Grade 9 Lines and Angles - PDF Free Download

ID : ww-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 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

(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?

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

Answers ID : ww-9-lines-and-angles [5] (1) 14 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 ] 50 + 22 + ACB = 180 72 + ACB = 180 ACB = 180-72 ACB = 108 ACB/2 = 108/2 ACE = 54...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = 180 50 + 90 + DCA = 180 DCA = 180-90 - 50 DCA = 40 Now, DCE = ACE - DCA DCE = 54-40 DCE = 14 Step 5 Hence, the value of angle DCE is 14.

(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 = 90-30 AOC = 60 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = 90 + 30-60 [Since BOD = 90, DOC = 30 and AOC = 60 ] = 60 Theref ore BOC - AOC = 60

(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 ] 90 + 45 + ABD = 180 135 + ABD = 180 ABD = 45 In ΔBEC, ABD + BEC + 32 = 180 45 + BEC + 32 = 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 Hence, the value of angle x is 77.

(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 = 180-81 = 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 = 180 9 Now, 4x = 4 180 9 = 80, 3x = 3 180 9 = 60 and 2x = 2 180 = 40. 9 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.

(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 = 180 100 + COE = 180 [Since AOC + BOE = 100 ] COE = 180-100 COE = 80 CD is also a straight line, theref ore COE + BOE + BOD = 180 80 + BOE + 70 = 180 [Since COE = 80 and BOD = 70 ] BOE + 150 = 180 BOE = 180-150 BOE = 30. Now BOE = 30.

(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 = 47 + 49 = 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 150 + a + b = 360 a + b = 360-150 a + b = 210 Theref ore the value of a + b is 210.

(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 = 180 - 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

(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

(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 = 180 - ABC CBG = 180-117 [Since ABC = 117 ] CBG = 63 -----(1) and CGB + CGF = 180 CGB = 180 - CGF CGB = 180-114 CGB = 66 -----(2) Step 5 The sum of all three angles of a triangle is 180. Now in triangle BCG, CBG + BGC + BCG = 180 63 + 66 + BCG = 180 [Since CBG = 63 and BGC = 66 ] 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.

(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.

(13) 25 ID : ww-9-lines-and-angles [15] 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 + 120 + 35 = 180...Using (1) 155 + x = 180 x = 180-155 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, 130 + y = 180 or y = 180-130 = 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]

(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 = 180 - 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.