 22 11 49 77 cm2 2 = = ×⎝ ⎠ = × = 4 7 4 14 4 8 219 Q.3. The length of the minute hand of a clock is 14 cm. F ind the area swept by the minute hand in 5 minutes. Ans. The angle described by the minute hand in 1 miunute = 66°. ∴ The angle described by 1 minute hand in 5 minutes = 30°. i.e., θ = 30° and r = 14 cm Now, area swept by the minute hand 2 =πr × θ 360° 30° ⎞ ⎛ 22 ⎟ cm2 = ⎜ × 14 × 14 × 7 360 °⎠ ⎝ = 51.3 cm2 Q.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) Ans. Here, we have r = 10 cm, and θ = 90° Area of sector (OAPBO) =πr × 2 θ 360° 90° = 3.14 ×10 ×10 × 360° = 3.14 × 5 × 5 = 78.5 cm2 Area of ΔAOB = 1 2 1 r sin θ = ×10 ×10 × sin 90° 2 2 = 1 × 10 ×10 × 1 = 50 cm2 2 Now, (i) Let APBA is the given minor segment, then Area of minor segment (APBA) = Area of sector (OAPBO) – area of Δ(AOB) = 78.5 cm2 – 50 cm2 = 28.5 cm2 (ii) Area of major sector = Area of circle – Area of sector OAPBO = (πr2 – 78.5) cm2 = (3.14 × 10 × 10 – 78.5) cm2 = (314 – 78.5) cm2 = 235.5 cm2 Q.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 Ans. Here, radius (r) = 21 cm angle (θ) = 60° (i) ∴ The length of arc (l) = π rθ 180° 22 × 21× 60° = 7 cm 180° = 22 cm (ii) Area of sector = 1 ×l × r 2 1 × 22 × 21 cm2 2 = 231 cm2 (iii) Area of required segment = 1 ⎛ πθ ⎞ − sin θ ⎟ = r2 ⎜ ⎝ 360° 2 ⎠ 60 1 ⎞ 2 ⎛ 22 − sin 60° ⎟ cm2 = (21) ⎜ × 7 360 ° 2 ⎝ ⎠ ⎛ 11 1 3 ⎞⎟ ⎜ = 441 ⎜ 21 − 2 × 2 ⎟ cm2 ⎝ ⎠ ⎛ 441 3 ⎞⎟ ⎜ = ⎜ 21×11 − 4 ⎟ cm2 ⎝ ⎠ ⎛ 441 3 ⎞⎟ ⎜ = ⎜ 231 − 4 ⎟ cm2 ⎝ ⎠ Q.6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Fi nd the areas of the corresponding minor and major segments of the √3 = 1.73) circle. (Use π = 3.14 and√3 Ans.Here, we have r = 15 cm, θ = 60° 220 Let OACBO be the given sector and OAB is a triangle. Then (i) Area of the sector (OACBO) 2 =πr × θ 360° 60 ⎞ ⎛ ⎟ cm2 = ⎜ 3.14 ×15 ×15 × 360° ⎠ ⎝ = 117.75 cm2 (ii) Area of the triangle (AOB) = 1 2 r sin θ 2 = 1 3 × 15 × 15 × 2 2 = 15 ×15 ×1.73 = 97.313 cm2 4 Now, Area of minor segment (ACBA) = Area of sector (OACBO) – Area of triangle (AOB) = (11.75 – 97.313) cm2 = 20.437 cm2 and Area of major segment (ABDA) = Area of circle – Area of minor segment = (πr2 – 20.437) cm2 = (3.14 × 15 × 15 – 20.437)cm2 = 686.063 cm2 Q.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) Ans. Here, radius (r) = 12 cm angle (θ) = 120° ∴ Required area of minor segement 1 ⎞ 2 ⎛ πθ − sin θ ⎟ =r ⎜ 360 ° 2 ⎝ ⎠ ⎞ 2 ⎛ 3.14 ×120 1 − sin 120° ⎟ cm2 = (12) ⎜ 360 ° 2 ⎝ ⎠ ⎛ 3.14 1 3 ⎞⎟ ⎜ = 144 ⎜ 3 − 2 × 2 ⎟ cm2 ⎝ ⎠ = 144 × 3.14 × 4 − 3 3 2 cm 12 = 12 × [12.56 – 3 × 1.73] cm2 = 12 × [12.56 – 5.19] cm2 = 12 × 7.37 cm2 = 88.44 cm2 Q.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. ). 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) Ans. Here, θ = 90° r = length of rope = 5. (i) The are aof part of field horse can graze = Area of sector = πr 2θ = 3.14 × 360° = 19.625m2 (ii) Now radius (r) = 10 m 5 × 5 × 90 360° 3.14 ×10 ×10 × 90 360° 2 = 78.5 m Increase in grazing area = 78.5m2 – 19.625 m2 = 58.875 m2 Q.