### Areas Related To A Circle Class 10th Mathematics Gujarat Board Solution

##### Question 1.The area of a circular park is 616 m2. There is a 3.5 m wide track around the park running parallel to the boundary. Calculate the cost of fencing on the outer circle at the rate of Rs. 5 per meter.Answer:Area of the circular park = 616 m2width of the track = 3.5 m Area of the circular park = 616 m2π × r2 = 616r2 = 196r = √196r = 14 m∴ the radius of the circular park (r1) = 14 mthe radius of the park with track (r2) = 14 + 3.5= 17.5 mCircumference of the track = 2π (r2)= 2× (17.5)= 110 mCost of fencing per meter = Rs. 5Cost of fencing 110 m = 110 × 5= 550Question 2.A man is cycling in such a way that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, then how much distance will he cover in 2 hours?Answer:Speed of the cycle = 140RPMDiameter of the wheel = 60 cmNo. of revolution in 2 hours = 120 × 140= 16,800Distance covered in 2 hours = Revolutions × Diameter= 16,800 × 60= 1008000 cm= 10080 m= 10.8 kmQuestion 3.If a chord of (O, 20) subtends right angle at O, find the area of the minor segment.Answer:Given radius = 20cmif we connect the chord to the center point that forms a segment.the centre measure angle (θ) =90owe can observe a right-angled triangle OAB Area of segment = = = 314 cm2Area of the triangle = OA× OB= × (20) × (20)= 200 cm2Area of minor segment = Area of segment - Area of the triangle= 314 – 200= 114 cm2Question 4.There are two arcs of (O, OA) and of (M, MA) as shown in figure 13.24. Find the area enclosed by two arcs. Figure 13.24Answer:from the figure we can notice all these1. OAB is an equilateral triangle(∵ All the sides are equal)2. OAPB is a minor segment with Radius 14 cm, the angle (θ) between them is 60o3. ABQ is a semi-circle with centre as M radius 7 cmArea of the segment = = = 102.62 CM2Area of the triangle = × OA2= × 142= 84.870 CM2Area of the semi-circle =Area of the segment - Area of the triangle= 102.62 - 84.870= 17.75 cm2Area of the semi-circle = × r2= × 72= 76.96 cm2area enclosed by two arcs =Area of the semi-circle - Area of the semi-circle= 76.96 – 17.75= 59.20 cm2 (approximately)Question 5.The length of a side of a square garden ABCD is 70 m. A minor segment of (O, OA) is drawn on each of two opposite sides for developing lawn, as shown in figure 13.25. Find the area of the lawn. figure 13.25Answer:side of a square garden ABCD (a) = 70 mFrom the fig. in the below we can see that1. OAXD is a segment2. OAD is an right-angled triangle.3. AXD is a minor segmentOA, OD are the radii of the segmentfrom the triangle OADAD2 = OA2 + OD2702 = r2 + r22r2 = 4900r2 = 2450r = √2450Area of minor segment = Area of segment - Area of triangle = - × OA2= - × = 1924.22 – 1225=699.22 cm2= 700 cm2since AXD, BYC are same∴ Area of lawn = 2 × area of minor segment= 2 × 700= 1400 cm2Question 6.ABCD is a square of side 20 cm. Find the area of blue coloured region formed by the semi-circles drawn on each side as shown in figure 13.26. ( = 3.14) figure 13.26.Answer:side of a square (a) = 20 cmwe can see the semi-circle in the square having 10 cm as radius (r)area of the colored region =area of 4 semi-circles - area of the square= 4 × ( × r2) - a2= 2 (π × 102) - 202= 228 cm2Question 7.On a circular table top of radius 30 cm a design is formed leaving an equilateral triangle inscribed in a circle. Find the area of the design. (π = 3.14)Answer:radius of the circle = 30 cmArea of the circle = π r2 = 3.14 × 302= 2826 cm2PQR is an equilateral triangle∠ QPR = 60O∠ QOR = 2 × ∠ QPR = 120oIn OQR triangle,let OS is perpendicular to QRIn triangle OQR as OQ = OR∠ QOS = ∠ QPR = 60OIn triangle OSQ ∠S = 90osin 60o =  = QS = 15 √3 cmNow,QR = 2 QS = 2 (15 √3) = 30√3 cmArea of equilateral triangle = × QR2= × (30√3)2= 1167.75 CM2Area of design = Area of circle - area of triangle PQR= 2826 – 1167.75= 1658.25 CM2Question 8.Figure 13.27 shows a kite formed by a square PQRS and an isosceles right triangle ARB whose congruent sides are 5 cm long. is an arc of a (R, 42). Find the area of the blue coloured region. Figure 13.27Answer:Side of square (r) = 42 cmIn triangle RAB,RA = RB = 5 cmIn minor sector RQCS angle is 90oArea of the shaded region = Area of minor sector RQCS + Area of RAB= + × RA × RB= + × 5 × 5= 1384.74 + 12.5= 1372.24 cm2Question 9.In figure 13.28, ABCD is a square with sides having length 8 cm. Find the area of the blue coloured region. (π = 3.14) Figure 13.28Answer:side of square (a) = 8 cmfrom the fig. we can see that of circle is placed at four parts of the square. if we join them a circle is formed with radius (r1) 2 cmAnother circle is placed at centre of the square with radius (r2) 2 cmArea of colored region = Area of square - Area of circles at 4 corners + Area of circle at centre= a2 – (4 × ( × r12) + (π r22))= a2 – (π r12 + π r22)= 82 – (3.14 (22 + 22))= 38.88 cm2Question 10.A circle is inscribed in APQR where m∠Q = 90, PQ = 8 cm and QR = 6 cm. Find the area of the blue coloured region shown in figure 13.29. ( = 3.14) Figure 13.29Answer:from the given figurePQR is a right-angle trianglePQ = 8 cmQR = 6 cmPR2 = PQ2 + QR2PR2 = 82 + 62PR = √100 = 10 cmRadius of the incircle (r) = = = = 2 CMarea of coloured region = Area of triangle - area of circle= × PQ × QR - π r2= × 8 × 6 - π 22= 24 – 12.566= 11.433 cm2Question 11.