Grade 9 Lines and Angles

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1 ID : ww-9-lines-and-angles [1] Grade 9 Lines and Angles For more such worksheets visit Answer t he quest ions (1) If CD is perpendicular to AB, and CE bisect angle ACB, f ind the angle DCE. (2) If OD is perpendicular to AB, and DOC = 30, f ind ( BOC - AOC). (3) If ADB is a right angle, f ind the value of angle x. (4) What is the value of the supplement of the complement of 9? Choose correct answer(s) f rom given choice (5) If angles of a triangle are in ratio 4:3:2, the triangle is a. an acute angled triangle b. an isosceles triangle c. an obtuse angled triangle d. a right triangle

2 (6) Lines AB and CD intersect at O. If AOC + BOE = 100 and BOD = 70, f ind BOE. ID : ww-9-lines-and-angles [2] a. 20 b. 45 c. 30 d. 40 (7) If AB and PQ are parallel, compute the angle Y, (8) a. 96 b. 89 c. 264 d. 84 Find the value of a+b. a. 150 b. 210 c. 200 d. 220 Fill in the blanks (9) ABC is an isosceles triangle with AB=AC. We extend the segment AB to D such that AB=AD. The value of BCD =? (10) The angle between hour and minute hands is, when clock shows 7:00 o'clock?

3 ID : ww-9-lines-and-angles [3] (11) If AB and DE are parallel to each other, the value of angle BCD =. (12) If AB and CD are parallel, value of angle x is. (13) If lines AB and CD intersects as shown below, the value of angle x =. (14) If two horizontal lines are parallel, value of angle x is. Check True/False (15) A triangle can have two obtuse angles. True False

5 Answers ID : ww-9-lines-and-angles [5] (1) 14 It is given that, CE bisect angle ACB. Theref ore, ACE = ACB/ (1) In triangle ABC, CAB + ABC + ACB = [Since the sum of all three angles of a triangle is 180 ] ACB = ACB = 180 ACB = ACB = 108 ACB/2 = 108/2 ACE = 54...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = DCA = 180 DCA = DCA = 40 Now, DCE = ACE - DCA DCE = DCE = 14 Step 5 Hence, the value of angle DCE is 14.

6 (2) 60 ID : ww-9-lines-and-angles [6] According to question DOC = 30 and OD is perpendicular to AB. Theref ore AOD = 90 and BOD = 90. DOC + AOC = AOD 30 + AOC = 90 [Since AOD = 90 and DOC = 30 ] AOC = AOC = 60 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = [Since BOD = 90, DOC = 30 and AOC = 60 ] = 60 Theref ore BOC - AOC = 60

7 (3) 77 ID : ww-9-lines-and-angles [7] According to the question ADB is the right angle, ADB = 90, DAB = 45 In right angled triangle ΔADB, ADB + DAB + ABD = 180 [Since, the sum of all the angles of a triangle is equal to 180 ] ABD = ABD = 180 ABD = 45 In ΔBEC, ABD + BEC + 32 = BEC + 32 = 180 BEC + 77 = 180 BEC = BEC = 103 x + BEC = 180 [Since, the sum of all angles on one side of a straight line is equal to 180.] x = BEC x = x = 77 Hence, the value of angle x is 77.

8 (4) 99 ID : ww-9-lines-and-angles [8] If you look at the question caref ully, you will notice that f irst of all we have to f ind the complement of 9, then f ind the supplement of the complement of 9. T he sum of the complementary angles is 90. Theref ore the complement of 9 = 90-9 = 81 T he sum of supplementary angles is 180. Theref ore, the supplement of 81 = = 99 Theref ore the value of the supplement of the complement of 9 is 99. (5) a. an acute angled triangle According to the question, all angles of the triangle are in ratio 4:3:2. We can assume three angles of the triangle to be 4x, 3x and 2x where x is common f actor. We know that the sum of the three angles of a triangle is 180. Theref ore, 4x + 3x + 2x = 180 9x = 180 x = Now, 4x = = 80, 3x = = 60 and 2x = = Theref ore, the three angles of the triangle are 80, 60 and 40. Since, all angles of the triangle are less than 90, the triangle is an acute angled triangle.