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. Find : (i) the total length of the silver wi re required. (ii) the area of each sector of the brooch. Ans. (i) Total length of the silver wire required = circumference of the circle + length of 5 diameters which divide the circle into 10 equal sectors. = 2 πr + 5 × 2r grazed area = 35 ⎞ ⎛ 22 35 = ⎜ 2 × × + 5 × 2 × ⎟ mm 7 2 2 ⎠ ⎝ = (110 + 175) mm = 285 mm 221 (ii) The area of each sector of the brooch = = of the blades 1 (area of circle) 10 2 =πr × 360° 115° ⎞ ⎛ 22 ⎟ cm2 = ⎜ × 25 × 25 × 360° ⎠ ⎝ 7 1 × πr2 10 ⎛ 1 22 35 35 ⎞ = ⎜ × × × ⎟ mm2 ⎝ 10 7 2 2 ⎠ 11× 35 385 = mm2 4 4 = 96.25 mm2 Q.10. An umbrella has 8 ribs which are equally spaced (see Fig.). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Ans. Total nuimber of ribs = 8 Therefore, angle (θ) made by two consecutive = 360° ribs of the centre = = 45° 8 i.e., θ = 45° We have, r = 45 cm and θ = 45° Therefore, the area between the two consecutive ribs of the umbrella 2 =πr × θ θ 360° 45 ⎞ ⎛ 22 ⎟ cm2 = ⎜ × 45 × 45 × 360° ⎠ ⎝ 7 ⎛ 1 22 ⎞ = ⎜ × × 45 × 45 ⎟ cm2 ⎝8 7 ⎠ 2 = 795.53 cm Q.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. Ans. Clearly, the length of the wiper will be radius and angle sweeping by the blades is θ. i.e., r = 25 cm, and θ = 115°. Therefore, the total area cleaned at each sweep ⎛ 625 × 253 ⎞ 158125 = ⎜⎜ 7 × 36 ⎟⎟ cm2 = cm2 252 ⎝ ⎠ = 627.45 cm2 Area swept by two vipers = 2 × 627.45 = 1254.90 cm2 Q.12. To warn ships for underwater rocks, a lighthouse spreads a red coloured 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) Ans. Clearly, distance of red coloured light which is thrown by lighthouse is radius (r) and area over light house is thrown by sector angle (θ) i.e., r = 16.5 km and θ = 80°. Now, area of the sea over which the ships are warned. 2 =πr × θ 360° 165 165 80° ⎞ ⎛ × × ⎟ km2 = ⎜ 3.14 × 10 10 360° ⎠ ⎝ 683892 km2 360 = 189.97 km2 Q.13. A round table cover has six equal designs as shown in Fig. 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 √3 = 1.7) Ans. Area of 1 design = Area of sector AOBC – Area of equilatral ΔAOB = ⎛ 60° 22 ⎞ 3 2⎟ ⎜ = ⎜ 360° × 7 × 28 × 28 – 4 × (28) ⎟ cm2 ⎝ ⎠ 222 = [410.6 – 33.2] cm2 = Q. 14. Tick the correct answer in the following: Area of a sector of angle p (in degrees ) of a circle with radius R is 232.4 2 cm 3 ⎛ 232.4 ⎞ ⎟ cm2 Total area covered by design = ⎜ 6 × 3 ⎠ ⎝ = 464.8 cm2 Cost of making the design = Rs. (0.35 × 464.8) = Rs. 162.68 (a) p × 2πR 180 (b) p × πR 2 180 (c) p × 2πR 360 (d) p × 2πR 2 720 Ans. (d) p × 2πR 2 720 EXERCISE 12.3 Q.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. Q.