ID : in-9-geometry-overall [1] Class 9 Geometry-Overall For more such worksheets visit www.edugain.com Answer t he quest ions (1) AB is diameter of the circle. If A and B are connected to E, circle is intersected at C and D respectively. If AB = 20 cm and CD = 10 cm, f ind AEB. Choose correct answer(s) f rom given choice (2) If OD is perpendicular to AB, and DOC = 25, f ind ( BOC - AOC). a. 40 b. 50 c. 45 d. 55 (3) If AB and CD are parallel, f ind the value of X a. 14 b. 24 c. 156 d. 34
(4) Lines AB and CD intersect at O. If AOC + BOE = 90 and BOD = 40, f ind BOE. ID : in-9-geometry-overall [2] a. 40 b. 65 c. 60 d. 50 (5) There is a circular park of radius 12 meters. Three f riends Surjeet, Gauri and Sneha 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. a. 24 m b. 12 2 m c. 18 m d. 12 3 m
(6) If BC is a diameter of the circle and BAO = 40º. Then f ind the value of ADC. ID : in-9-geometry-overall [3] a. 40º b. 50 c. 45 d. 35 (7) If ADC = 110º and chord BC = chord BE. Find CBE. a. 135 b. 145 c. 140º d. 130
ID : in-9-geometry-overall [4] Fill in the blanks (8) If APB = 100 and DAC = 20, f ind angle ADB =. (9) If AD and BD are bisectors of CAB and CBA respectively, sum of angle x and y =. (10) If DAB = 71 and ABD = 59, the value of ACB = (11) The value of the supplement of the complement of 68 is
(12) ID : in-9-geometry-overall [5] a+b = (13) If AD and BD are bisectors of CAB and CBA respectively, the value of ADB =. (14) Two chords AB and AC of a circle subtends angles equal to 50º and 100º, respectively at the centre. if AB and AC lie on the opposite sides of the centre, the BAC = º. Check True/False (15) If two arcs of a circle are congruent, then their corresponding chords are equal. True False 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : in-9-geometry-overall [6] (1) 60 (2) b. 50 According to question DOC = 25 and OD is perpendicular to AB. Theref ore AOD = 90 and BOD = 90. DOC + AOC = AOD 25 + AOC = 90 [Since AOD = 90 and DOC = 25 ] AOC = 90-25 AOC = 65 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = 90 + 25-65 [Since BOD = 90, DOC = 25 and AOC = 65 ] = 50 Step 4 Theref ore BOC - AOC = 50 (3) b. 24 Parallel line AB and CD are intersected by a transversal as shown below, Here angle P and Q are alternate exterior angles. i.e. P = Q On comparing given angles with P and Q, 2x + 3x = 120 5x = 120 x = 24
(4) d. 50 ID : in-9-geometry-overall [7] If you look at the given f igure caref ully, you will notice that AB and CD are straight lines. AOC + BOE = 90 and BOD = 40. The angles of straight line add up to 180. Line AB is a straight line, theref ore we can say that AOC + COE + BOE = 180 AOC + BOE + COE = 180 90 + COE = 180 [Since AOC + BOE = 90 ] COE = 180-90 COE = 90 CD is also a straight line, theref ore COE + BOE + BOD = 180 90 + BOE + 40 = 180 [Since COE = 90 and BOD = 40 ] BOE + 130 = 180 BOE = 180-130 BOE = 50. Step 4 Now BOE = 50.
(5) d. 12 3 m ID : in-9-geometry-overall [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 f riends Surjeet, Gauri and Sneha 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, f or instance BAC = 60, and the angle subtended by a chord to the center is double the angle subtended to the angle at the circumf erence) We also draw perpendiculars f rom 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 f or triangles ROC, COQ, QOA, AOP and BOP) For triangle BOR, ORB = 90, RBO = 30, BOR = 60 (since it bisects BOC), and theref ore 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 T his means RO:RB:OB=1: :2 We know OB = radius = 12 m Theref ore RB = 12 3 2 Length of the string between A and B = 2 x RB = 12 3 m
(6) a. 40º ID : in-9-geometry-overall [9] 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 = 40º (7) 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 (8) 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 = 100-20 = 80
(9) 45 ID : in-9-geometry-overall [10] It is given that AD and BD are bisectors of CAB and CBA respectively. Theref ore, x = CAB/2 -----(1) y = CBA/2 -----(2) In triangle ABC, CAB + CBA + ACB = 180...[The sum of all three angles of a triangle is 180 ] CAB + CBA + 90 = 180 CAB + CBA = 180-90 CAB + CBA = 90 CAB/2 + CBA/2 = 90/2 x + y = 45...[From equation (1) and (2)] Hence, the sum of the angles x and y is 45. (10) 50 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 = 180 71 + 59 + BDA = 180 BDA = 180 (71 + 59) BDA = 50
(11) 158 ID : in-9-geometry-overall [11] If you look at the question caref ully, you will notice that f irst of all we have to f ind the complement of 68, then f ind the supplement of the complement of 68. T he sum of the complementary angles is 90. Theref ore the complement of 68 = 90-68 = 22 T he sum of supplementary angles is 180. Theref ore, the supplement of 22 = 180-22 = 158 Step 4 Theref ore the value of the supplement of the complement of 68 is 158. (12) 210 If you look at the f igure caref ully, you will notice that angles a, b and 150 are made around a point. The sum of all the angles around a point is 360. theref ore a + b + 150 = 360 a + b = 360-150 a + b = 210 Theref ore the value of a + b is 210.
(13) 140 ID : in-9-geometry-overall [12] It is given that AD and BD are bisectors of CAB and CBA respectively. Theref ore, BAD = CAB/2------(1) ABD = CBA/2-------(2) In triangle ABC, CAB + CBA + ACB = 180...[The sum of all three angles of a triangle is 180 ] CAB + CBA + 100 = 180 CAB + CBA = 180-100 CAB + CBA = 80 CAB/2 + CBA/2 = 80/2 = 40 -------(3) Now, In triangle ABD, BAD + ABD + ADB = 180 CAB/2 + CBA/2 + ADB = 180...Using (1) &(2) 40 + ADB = 180...Using (3) ADB = 180-40 = 140 Step 4 Hence, ADB = 140 (14) 105 If we consider that the center of the circle is O, then - AOB is the angle subtended by the chord AB. AOB = 50º - AOC is the angle subtended by the chord AC. AOC = 100º In the triangle OAB, we know that BAO = ABO, and BAO + ABO + AOB = 180º 2 x BAO + AOB = 180º BAO = 0.5(180º - AOB) = 0.5(180º - 50º) = 65º By similar reasoning, we see that CAO = 0.5(180º - AOC) = 0.5(180º - 100º) = 40º Step 4 Now, BAC = BAO + CAO = 65º + 40º = 105º (15) True