ID : ae-9-circles [1] Grade 9 Circles For more such worksheets visit www.edugain.com Answer the questions (1) Two circles with centres O and O intersect at two points A and B. A line PQ is drawn parallel to OO through A(or B) intersecting the circles at P and Q. Find the ratio PQ:OO. (2) The bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle circumscribing it at the points P and Q. If radius of the circle is 4 cm, find the distance between points P and Q. (3) A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in the minor segment. (4) If ACB = 25 and ABD = 40, find angle BAD. (5) ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Find value of AE. AD (6) There is a circular park of radius 15 meters. Three friends Atifa, Akila and Ginton are sitting at equal distance on its boundary each having a toy telephone (connected using strings) in their hands to talk each other. Find the length of the string between a pair of the telephones.
ID : ae-9-circles [2] (7) The Ferris Wheel at the school fair has radius of 12 metres. It revolves at the rate of one revolution per 2 minutes. How many seconds does it take a rider to travel from the bottom of the wheel to a point 6 vertical metres above the bottom? (8) AB is diameter of the circle. If A and B are connected to E, circle is intersected at C and D respectively. If AB = 18 cm and CD = 9 cm, find AEB. (9) If ADC = 125 and chord BC = chord BE. Find CBE. (10) Two chords AB and AC of a circle subtends angles equal to 50 and 30, respectively at the centre. Find BAC, if AB and AC lie on the opposite sides of the centre. (11) If BC is a diameter of the circle and BAO = 30. Then find the value of ADC. (12) AB and AC are two chords of a circle such that AB = 2AC. If distances of AB and AC from the centre are 3 cm and 4 cm respectively, find the area of circle. (Assume π =3) (13) Two circles with radii of 9 and 14 are drawn with the same center. The smaller inner circle is painted red, and the part outside the smaller circle and inside the larger circle is painted green. What is the ratio of the areas painted green to the area painted red?
ID : ae-9-circles [3] Check True/False (14) Two congruent circles with centres O and O intersect at two points A and B. Then, AOB = AO'B. True False (15) If two arcs of a circle are congruent, then their corresponding chords are equal. True False 2017 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ae-9-circles [4] (1) 2:1 Consider the image below: We see that the line drawn at point O perpendicular to OO meets the line PQ at a point R. A similar line drawn at O meets PQ at a point S. Now, PA is a chord, and OR is a line perpendicular to it drawn from the centre of the circle. We know that a perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore, OR bisects PA, i.e. PR = RA or PA = 2RA. Similarly, AQ = 2AS. We know, PQ = PA + AQ PQ = 2RA + 2AS PQ = 2(RA + AS) PQ = 2OO (RA + AS is the same as OO ) The ratio of PQ:OO is 2:1.
(2) 8 cm ID : ae-9-circles [5] Look at the image below: We consider the cyclic quadrilateral ABCD. Bisecting the angle DAB, we get the bisector which intersects the circle at P. Bisecting the opposite angle DCB, we get the bisector intersecting the circle at Q. Given the radius of the circle, we need to find the length of the segment PQ. Since, ABCD is a cyclic quadrilateral, DAB + BCD = 180. 1 ( DAB + BCD) = 90 2 PAD + QCD = 90 (Since AP bisects DAB and QC bisects BCD.) Now, we see that QCD = QAD (Angles subtended by the same segment, QD in this case, to points on the circumference are equal.) Using the equation obtained in step 2. We have, PAD + QAD = 90 PAQ= 90 This means that PAQ is the angle in a semi-circle. PQ is the diameter of the circle. As, PQ is the diameter of the circle, it is twice the radius. PQ = 2 4 cm = 8 cm
(3) 150 ID : ae-9-circles [6] Look at the image below: The chord AB has a length equal to the radius of the circle. This means that ΔOAB is an equilateral triangle. (all the sides and angels are equal and each angle measure 60 ) So, AOB = 60. We know that the angle subtended by a chord at the center is twice the angle subtended by the chord at a point in the major segment. Consider a point R on the major segment. AOB = 2 ARB. Therefore, ARB = 30 Now, consider a point S on the minor segment. ARBS is a cyclic quadrilateral, and the opposite angles of such a quadrilateral sum up to 180. ARB + ASB = 180 ASB = 180 - ARB = 180-30 = 150
(4) 115 ID : ae-9-circles [7] Angle ADB = ACB [ Angels inscribed by same chord AB] Angle ADB = 180 - ( BAD + ABD ) [ Angels of triangle ABD] On equating RHS of above equations ACB = 180 - ( BAD + ABD ) Now replace the values of ACB and ABD in above equations and solve for BAD BAD = 115 (5) 1 ABC = 180 - AEC Since ABC = ADE ADE = 180 - AEC [ Opposite angels of cyclic quadrilateral are supplementary.] [ Opposite angles of parallelogram are equal.] ADE = AED [ 180 - AEC = AED] Step 5 Since ADE is isosceles with angles ADE = AED, AD = AE, therefore AE AD = 1.