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°. Ans. Here, r1 = (OA) = 14 cm r2 = (OB) = 7 cm θ = 40° Ans. Here, PQ = 24 cm, PR = 7 cm ∠ QPR = 90° (QR is diameter) RQ2 = PQ2 + PR2 = (24)2 + (7)2 = 576 + 49 = 625 ⇒ RQ = Area of sector AOC = = RQ 25 Radius of circle = OQ = = cm 2 2 360° Area of sector BOD = 40 22 154 × × 7×7 = cm2 360° 7 9 ⎛ 616 154 ⎞ − ⎟ cm2 Area of shaded region BDCA = ⎜ 9 ⎠ ⎝ 9 1 2 1 22 25 25 Area of semi-circle = πr = × × × 2 2 7 2 2 = 6875 cm2 28 ⎛ 6875 ⎞ – 84 ⎟ cm2 ∴ Area of shaded region = ⎜ 28 ⎝ ⎠ = 4523 cm2 28 ×π r 2 40 22 616 × ×14 ×14 cm2 = cm2 360° 7 9 625 = 25 cm 1 1 Area of ΔQPR = × PQ × PR = × 24 × 7 = 84 cm2 2 2 θ Q.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. = 462 cm2 9 = 154 cm2 3 223 Ans. Area of square = (side)2 = (14)2cm2 = 196cm2 Area of 2 semi-circle = 2 × 1 2 πr 2 1 22 14 14 = 2 × × × × cm2 2 7 2 2 = 154 cm2 Area of shaded region= (196 – 154) cm2 = 42cm2 Q.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. Now, required area = Area of ΔOAB + Area of circle – Area of sector 792 132 ⎞ ⎛ − ⎟ cm2 = ⎜ 36 3 + 7 7 ⎠ ⎝ 660 ⎞ ⎛ ⎟ cm2 = ⎜ 36 3 + 7 ⎠ ⎝ Q.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. Ans. Area of remaining portion of square = Area of square – (Area of 4 quadrants of circle + Area of circle) 1 22 22 2 ⎛ 2 2⎞ = ( 4) − ⎜ 4 × × × (1) + × (1) ⎟ 7 ⎝ 4 7 ⎠ Ans. It is given that side of equilateral triangle =12 cm Therefore, area = ⎛ 22 22 ⎞ = 16 − ⎜ + ⎟ 7 ⎠ ⎝ 7 3 side2 4 ⎛ 3 ×12 ×12 ⎞ ⎜ ⎟ 2 =⎜ ⎟ cm 4 ⎝ ⎠ = 36√ 3cm2 Let radius of the circle be r, then r = 6 cm Therefore, area = πr2 22 792 ×6× 6 = cm2 7 7 Now, Area of a sector of a circle of radius 6 cm and of angle 60°. = 2 =πr × θ 360° [Triangle is equilateral, so θ = 60°] 60° ⎞ ⎛ 22 ⎟ = ⎜ ×6× 6 = 360° ⎠ ⎝ 7 132 = cm2 7 44 68 = cm2 7 7 Q.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. Ans. Now, area of shaded part = Area of circle – Area of ΔABC = 16 – = πr2 – 3 × = 1 OB × OC × sin 120° 2 22 1 3 × 32 × 32 – × 32 × 32 × 7 2 2 ⎛ 22528 ⎞ − 768 3 ⎟ cm2 =⎜ ⎝ 7 ⎠ 224 Q.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. Ans. Area of the shaded region = Area of the sqaure of side 14 cm– 4[Area of a sector of centre angle 90°] 90° ⎛ 14 ⎞ π⎜ ⎟ = 14 × 14 – 4 × 360° ⎝ 2 ⎠ 2 22 2 (7) 7 = 196 – 154 = 42 cm2 Q.8. Fig. 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. = 196 – = 2120 + 22 (40 + 30) (40 – 30) 7 22 × 70 × 10 7 = 2120 + 2200 = 4320 m2 Hence, the area of track is 4320 m2 Q.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. Ans. Area of shaded region = Area of small circle + [Area of semi-circle (with centre O) – Area of ΔABC] = 2120 + ⎛7⎞ ⎝2⎠ 2 1 ⎛ 1 22 ⎞ × 7 × 7 − ×14 × 7 ⎟ 2 ⎝2 7 ⎠ = π ×⎜ ⎟ + ⎜ × 22 49 × + [77 – 49] 7 4 = 38.5 + 28 = 66.5 cm2 Q.