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

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1 ID : pk-9-circles [1] Grade 9 Circles For more such worksheets visit Answer t he quest ions (1) AB and AC are two chords of a circle such that AB = 2AC. If distances of AB and AC f rom the centre are 2 and 3 cm respectively, f ind the area of circle (Assume π =3 f or this question). (2) 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 19 cm, f ind the distance between points P and Q. (3) 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. (4) if CD = 2 cm, AB = 8 cm and OD is perpendicular to AB, then OA =? (5) A square of area 64 cm 2 is inscribed into a semi-circle. What is the area of the semi-circle? (6) A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in minor segment.

2 ID : pk-9-circles [2] Choose correct answer(s) f rom given choice (7) If DAB = 85 and ABD = 65, f ind the value of ACB. a. 25 b. 30 c. 40 d. 20 (8) If O is center of the circle and ABC = 35, f ind the value of AOC a. 80 b. 75 c. 70 d. 60 (9) ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 110º, then f ind the value of BAC. a. 25 b. 15 c. 10 d. 20º

3 (10) Given a circle, a circumscribed square is one that completely encloses the circle, and an inscribed square is completely enclosed by the circle, as shown below ID : pk-9-circles [3] If the the dif f erence in the areas of the two squares is 42 sq. m., then what is the area of the circle in sq. m.? Assume π = a. 66 b. 77 c. 88 d. 50 (11) If BC is a diameter of the circle and BAO = 50º. Then f ind the value of ADC. a. 40 b. 55 c. 50º d. 45 (12) Two circles with radii of 5 and 9 are drawn with the same center. The smaller inner circle is painted white, and the part outside the smaller circle and inside the larger circle is painted purple. What is the ratio of the areas painted purple to the area painted white? a. 57:25 b. 56:25 c. 57:26 d. 28:13 (13) AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, f ind the perpendicular distance of AB f rom the centre of the circle. a. 7cm b. 9 cm c. 10 cm d. 8 cm

4 ID : pk-9-circles [4] Fill in the blanks (14) The chords AB and CD of a circle are perpendicular to each other. If radius of the circle is 35 cm, and length of the arc BPC is 25 cm, the length of arc AQD = cm (Assume π = 22/7). Check 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

5 Answers ID : pk-9-circles [5] (1) 32 cm 2 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 = 3 cm Similarly, OP = 2 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 = 3 OQC is a right angle triangle, so OQ 2 + CQ 2 = r x 2 = r x 2 = r 2 : Eq (1) Step 4 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 (4-9) + (4x 2 - x 2 = 0 3x 2 = 5

6 3x = 5 ID : pk-9-circles [6] x 2 = 5 3 Step 6 Substituting in equation = r 2 r 2 = 3x = 32 3 Step 7 Area of the circle = πr 2 = 3 x 32 3 = 32 cm 2

7 (2) 38 cm ID : pk-9-circles [7] 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 Step 4 As PQ is the diameter of the circle, it is twice the radius PQ = 2 x 19 = 38 cm

8 (3) 2:1 ID : pk-9-circles [8] 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

9 (4) 5 cm ID : pk-9-circles [9] We know the f ollowing OA and OD are the radii of the circle. Theref ore OA=OD OD is perpendicular to AB. Segment OC is also theref ore perpendicular to AB. OCA = 90 Also, the point C divides AB into two. So AC=CB = AB 2 We also know that CD = OD - OC = OA - OC OAC is a right-angle triangle - theref ore AC 2 + OC 2 = OA 2 From this, the relation between AB, CD and OA can be given as - ( AB ) 2 + (OA - CD) 2 = OA 2 2 Here we know CD = 2 and AB = 8 Substituting and simplif ying in the equation we get the value of OA = 5 cm

10 (5) 40π cm 2 ID : pk-9-circles [10] 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 Step 4 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.