(6) 15 3 m ID : ae-9-circles [8] Take a look at the image below, which represents the scenario outlined in the question. The corners of the triangle A, B and C represent the three friends Atifa, Akila and Ginton O is the center of the cirlce ABC is an equilateral triangle, and we connect A, B and C to the center O. We can see that BOC = COA = AOB = 120 (remember, for instance BAC = 60, and the angle subtended by a chord to the center is double the angle subtended to the angle at the circumference) We also draw perpendiculars from the center to AB, CA and BC, meeting the lines at points P, Q and R respectively Take the triangle BOR (and remember that the same will hold true for triangles ROC, COQ, QOA, AOP and BOP) For triangle BOR, ORB = 90, RBO = 30, BOR = 60 (since it bisects BOC), and therefore RBO = 30 So BOR is a 30-60 -90 triangle We know that the sides of a 30-60 -90 triangle are in the proportion 1: :2 This means RO:RB:OB=1: :2 We know OB = radius = 15 m Therefore RB = 15 3 2 Length of the string between A and B = 2 x RB = 15 3 m
(7) 20 seconds ID : ae-9-circles [9] After travelling 6 vertical metres from the point of start C, its new position is B. Let us consider triangle ABO and ABC: We have: AB is common AC = AO = 6 m (AC = 12-6 = 6 m) Angle BAO = Angle BAC (right angles) This means triangles ABO and ABC are congruent. From step 2 we have BC = OB = 12 m. Now we have BC = OB = CO = 12 m. This means that triangle BCO is an equilateral triangle. We know that all interior angles of an equilateral triangle are equal to 60 degrees, we have Angle COB = 60. Step 5 One revolution is equal to 360 degrees, which, according to the question, is completed in 2 minutes. 360º in 2 minutes Or, 360º in 2 60 = 120 seconds 60º in 120 60 = 20 seconds. 360 Step 6 It takes rider 20 seconds to travel 6 vertical metres from the point of start. (8) 60
(9) 110 ID : ae-9-circles [10] ABCD is a cyclic quadrilateral since all the four points A, B, C and D lie on the circumference of a circle. We know, the opposite angles of a cyclic quadrilateral add up to 180. So, ADC + CBA = 180 CBA = 180 - ADC CBA = 180-125 CBA = 55 We know that chord BC = chord BE. Join the points C and E to the centre of the circle. Consider ΔCOE and ΔBOE, BO = BO (common) BC = BE (given) OC = OE (radius of the circle) So, ΔCOE ΔBOE by the property SSS. Hence, OBC = OBE by CPCT We have, OBC = OBE = CBA = 55 Therefore, CBE = OBC + OBE = 55 + 55 = 110.