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) = Ans. We have, The distance between the two inner parallel line segments = 60 m. Length of each parallel lines = 106 m Width of track = 10 m (i) The distance around the track along its inner edge = sum of length of inner parallel lines + 2( semi perimeter of inner circular ends) = (106 + 106) + 2(π × 30) 22 × 30 = 212 + 188.57 = 405.57 cm 7 (ii) OD = 30 m and Ad = 10 m, OA = OD + DA = 30 + 10 = 40 m Area of the track = Area of rectangle ABCD + Area of rectangle EFGH + 2 (Area of semicircle with radius 40 m – Area of semi-circle with radius 30m) = 212 + 2 × ⎛1 2 1 2⎞ = 106 × 10 + 106 × 10 + 2 ⎜ π (40) − π (30) ⎟ 2 ⎝2 ⎠ = 1060 + 1060 + π [(40)2 – (30)2] Ans. Let the side of the equilateral triangle be 'a' cm, then, Area = 17320.5 cm2 ⇒ 3 2 a = 17320.5 4 225 [Area = ⇒ a2 = 3 (side)2] 4 17320.5 × 4 3 17320 .5 × 4 = 40000 1.73205 ⇒ a = 200 Thus radius of each sctor is 100 cm Now required Area = Area of ΔABC – 3 × Area of sector with central angle 60° and r = 100 cm ⇒ a2 = ⎡ ⎛ 60 ⎞⎤ = ⎢17320.5 – 3 ⎜ 360 × 3.14 ×100 ×100 ⎟⎥ cm2 ⎝ ⎠⎦ ⎣ = (17320.5 – 15700) cm2 = 1620.5 cm3 Q.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. Ans. Radius of a circle = 7 cm Length of side of the sqaure = 3(diameter) = 3(14) = 42 cm Area of the square = side2 = 422 = 1764 cm2 22 × (7)2 = 154 7 Area of a circle = 9 × 154 = 1386 cm2 Remaining portion (without design)= Area of the handkerchief – Area of design = 1764 – 1386 = 378 cm2 Hence, the area of the remaining portion is 378 cm2. ≅CB is a quadrant of a circle Q.12. In Fig. , OA≅ with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region. Area of one circle = πr2 = Ans. (i) Area of quadrant OACB π r2 1 22 = × × (3.5) 2 cm2 4 4 7 = 9.625 cm2 = (ii) And, 1 × 3.5 × 2 cm2 = 3.5 cm2 2 ∴ Area of the shaded portion = 9.625 cm2 – 3.5 cm2 = 6.125 cm2 Q.13. In Fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of t he shaded region. (Use π = 3.14) Area of ΔBOD = Ans.Since OABC is a square ∴ Diagonal of a square = √2 × side = √2 × 20 cm [OA (side) = 20 cm] ∴ Radius of circle, r = 20√2 ∴ Area of quadrant = π r2 4 = 3.14 × (20 2 ) 2 4 3.14 × 800 = 628 cm2 4 Now,Area of square = (OA)2 = (20)2 = 400 cm2 ∴ Area of shaded region = Area of quadrant – Area of square = 628 – 400 = 228 cm2 = 226 Since, BC is a diameter for semicircle BCQB ∴ Therfore, radius (r) = 7 2 cm Now, area of semicircle (BCQB) Q.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. Ans. Given, ∠COD = 30° r1 = OC = 7 cm, and r2 = OB = 21 cm Area of sector COD = = 360 θπr12 360 = 30 22 × × (7 ) 2 360 7 = 11× 63 693 = = 115.5 cm2 6 6 Area of shaded region = Area of sector OAB – Area of sector COD = 115.5 – 12.83 = 102.67 cm2 = 308 2 cm 3 Q.