9 (6) c. 30 ID : ww-9-lines-and-angles [9] If you look at the given f igure caref ully, you will notice that AB and CD are straight lines. AOC + BOE = 100 and BOD = 70. 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 = COE = 180 [Since AOC + BOE = 100 ] COE = COE = 80 CD is also a straight line, theref ore COE + BOE + BOD = BOE + 70 = 180 [Since COE = 80 and BOD = 70 ] BOE = 180 BOE = BOE = 30. Now BOE = 30.

10 (7) a. 96 ID : ww-9-lines-and-angles [10] Draw a line MN which is parallel to line PQ and line AB and angle Y = angle Y1 + angle Y2 Angle Y1 and angle Q are alternate angles. Theref ore angle Y1 = angle Q angle Y1 = 47 Similarly angle Y2 and angle B are alternate angles. Theref ore angle Y2 = angle B angle Y2 = 49 Now angle Y = angle Y1 + angle Y2 = = 96 (8) b. 210 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 a + b = 360 a + b = a + b = 210 Theref ore the value of a + b is 210.

11 (9) 90 ID : ww-9-lines-and-angles [11] Take a look at the image below ABC is an isosceles triangle with AB = AC, theref ore ABC = ACB = x Let CAB = y. Then y + x + x = 180 y+2x = 180 ACB = x = 180-y 2 Since we extended BA to f orm line BD, BAC + DAC = 180 y + DAC = 180 DAC = y CAD is also an isosceles triangle since AD = AC (remember AD = AB, and AB = AC) So CDA = DCA = z CDA + DCA + DAC = 180 z + z + (180-y) = 180 2z = y DCA = z = y 2 Now DCB = DCA + ACB = 180-y 2 + y 2 = 90

12 (10) 150 ID : ww-9-lines-and-angles [12] At 7:00 o'clock, hour hand of the clock will be at 7 and minute hand will be at 12. In clock a whole circle is divided into 12 parts, where each part represents an hour. T heref ore, angle between consecutive numbers on clock, = 360 /12 = 30 At 12 o'clock the angle between hour and minute hands is 0 and the angle increases by 30 till 6 o'clock f or every hour. Af ter 6 o'clock the angle decreases by 30 f or every hour. Theref ore the angle between hour and minute hands, when clock shows 7:00 o'clock = 30 (12-7) = 150

13 (11) 51 ID : ww-9-lines-and-angles [13] Lets increase the line AB till point F and AF is also parallel to line DE. According to question ABC = 117 and CDE = 114. 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 = 114 [Since CDE = 114 ] The angles of straight line add up to 180. Line AF and is a straight line Theref ore ABC + CBG = 180 CBG = ABC CBG = [Since ABC = 117 ] CBG = (1) and CGB + CGF = 180 CGB = CGF CGB = CGB = (2) Step 5 The sum of all three angles of a triangle is 180. Now in triangle BCG, CBG + BGC + BCG = BCG = 180 [Since CBG = 63 and BGC = 66 ] BCG = 180 BCG = 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.

14 (12) 105 ID : ww-9-lines-and-angles [14] 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, a = 105 and e = x As a is equal to e, x is 105. Theref ore, the value of x is 105.

15 (13) 25 ID : ww-9-lines-and-angles [15] In triangle BCE, CBE + BCE + BEC = [Since the sum of all three angles of a triangle is 180 ] BEC = BEC = 180 BEC = 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 = (1) Now, in triangle ADE, DAE + ADE + AED = [Since the sum of all three angles of a triangle is 180 ] x = Using (1) x = 180 x = x = 25 Hence, the value of x is 25. (14) 50 We know that the angle made by a straight line is 180. Theref ore, we can write, y = 180 or y = = 50 When a straight line cuts any two parallel lines, its Alternate Angles are equal. Since angles x and y are Alternate angles. Theref ore, x = y = 50...[Alternate angles of two parallel lines are equal]

16 (15) False ID : ww-9-lines-and-angles [16] Let's consider the triangle ABC in the f igure above. Since we know that the sum of all three angles of a triangle is 180, in ΔABC: A + B + C = 180. Let's assume that A of the ΔABC is an obtuse angle. That is, A > 90. Now, A + B + C = 180 B + C = A B + C < 90 (Since A > 90 ) We just saw that the sum of B and C of the ΔABC is less than 90. Theref ore, we can say that the B and the C must be acute angles and the statement "A triangle can have two obtuse angles" is False.

Grade 7 Lines and Angles

ID : ph-7-lines-and-angles [1] Grade 7 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) If CD is perpendicular to AB, and CE bisect angle ACB, f ind the angle