Grade 9 Lines and Angles

ID : sg-9-lines-and-angles [1] Grade 9 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) If AB and PQ are parallel, compute the angle Z. (2) Find the value of a+b. Choose correct answer(s) f rom given choice (3) If angles of a triangle are in ratio 4:5:11, the triangle is a. an isosceles triangle b. a right triangle c. an acute angled triangle d. an obtuse angled triangle (4) If lines AB and CD intersects as shown below, f ind value of angle x. a. 35 b. 25 c. 45 d. 30

ID : sg-9-lines-and-angles [2] (5) WXYZ is a quadrilateral whose diagonals intersect each other at the point O such that OW = OX = OZ. If OWX = 40, then f ind the measures of OZW. a. 45 b. 60 c. 50 d. 40 (6) If AB and PQ are parallel, compute the angle Z. a. 265 b. 450 c. 95 d. 445 (7) If AB and CD are parallel, f ind the value of angle x. a. 50 b. 60 c. 130 d. 40 (8) If AB and DE are parallel to each other, f ind value of angle BCD. a. 56 b. 41 c. 46 d. 51

(9) If AB and CD are parallel, f ind the value of angle x. ID : sg-9-lines-and-angles [3] a. 95 b. 75 c. 115 d. 105 (10) If AB and CD are parallel, f ind the value of angle x. a. 105 b. 65 c. 75 d. 85 Fill in the blanks (11) If OD is perpendicular to AB, and DOC = 20, ( BOC - AOC). =.

ID : sg-9-lines-and-angles [4] (12) If AD and BD are bisectors of CAB and CBA respectively, the value of ADB =. (13) If CD is perpendicular to AB, and CE bisect angle ACB, the angle DCE =. (14) Value of angle q is (15) If ADB is a right angle, the value of angle x =. 2016 Edugain (www.edugain.com). Many more such worksheets can be

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Answers ID : sg-9-lines-and-angles [6] (1) 113 Construction: Draw a line MN which is parallel to the line PQ and the line AB. If you look at the given f igure caref ully, you will notice that, the addition of the angle Z1, angle Z2 and angle Z makes f ull angle and f ull angle is exactly 360. Theref ore, angle Z1 + angle Z2 + angle Z = 360, angle Z = 360 - angle Z1 - angle Z2. Angle P = 55, angle Z1 and angle P are alternate angles, angle Z1 = angle P, angle Z1 = 55. Similarly, angle A = 58, angle Z2 and angle A are alternate angles, angle Z2 = angle A angle Z2 = 58. Step 4 Angle Z = angle Z1 + angle Z2 = 55 + 58 = 113. Step 5 Theref ore, the angle Z is 113.

(2) 210 ID : sg-9-lines-and-angles [7] If you look at the f igure caref ully, you will notice that angles 150, a and b are made around a point. The sum of all the angles around a point is 360. theref ore 150 + a + b = 360 a + b = 360-150 a + b = 210 Theref ore the value of a + b is 210. (3) d. an obtuse angled triangle According to the question, all angles of the triangle are in ratio 4:5:11. We can assume three angles of the triangle to be 4x, 5x and 11x where x is common f actor. We know that the sum of the three angles of a triangle is 180. Theref ore, 4x + 5x + 11x = 180 20x = 180 x = 180 20 Now, 4x = 4 180 20 = 36, 5x = 5 180 20 = 45 and 11x = 11 180 = 99. 20 Theref ore, the three angles of the triangle are 36, 45 and 99. Since, one of the angle of the triangle is greater than 90, the triangle is an obtuse angled triangle.

(4) a. 35 ID : sg-9-lines-and-angles [8] In triangle BCE, CBE + BCE + BEC = 180...[Since the sum of all three angles of a triangle is 180 ] 30 + 115 + BEC = 180 145 + BEC = 180 BEC = 180-145 BEC = 35 Since AED and BEC are the opposite angles of intersecting lines AB and CD and we know that the opposite angles are congruent Theref ore, AED = BEC = 35 ------(1) Now, in triangle ADE, DAE + ADE + AED = 180...[Since the sum of all three angles of a triangle is 180 ] x + 110 + 35 = 180...Using (1) 145 + x = 180 x = 180-145 x = 35 Step 4 Hence, the value of x is 35. (5) c. 50

(6) a. 265 ID : sg-9-lines-and-angles [9] Draw a line MN which is parallel to line PQ and line AB If you look at the given f igure caref ully, you will notice that angle Z1 + angle Z2 + angle Z makes f ull angle and the f ull angle is exactly 360. Theref ore angle Z1 + angle Z2 + angle Z = 360 angle Z = 360 - angle Z1 - angle Z2 Angle Z1 and angle Q are alternate angles. Theref ore angle Z1 = angle Q angle Z1 = 44 Similarly angle Z2 and angle B are alternate angles. Theref ore angle Z2 = angle B angle Z2 = 51 Step 4 Theref ore angle Z = 360 - angle Z1 - angle Z2 = 360-44 - 51 = 360-95 = 265