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. = Ans. Let points P and Q on arcs BPC and BQC respectively with BC as diameter Now in ΔABC, we have BC2 = AB2 + AC2 (Using pythagoras theorem) ⇒ BC2 = 142 + 142 BC = 196 + 196 ⇒ BC = 14 2 cm 2 22 1 × ×7 2 ×7 2 7 2 = 152 cm2 Area of sector (ACPBA) = 1 2 πr 4 1 22 × ×14 ×14 4 7 = 154 cm2 = 30 22 × × (21) 2 360 7 22 × 3 × 21 12 ⇒ π r2 = 22 × 7 77 = = 12.83 cm2 12 6 Area of sector OAB = = θπr12 = And, ar (ΔACB) = 1 × AC × AB 2 1 × 14 × 14 = 98 cm2 2 Hence, required area = [ar (ΔABC) +ar (BCQB)] – ar [(ACPBA)]cm2 = [(98 + 154) – (154)] cm2 = 98 cm2 Q.16. Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each. = Ans. Area of designed region = 2 [ Area of quadrant – Area of triangle] 1 ⎛ 1 22 ⎞ = 2 ⎜ × × 8× 8 − × 8× 8 ⎟ 2 ⎝4 7 ⎠ ⎛ 352 ⎞ − 32 ⎟ = 2⎜ ⎝ 7 ⎠ = 256 cm2 7 227 Additional Questions Q.1. How many times will the wheel of a car rotate in journey of 88 km, if it is given that the diameter of the wheel is 56 cm? (Take π = 22/7]. Ans. Diameter of wheel = 56 cm 56 = 28 cm 2 ∴ Circumference of wheel ∴ r= = 2πr ⎛ 22 ⎞ = ⎜ 2 × × 28 ⎟ cm = 176 cm 7 ⎝ ⎠ In 1 revolution, wheel will cover distance = 176 cm Total distance = 88 km = 88 × 1000 × 100 cm 88 ×1000 × 10 = 50,000 176 Hence, the number of times the wheel will rotate is 50,000. Q.2. A wire is in the form of a circle of radius 42 cm. It is bent into a square. Determine the side of the square. Ans. Radius of the circle = 42 cm Length of the wire = circumference of the circle ∴ Number of revolutions = 22 = 2π r = 2 × × 42 cm = 264 cm 7 Let the side of the square be x cm. Now, the wire is bent into a square. ∴ Perimeter of the square = Length of the wire ⇒ 4x = 264 ⇒ x = 66 Hence, the length of the side of the square is 66 cm. Q.3.If the perimeter of a semi-circular protractor is 36 cm, find its diameter. (Take π = 22/ 7] Ans. Let r be the radius of the protractor, then Perimeter = πr + 2r ⇒ 36 = r (π + 2) ⇒ ⇒ ⇒ 36 = 36 = ⎛ 22 ⎞ r ⎜ + 2⎟ ⎝ 7 ⎠ ⎛ 22 + 14 ⎞ 36r ⎟= r⎜ 7 ⎝ 7 ⎠ ⎛ 36 × 7 ⎞ ⎟ cm = 7 cm =⎜ ⎝ 36 ⎠ ∴ Diameter (d) = 2r = (2 × 7) cm = 14 cm Hence, the diameter is 14 cm. r Q.4.Find the radius of a circle where circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm. Ans. For first circle, r = 15 cm. ∴ Its circumference = 2π (15) = 30π For second circle r = 18 cm ∴ Its circumference = 2π (18) = 36π Let R be the radius of the required circle, ∴ 2πR = 30π + 36π ⇒ 2R = 66 ⇒ R = 33 Hence radius of the required circle = 33 cm. Q.5. Find the area of a sector of a circle of radius 28 cm and central angle 45°. Ans. r = 28 cm, θ = 45° ∴ Area of sector = = θ 360 ×π r 2 45 22 × × 28 × 28 360 7 1 22 × × 28 × 28 8 7 = 308 cm2 Q.6. The wheel of motor cycle is of radius 35 cm. How many revolutions for minute must the wheel make so as to keep a speed of 66 km/h? = 66 ×1000 × 100 cm/minute 60 For wheel, r = 35 cm ∴ Circumference = 2πr Ans. Speed = 22 × 35 7 = 220 cm = 2.2 cm. Number of revolutions = 2× 66 ×1000 × 100 1 × 60 220 = 500 revolutions/minute Q.7. A piece of 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at the centre. Find the radius of the