(10) 140 ID : ae-9-circles [11] If we consider that the center of the circle is O, then AOB = 50 and AOC = 30 In ΔOAB, we know that BAO = ABO, (As, OB = OA and angle opposite to equal sides are equal) and BAO + ABO + AOB = 180 (Angle sum property) 2 BAO + AOB = 180 BAO = 1 2 (180 - AOB) = 1 2 (180-50 ) = 65 Similarly, CAO = 1 2 (180 - AOC) = 1 2 (180-30 ) = 75 Now, BAC = BAO + CAO = 65 + 75 = 140
(11) 30 ID : ae-9-circles [12] As,OA and OB are the radius of the circle, OA = OB. This means ΔAOB is an isosceles triangle. So, ABO = BAO = 30. Also, ABO = ABC Considering the chord AC, ABC and ADC are the angles subtended by the chord AC in the same segment of the circle. We know that the angle subtended by a chord in the same segment of a circle are equal. So, ABC = ADC Therefore, ADC = ABC = 30. (12) 55 cm 2 Take a look at the representative image below: We are told that AB = 2AC. Also, if the perpendicular from O to AC meets the chord at Q, then OQ = 4 cm. Similarly, OP = 3 cm. As the perpendicular from the centre on the chord bisects the chord, OQ bisects AC, and OP bisects AB. From the earlier relation AB = 2AC. Therefore, BP = 2CQ
Let us assume CQ = x. Then, BP = 2x ID : ae-9-circles [13] Now consider ΔOQC, OC = r, the radius of the circle, and OQ = 4 cm As, the distance of a chord from the centre is always the perpendicular distance. ΔOQC is a rightangled triangle. By using pythagoras theorum, OQ 2 + CQ 2 = r 2 4 2 + x 2 = r 2 or, 16 + x 2 = r 2 ------(1) Similarly, ΔOPB is a right-angled triangle, OP 2 + BP 2 = r 2 3 2 + (2x) 2 = r 2 or, 9 + 4x 2 = r 2 ------(2) Step 5 Subtracting equation (1) from equation (2), we get: (9-16) + (4x 2 - x 2 ) = 0 or, 3x 2 = 7 or, x 2 = 7 3 Step 6 On substituting x 2 = 7 3 in equation (1), we get: 16 + 7 3 = r 2 or, r 2 = 3 16 + 7 3 = 55 3 Step 7 Therefore, area of the circle = πr 2 = 3 55 3 = 55 cm 2
(13) 115:81 ID : ae-9-circles [14] Following figure shows the circles with radii 9 and 14 are drawn with the same center, We know that the area of a circle = π(r) 2 According to the question, the smaller inner circle is painted red, and the part outside the smaller circle and inside the larger circle is painted green. The area painted red = The area of the smaller inner circle = π(9) 2 = 81π The area painted green = The area of the larger circle - The area of the smaller inner circle = π(14) 2 - π(9) 2 = π(14 2-9 2 ) = π(196-81) = 115π Thus, the ratio of the areas painted green to the area painted red = 115π 81π = 115 81 = 115:81.
(14) True ID : ae-9-circles [15] We are given that the circles are congruent, which means they have the same radius. We can say that OA = O'A = OB = O'B = r, where r is the radius of the circle. Now, consider ΔOAB and ΔO'AB, we have OA = O'A (Radius) OB = O'B (Radius) AB = AB (common side) Hence, ΔOAB ΔOAB by the property of SSS. As, ΔOAB ΔOAB, we can say that AOB = AO'B by CPCT. Hence, the statement is true.
(15) True ID : ae-9-circles [16] In the figure given below PQ and RS are the two chords that are equal, OC and OD are the perpendiculars drawn to the chords from the centre. Consider ΔPCA and ΔRDB. We know that the perpendicular to the chord from the centre bisects the chord. Since, the chords PQ and RS are equal. This means PC = RD and CQ = DS. As, we know that equal chords are equidistant from the center, OC = OD. We also know that OA = OB, as both are the radius. Subtracting the two equations, OA - OC = OB - OD AC = BD. So, in ΔPCA and ΔRDB. We have, PC = RD (From step 2) AC = BD (From step 3) PCA = RDB = 90 (By construction) Hence, ΔPCA ΔRDB by SAS So, RB = PA by CPCT Step 5 Similary, ΔDBS ΔCAQ. Hence, SB = QA by CPCT. Step 6 Now, we have RB = PA and SB = QA, on adding both we have RB + SB = PA + QA. The lengths PA + QA and RB + SB can be taken to be an approximate measure of the length of the arc PAQ and arc RBS respectively. Step 7 This means if two chords of a circle are equal, their corresponding arcs are congruent. Hence, the statement is true.