(7) a. 50 ID : sg-9-lines-and-angles [10] It is given that line AB and CD and parallel lines and the third line (say EF) cuts them as shown in the f igure. a = c (vertically opposite angles) c = e (alternate interior angles) Theref ore we can write, a = c = e = g Again, b = d (vertically opposite angles) d = f (alternate interior angles) Theref ore we can write, b = d = f = h We know that sum of two adjacent angle is equal to 180. Theref ore, f rom the diagram, you can write, a + b = 180, b + c = 180, c + d = 180, d + a = 180 Given, f = 50 and b = x As b is equal to f So, x is 50. Theref ore, the value of x is 50.

(8) d. 51 ID : sg-9-lines-and-angles [11] Lets increase the line AB till point F and AF is also parallel to line DE. According to question ABC = 124 and CDE = 107. If you look at the given f igure caref ully, you will notice that CGF and CDE are corresponding angles. theref ore CGF = CDE [Corresponding angles] CGF = 107 [Since CDE = 107 ] Step 4 The angles of straight line add up to 180. Line AF and is a straight line Theref ore ABC + CBG = 180 CBG = 180 - ABC CBG = 180-124 [Since ABC = 124 ] CBG = 56 -----(1) and CGB + CGF = 180 CGB = 180 - CGF CGB = 180-107 CGB = 73 -----(2) Step 5 The sum of all three angles of a triangle is 180. Now in triangle BCG, CBG + BGC + BCG = 180 56 + 73 + BCG = 180 [Since CBG = 56 and BGC = 73 ] 129 + BCG = 180 BCG = 180-129 BCG = 51 Step 6 If you look at the given f igure caref ully, yoou will notice that BCD = BCG. Theref ore the value of angle BCD = 51.

(9) d. 105 ID : sg-9-lines-and-angles [12] It is given that line AB and CD and parallel lines and the third line (say EF) cuts the lines AB and CD at certain angle as shown in the f igure above. Let us redraw the f igure as below: a = c (vertically opposite angles) c = e (alternate interior angles) Theref ore we can write, a = c = e = g Again, b = d (vertically opposite angles) d = f (alternate interior angles) Theref ore we can write, b = d = f = h We know that sum of two adjacent angle is equal to 180. Theref ore, f rom the diagram, you can write, a + b = 180, b + c = 180, c + d = 180, d + a = 180 Here, d = 75 and e = x d + a = 180 75 + a = 180 a = 180-75 a = 105 As a is equal to e, x is 105. Theref ore, the value of x is 105.

ID : sg-9-lines-and-angles [13]

(11) 40 ID : sg-9-lines-and-angles [14] According to question DOC = 20 and OD is perpendicular to AB. Theref ore AOD = 90 and BOD = 90. DOC + AOC = AOD 20 + AOC = 90 [Since AOD = 90 and DOC = 20 ] AOC = 90-20 AOC = 70 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = 90 + 20-70 [Since BOD = 90, DOC = 20 and AOC = 70 ] = 40 Step 4 Theref ore BOC - AOC = 40 (12) 140 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

(13) 15 ID : sg-9-lines-and-angles [15] It is given that, CE bisect angle ACB. Theref ore, ACE = ACB/2 -----(1) In triangle ABC, CAB + ABC + ACB = 180...[Since the sum of all three angles of a triangle is 180 ] 60 + 30 + ACB = 180 90 + ACB = 180 ACB = 180-90 ACB = 90 ACB/2 = 90/2 ACE = 45...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = 180 60 + 90 + DCA = 180 DCA = 180-90 - 60 DCA = 30 Step 4 Now, DCE = ACE - DCA DCE = 45-30 DCE = 15 Step 5 Hence, the value of angle DCE is 15. (14) 54 If you look at the angles 107 and q + 53, these are opposite angles We know that opposite angles are equal. Theref ore q + 53 = 107 q = 107-53 q = 54

(15) 77 ID : sg-9-lines-and-angles [16] According to the question ADB is the right angle, ADB = 90, DAB = 35 In right angled triangle ΔADB, ADB + DAB + ABD = 180 [Since, the sum of all the angles of a triangle is equal to 180 ] 90 + 35 + ABD = 180 125 + ABD = 180 ABD = 55 In ΔBEC, ABD + BEC + 22 = 180 55 + BEC + 22 = 180 BEC + 77 = 180 BEC = 180-77 BEC = 103 x + BEC = 180 [Since, the sum of all angles on one side of a straight line is equal to 180.] x = 180 - BEC x = 180-103 x = 77 Step 4 Hence, the value of angle x is 77.