circle. Ans. Length of arc = 20 cm θ = 60° = 228 Let radius = r Length opf arc = 20 = ⇒ 20 = ⇒ r 360° × 2πr 60° × 2×π × r 360° 1 ×π × r 3 60 cm π Q.8. What is the area of the circle, the circumference of which is equal to the perimeter of a square of side 11 cm. Ans. It is given that Side of square = 11 cm ∴ Perimeter = 4 × side = 4 × 11 = 44 cm Now, Circumference of circle = Perimter of square ⇒ 2π r = 44 ⇒ = θ r 44 44 × 7 = 2π = 2 × 22 = 7 cm r2 ⇒ ⇒ r2 r2 ⇒ r2 ⎛ 22 ⎞ = ⎜ × 7 × 7 ⎟ cm2 ⎝ 7 ⎠ 2 = 154 cm Q.9. The area of a circular playgound is 22176 m2. Find the cost of fencing this ground at the rate of Rs. 50 per meter. Ans. Let radius of the playground = r m ∴ πr2 = 22176 22 2 × r = 22176 7 22176× 7 22 = 1008 × 7 = 7056 = = 7056 = 844m Circumference of the circle = 2πr 22 × 84 7 = 528 m Cost of fencing = 528 × 50 = Rs. 26400 Q.10. Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm. Ans. r = 5 cm Length of the arc = 3.5 cm = 2× θ ∴ ⇒ ⇒ Now, area of the circle = πr2 ⇒ ⇒ 360° θ 360° × 2× × 2πr = 3.5 22 × 6 = 3.5 7 θ 360° Area of sector 3.5 = 10 × 22 × 7 = θ 360° × 2πr 3.5 × 7 22 = 10 × 22 × 7 × (5)2 = 3 .5 × 25 10 = 87.5 = 8.75 cm2 10 Multiple Choice Questions Q.1. The area of the sector of a circle of radius r and central angle θ, is : (a) 1/2 l.r (b) 2πr2 θ/720 2 (c) 2πr θ/360 (d) πr θ/360 Ans. (b) Q.2. If the perimeter of a circle is equal to that of a square, then the ratio of their area is : (a) 2 : 7 (b) 14 :11 (c) 7 : 20 (d) 11 : 14 Ans. (b) Q.3. It is proposed to build a single circular park equal is area of the sum of areas of two circular parks of diameters 165 m and 12 m in a locality. The radius of the new park would be : 229 (a) 10 m (b) 15 m (c) 20 m (d) 24 m Ans. (a) Q.4. The area of the circle that can be inscribed in a square of side 6 cm is (a) 36 π cm2 (b) 18 π cm2 (c) 12 π cm2 (d) 9 π cm2 Ans. (d) Q.5. The area of the sqaure that can be inscribed in a circle of radius 8 cm is : (a) 256 cm2 (b) 128 cm2 (c) 64√2 cm2 (d) 64 cm2 Ans. (b) Q.6. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is : (a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm Ans. (d) Q.7. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameter 36 cm and 20 cm is : (a) 56 cm (b) 42 cm (c) 28 cm (d) 16 cm Ans. (c) Q.8. If the circumference of a circle and the perimeter of a square are equal, then : (a) area of circle = Area of the square (b) Area of circle > Area of the square (c) Area of circle < Area of the square (d) Nothing definite can be said about the relation between the areas of the circle and square. Ans. (b) Q.9. The diameter of a wheel is 2.8 m. How far it will travel is 1000 revolutions. (a) 880 m (b) 8800 m (c) 88m (d) 8700 Ans. (b) Q.10. The perimeter of a square circumscribing a circle of a radius a unit is : (a) 2 a unit (b) 4 a unit (c) 8 a unit (d) 16 a unit Ans. (c)