# Grade 9 Circles. Answer t he quest ions. For more such worksheets visit

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1 ID : th-9-circles [1] Grade 9 Circles For more such worksheets visit Answer t he quest ions (1) ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 115º, then f ind the value of BAC. (2) The chords AB and CD of a circle are perpendicular to each other. If radius of the circle is 7 cm, and length of the arc AQD is 17 cm, f ind the length of arc BPC (Assume π = 22/7). (3) A square of area 64 cm 2 is inscribed into a semi-circle. What is the area of the semi-circle? (4) ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Find value of AE/AD. (5) T he 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 8 cm, f ind the distance between points P and Q. (6) A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. (7) AB is diameter of the circle. If A and B are connected to E, circle is intersected at C and D respectively. If AB = 26 cm and CD = 13 cm, f ind AEB.

2 (8) If ACB = 60 and BAD = 95, f ind angle ABD. ID : th-9-circles [2] Choose correct answer(s) f rom given choice (9) Two circles with radii of 3 and 7 are drawn with the same center. The smaller inner circle is painted blue, and the part outside the smaller circle and inside the larger circle is painted pink. What is the ratio of the areas painted pink to the area painted blue? a. 41:9 b. 4:1 c. 40:9 d. 41:10 (10) If APB = 110 and DAC = 30, f ind angle ADB. a. 75 b. 80 c. 85 d. 90 (11) 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. a. 3:2 b. 1:2 c. 2:1 d. 1:1

3 ID : th-9-circles [3] (12) AB and AC are two chords of a circle such that AB = 2AC. If distances of AB and AC f rom the centre are 3 and 5 cm respectively, f ind the area of circle (Assume π =3 f or this question). a. 95 cm 2 b. 86 cm 2 c. 89 cm 2 d. 91 cm 2 (13) If ADC = 110º and chord BC = chord BE. Find CBE. a. 130 b. 150 c. 140º d. 145 Check True/False (14) The angles subtended by a chord at any two points of a circle are equal. True False (15) The angle subtended by an arc at the centre is same as the angle subtended by it at any point on the remaining part of the circle, True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : th-9-circles [4] (1) 25º Consider a cyclic quadrilateral We know the f ollowing - The sum of the opposite angles is 180º i.e ADC + ABC = 180º - The angle subtended by a diameter to a point on the circumf erence = 90º We know that ADC = 115º ADC + ABC = 115º + ABC = 180º ABC = 65º Now consider the triangle f ormed by points A,B,C. We need to f ind BAC We know ABC + BAC + ACB = 180º (as it is a triangle) ABC = 65, and ACB is the angle subtended by the diameter to a point at the circumf erence, C Theref ore ACB = 90º From these, we get 65º + BAC + 90º = 180º BAC = 180º - (65º + 90º) = 25º

5 (2) 5 cm ID : th-9-circles [5] Consider the representative image below We have drawn a diameter RS parallel to AB Now RS is a diameter and perpendicular to CD (whence RS is parallel to AB and AB is perpendicular to CD) Theref ore RS bisects arc CD Arc RC = arc RD Since RS is a diameter Arc RS = Arc RC + Arc CS = 180 Arc RD + Arc CS = 180 This implies that the remaining arcs should add up to the remaining 180 Arc RC + Arc SD = 180 Now looking at the question We know length of arc AQD is 17 cm, and need to f ind length of arc BPC From the previous analysis, we know that arc AQD and arc BPC cover a semi-circle, so the total length of the two arcs is half the circumf erence Circumf erence of the circle = 2 x 22 x 7 = 44 7 length of arc AQD + length of arc BPC = 0.5 x 44 = 22 length of arc BPC = 22 - length of arc AQD = = 5 cm

6 (3) 40π cm 2 ID : th-9-circles [6] Following f igure shows the square inscribed into a semi-circle, Let's assume, a is the length of the side of the square. Theref ore, AB = BC = CD = DA = a, The area of the square = a 2 According to the question, the area of the square is 64 cm 2. Theref ore, a 2 = (1) If we look at the f igure caref ully, we notice the OC is the radius of the semi-circle and 'O' is the center of the semi-circle. Theref ore, OA = OB = a/2 In right angled triangle OBC, OC 2 = OB 2 + BC 2 [By the Pythagorean theorem] = (a/2) 2 + a 2 = (a 2 /4 + a 2 ) = 5a 2 /4 = (5 64)/4 [From equation (1)] = 80 cm 2 Step 5 Now, the area of the semi-circle = π(oc) 2 /2 = (π 80)/2 = 40π Step 6 Hence, the area of the semi-circle is 40π cm 2.