11 (6) 150 ID : pk-9-circles [11] 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 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 = ARB = = 150

12 (7) b. 30 ID : pk-9-circles [12] The angle subtended by a chord to two points on the circumf erence are equal, if the two points are on the same side of the chord This means that ACB = BDA Now, ABC and ABD are triangles, so the sum of the internal angles is 180 Step 4 For triangle ABD, this means DAB + ABD + BDA = BDA = 180 BDA = 180 ( ) BDA = 30 (8) c. 70 The key point to note here is that AC is a chord of the circle B is a point on the circumf erence, and O is the center We also know that the angle subtended by a chord at the centre of a circle is double the size of the angle subtended by the same chord at the circumf erence of the circle. Step 4 The angle subtended by the chord AC to the point on the circumf erence is 35, so going by the above, the angle subtended by the chord AC to the center is 70

13 (9) d. 20º ID : pk-9-circles [13] 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 = 110º ADC + ABC = 110º + ABC = 180º ABC = 70º 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 = 70, 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 70º + BAC + 90º = 180º BAC = 180º - (70º + 90º) = 20º

14 (10) a. 66 ID : pk-9-circles [14] Let's name the outer square as ABCD, and the inner square as PQRS as shown below: If the radius of the circle is R, the side of the outer square will be double of circle's radius: AB = 2 R Area of outer square, Area(ABCD) = 2R 2R = 4R 2 Step 4 Side of the inner square PQRS can be calculated using Pythagoras T heorem PQ = (R 2 + R 2 ) PQ = 2 R Step 5 Area of inner square, Area(PQRS) = 2R 2R = 2R 2 Step 6 The dif f erence in areas of the two squares is given to be 42 sq m. Thus: Area(ABCD) - Area(PQRS) = 42 4R 2-2R 2 = 42 2R 2 = 42 R 2 = 42/2 Step 7 Now that we know the radius of the circle, we can calculate the area of the circle using the f ormula, Area(circle) = π R 2 Area(circle) = (22/7) (42/2) Area(circle) = 66 sq. m.

15 (11) c. 50º ID : pk-9-circles [15] Since BC is a chord, and O is the center of the circle, ABO = BAO Also ABO is a triangle, and the AOB = 180º - ( ABO + BAO) We also see that AOB = 180º - AOC Theref ore AOC = ABO + BAO = 2 BAO Also, considering the chord AC, AOC is the angle subtended to the center and ADC is the angle subtended by the arc AC to a point D on the circumf erence Theref ore AOC = 2 ADC From these, we get that BAO = ADC = 50º

16 (12) b. 56:25 ID : pk-9-circles [16] Following f igure shows the circles with radii 5 and 9 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 white, and the part outside the smaller circle and inside the larger circle is painted purple. The area painted white = The area of the smaller inner circle = π(5) 2 = 25π The area painted purple = The area of the larger circle - The area of the smaller inner circle = π(9) 2 - π(5) 2 = π( ) = π(81-25) = 56π Step 4 Thus, the ratio of the areas painted purple to the area painted white = 56π 56:25. 25π = =

17 (13) d. 8 cm ID : pk-9-circles [17] Let's consider the problem in terms of the image below We are told that the diameter AD=34 cm and the length of the chord AB=30 cm We are asked to f ind the perpendicular distance of AB f rom the center O This is the length of the line segment OC We can see that OAC is a right angle triangle We also know that OA is the radius, i.e half of the diameter AD. Theref ore AO = 17 cm We can also see that the perpendicular to the chord f rom the center - OC, divides the chord into two equal parts Theref ore AC = CB = AB = 30 = Step 4 Since OAC is a right angle triangle, OC 2 + AC 2 = OA 2 OC = 17 2 OC 2 = = 64 OC = 8 cm

18 (14) 85 ID : pk-9-circles [18] 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 Step 4 Now looking at the question We know length of arc BPC is 25 cm, and need to f ind length of arc AQD From the previous analysis, we know that arc BPC and arc AQD 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 35 = length of arc BPC + length of arc AQD = 0.5 x 220 = 110 length of arc AQD = length of arc BPC = = 85 cm (15) False