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:If an arc of a circle subtends an angle of measure 0 at the centre, then the area of the minor sector is…..A. B. C. D. Answer:we know that area of minor sector = So, option(d) is the correct answer.Question 12.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The area of a sector is given by the formula ….(r is the radius and 1 is the length of an arc.)A. rlB. r2/ lC. rlD. rlAnswer:we know that when arc length (l) and radius (r) of sector is givenArea of the sector is = rlSo, option(a) is the correct answer.Question 13.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct: and are the two mutually perpendicular radii of a circle having radius 9 cm. The area of the minor sector corresponding to AOB is cm2( = 3.14)A. 63.575B. 63.585C. 63.595D. 63.60Answer:Given radius (r)= 9 cm∵ They are perpendicular to each other θ = 90oarea of the minor sector = = = = 63.585 cm2So, option(b) is the correct answer.Question 14.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:A sector subtends an angle of measure 120 at the centre of a circle having radius of 21 cm. The area of the sector is ……….. cm2.A. 462B. 460C. 465D. 470Answer:Given radius (r)= 21 cmangle(θ) = 120oarea of the minor sector = = = = 461.58 cm2= 462 cm2So, option(a) is the correct answer.Question 15.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:If the area and the circumference of a circle are numerically equal, then r = …………A. πB. C. 1D. 2Answer:Area of the circle = πr2circumference of the circle = 2πrgiven, Area = Circumference2πr = πr22 = rSo, option(d) is the correct answer.Question 16.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The length of an arc subtending an angle of measure 60 at the centre of a circle whose area is 616 is ……….A. B. 66C. D. 33Answer:Area of circle = 616cm2πr2 = 616r2 = 196r = 14 cmGiven θ = 60oarea of the minor sector = = = = cm2So, option(c) is the correct answer.Question 17.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The area of a minor sector of (O, 15) is 150. The length of the corresponding arc is (π = 3.14)A. 30B. 20C. 90D. 15Answer:Radius of the minor sector= 15 cmArea of minor sector = 150Area of a minor sector = r l150 = × 15 × ll = 20 cmSo, option(b) is the correct answer.Question 18.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:If the radius of a circle is increased by 10 %, then corresponding increase in the area of the circle is ………. (π = 3.14)A. 19 %B. 10 %C. 21 %D. 20 %Answer:r be the radius of the circleArea of the circle = πr210% increase in radius = 1.1rArea of circle with new radius = π(1.1r)2= 1.21πr2increase in area = 1.21πr2 - πr2= 0.21 πr2∴ 21% increase in the area of circleSo, option(c) is the correct answer.Question 19.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:If the ratio of the area of two circles is 1: 4, then the ratio of their circumference …..A. 1: 4B. 1: 2C. 4: 1D. 2: 1Answer:let r1, r2 are the radii of two circlesgiven,πr12: πr22 = 1 : 4r12: r22 = 1 : 4r1 : r2 =1 : 2r2 = 2 r1circumference of 1st circle = 2πr1circumference of 2nd circle = 2πr2 = 2π(2r1) = 4πr1ratio of circumference = 2πr1 : 4πr1 = 1 : 2So, option(b) is the correct answer.Question 20.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The area of the largest triangle inscribed in a semi-circle of radius 8 is ……A. 8B. 16C. 64D. 256Answer:Given radius of semi-circle = 8 cmFor a largest triangle in semi-circlebase will be its diameter of semi-circleheight will be radius of semi-circleso, its base = 2 × radius = 2 × 8 = 16 cmheight = 8 cmArea of the triangle = × base × height= × 16 × 8= 64 cm2So, option(c) is the correct answer.Question 21.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:If the circumference of a circle is 44 then the length of a side of a square inscribed in the circle is ….A. B. C. D. Answer:Given circumference = 44 cm2πr = 44πr = 22r = 7 cmGiven that square inscribed in a circle.so, diagonal of square will be equal to diameter of circle.diagonal = 2 × radius = 2 × 7 = 14 cmdiagonal = √2 × side of square14 = √2 × sideside = 7√2 cmQuestion 22.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The length of minute hand of a clock is 14 cm. If the minute hand moves from 2 to 11 on the circular dial, then area covered by it is cm2.A. 154B. 308C. 462D. 616Answer:The angle moved by minute hand = 90olength of minute hand (r) = 14 cmArea of covered by minute hand = = = = 153.93 cm2= 154 cm2So, option(a) is the correct answer.Question 23.Select a proper option (a), (b), (c) or (d) from given options and write in the box given on the right so that the statement becomes correct:The length of minute hand of a clock is 15 cm. If the minute hand moves for 20 minutes on a circular dial of a clock, the area covered by it is …….. cm2. (= 3.14)A. 235.5B. 471C. 141.3D. 706.5Answer:The angle moved by minute hand = 120olength of minute hand (r) = 15 cmArea of covered by minute hand = = = = 235.619 cm2= 235.5 cm2So, option(a) is the correct answer.

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