7 (4) 1 ID : th-9-circles [7] ABC = AEC [ Opposite angels of cyclic quadrilateral are supplementary ] Since ABC = ADE [ Opposite angles of parallelogram are equal ] ADE = AEC ADE = AED [ AEC = AED ] Step 5 Since ADE is isosceles with angles ADE = AED, AD = AE, theref ore AE/AD = 1

8 (5) 16 cm ID : th-9-circles [8] Take a 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 f ind the length of the segment PQ Since ABCD is a cyclic quadrilateral, DAB + BCD = ( DAB + BCD) = 90 2 PAD + QCD = 90 (since AP bisects DAB and QC bisects BCD) Now consider triangles QCD And QAD We see that QCD = QAD (angles subtended by the same segment - QD in this case - to points on the circumf erence are equal) 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 x 8 = 16 cm

9 (6) 30 ID : th-9-circles [9] Take a look at the image below The chord AB has a length equal to the radius of the circle This means that the triangle OAB is an equilateral triangle (all three sides are of the same length) AOB = 60 The angle subtended by a chord to the center is twice the angle subtended by the chord to a point in the major segment Consider a point R on the major segment AOB = 2 ARB. Theref ore ARB = 30 (7) 60 (8) 25 Angle ADB = ACB [ Angels inscribed by same chord AB] Angle ADB = ( BAD + ABD ) [ Angels of triangle ABD] On equating RHS of above equations ACB = ( BAD + ABD ) Now replace the values of ACB and BAD in above equations and solve f or ABD ABD = 25

10 (9) c. 40:9 ID : th-9-circles [10] Following f igure shows the circles with radii 3 and 7 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 blue, and the part outside the smaller circle and inside the larger circle is painted pink. The area painted blue = The area of the smaller inner circle = π(3) 2 = 9π The area painted pink = The area of the larger circle - The area of the smaller inner circle = π(7) 2 - π(3) 2 = π( ) = π(49-9) = 40π Thus, the ratio of the areas painted pink to the area painted blue = 40π 9π = 40 9 = 40:9. (10) b. 80 We know that a chord subtends the same angle to points on the circumf erence of the circle that are on the same side Here, we know ADB = ACB Also, in triangle APC, APC + ACP + PAC = 180 We need to f ind DAC DAC = PAC = APB - ACB = APB - DAC = = 80

11 (11) c. 2:1 ID : th-9-circles [11] 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 f rom the circle OR theref ore bisects OA, and PR = RA PA = 2 RA By the same reasoning AQ = 2 AS PQ = PA + AQ PQ = 2RA + 2 AS PQ = 2 (RA + AS) But RA + AS is the same as OO Theref ore PQ = 2 OO The ratio of PQ:OO is theref ore 2:1

12 (12) d. 91 cm 2 ID : th-9-circles [12] Take a look at the representative image below We are told that AB = 2AC. Also, if the perpendicular f rom O to AC meets the chord at Q, then OQ = 5 cm Similarly, OP = 3 cm Now OQ bisects AC, and OP bisects AB. From the earlier relation AB = 2AC. Theref ore BP = 2 CQ Let's assume CQ = x. Then BP = 2x Now consider the triangle OQC OC = r, the radius of the circle, and OQ = 5 OQC is a right angle triangle, so OQ 2 + CQ 2 = r x 2 = r x 2 = r 2 : Eq (1) Similarly, since OPB is a right angle triangle, OP 2 + BP 2 = r (2x) 2 = r x 2 = r 2 : Eq (2) Step 5 Subtracting equation (1) f rom equation (2), we get (9-25) + (4x 2 - x 2 = 0 3x 2 = 16

13 x 2 = 16 3 ID : th-9-circles [13] Step 6 Substituting in equation = r 2 3 r 2 = 3x = 91 3 Step 7 Area of the circle = πr 2 = 3 x 91 3 = 91 cm 2 (13) c. 140º ABCD is a cyclic quadrilateral since all 4 points A, B, C and D lie on the circumf erence The opposite angles of a cyclic quadrilateral add up to 180º From this, we f ind ADC + CBA = 180º CBA = 180º - ADC CBA = 180º - 110º CBA = 70º We know that chord BC = chord BE We have also learnt that the center of the circle lies on the bisector of CBE This means that BA is the bisector of CBE Theref ore CBE = 2 CBA = 2 x 70º = 140º (14) False (15) False

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