ID : ae-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 lines AC and BD intersects at point O such that AOB: BOC = 3:2, f ind AOD. (2) If AB and CD are parallel, f ind the value of angle x. (3) What is the angle between hour and minute hands, when clock shows 8:00 o'clock? (4) What is the value of the supplement of the complement of 77?
ID : ae-7-lines-and-angles [2] Choose correct answer(s) f rom given choice (5) If OD is perpendicular to AB, and DOC = 50, f ind ( BOC - AOC). a. 95 b. 105 c. 100 d. 110 (6) If AB and PQ are parallel, compute the angle Y, a. 263 b. 83 c. 97 d. 92 (7) If angles of a triangle are in ratio 5:3:2, the triangle is a. an obtuse angled triangle b. an acute angled triangle c. a right triangle d. an isosceles triangle (8) If AP and BP are bisectors of angles CAB and CBD respectively, f ind the angle APB. a. 40 b. 50 c. 30 d. 45
(9) If AB and CD are parallel, f ind the value of angle x. ID : ae-7-lines-and-angles [3] a. 80 b. 90 c. 70 d. 100 (10) If AB and DE are parallel, f ind the value of ACB a. 119 b. 114 c. 109 (11) If AC and EF are parallel, f ind ADB. a. 120 b. 110 c. 90 d. 100 (12) Which of f ollowing is f alse f or a triangle? a. Two angles are acute angles b. Two angles are obtuse angles c. One angle is obtuse angle d. Each angle is equal to 60 (13) If CD is perpendicular to AB, and CE bisect angle ACB, f ind the angle DCE. a. 12 b. 15 c. 17 d. 14
ID : ae-7-lines-and-angles [4] Fill in the blanks (14) If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 1:2, then the smaller of the two angles =. (15) a+b = 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ae-7-lines-and-angles [5] (1) 72 (2) 100 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, h = 100 and b = x As b is equal to h, x is 100. Theref ore, the value of x is 100.
(3) 120 ID : ae-7-lines-and-angles [6] At 8:00 o'clock, hour hand of the clock will be at 8 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 8:00 o'clock = 30 (12-8) = 120 (4) 167 If you look at the question caref ully, you will notice that f irst of all we have to f ind the complement of 77, then f ind the supplement of the complement of 77. T he sum of the complementary angles is 90. Theref ore the complement of 77 = 90-77 = 13 T he sum of supplementary angles is 180. Theref ore, the supplement of 13 = 180-13 = 167 Theref ore the value of the supplement of the complement of 77 is 167.
(5) c. 100 ID : ae-7-lines-and-angles [7] According to question DOC = 50 and OD is perpendicular to AB. Theref ore AOD = 90 and BOD = 90. DOC + AOC = AOD 50 + AOC = 90 [Since AOD = 90 and DOC = 50 ] AOC = 90-50 AOC = 40 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = 90 + 50-40 [Since BOD = 90, DOC = 50 and AOC = 40 ] = 100 Theref ore BOC - AOC = 100 (6) c. 97 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 = 50 Similarly angle Y2 and angle B are alternate angles. Theref ore angle Y2 = angle B angle Y2 = 47 Now angle Y = angle Y1 + angle Y2 = 50 + 47 = 97
(7) c. a right triangle ID : ae-7-lines-and-angles [8] According to the question, all angles of the triangle are in ratio 5:3:2. We can assume three angles of the triangle to be 5x, 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, 5x + 3x + 2x = 180 10x = 180 x = 180 10 Now, 5x = 5 180 10 = 90, 3x = 3 180 10 = 54 and 2x = 2 180 = 36. 10 Theref ore, the three angles of the triangle are 90, 54 and 36. Since, one of the angle of the triangle is 90, the triangle is a right triangle.
(8) a. 40 ID : ae-7-lines-and-angles [9] As per the question CBD is exterior angle of the triangle ABC and we know that an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. Theref ore, CBD = CAB + 80 ----(1) In triangle ABC, CAB + ABC + ACB = 180 CAB + ABC = 180-80 CAB + ABC = 100 -----(2) It is given that AP and BP are bisectors of angles CAB and CBD respectively. Theref ore, PAB = CAB/2 -----(3) CBP = CBD/2 -----(4) Now, in triangle ABP, PAB + ABP + APB = 180...[Since the sum of all three angles of a triangle is 180 ] APB = 180 - PAB - ABP APB = 180 - CAB/2 - ABP..[From equation (3), PAB = CAB/2] APB = 180 - CAB/2 - ( ABC + CBP) APB = 180 - CAB/2 - ( ABC + CBD/2)...[From equation (4), CBP = CBD/2] APB = 180 - CAB/2 - { ABC + ( CAB + 80 )/2}...[From equation (1)] APB = 180 - CAB/2 - ( ABC + CAB/2 + 40 ) APB = 180 - CAB/2 - ABC - CAB/2-40 APB = 140 - ( CAB + ABC) APB = 140-100...[From equation (2)] APB = 40 Step 5 Hence, the value of angle APB is 40.
(9) a. 80 ID : ae-7-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, c = 100 and h = x and c + d = 180 100 + d = 180 d = 180-100 d = 80 As d is equal to h, x is 80. Theref ore, the value of x is 80.
(10) b. 114 ID : ae-7-lines-and-angles [11] If you look at the f igure caref ully, you will notice that the angle BAC = 31, BDE = 35. According to question AB and DE are parallel. T heref ore the angles ABC and BDE are alternate interior angles. ABC = BDE [Alternate interior angles] ABC = 35 The sum of all three angles of a triangle is 180. Now in triangle ABC, ABC + BAC + ACB = 180 35 + 31 + ACB = 180 [Since ABC = 35 and BAC = 31 ] 66 + ACB = 180 ACB = 180-66 ACB = 114 Theref ore the value of ACB is 114.
(11) d. 100 ID : ae-7-lines-and-angles [12] If you look at the f igure caref ully, you will notice that ADE = 50 and DBC = 150. According to question AC and EF are parallel. Theref ore we can say that EDB and DBC are alternate interior angles. Now EDB = DBC [Alternate interior angles] ADE + ADB = DBC [Since EDB = ADE + ADB] ADB = 150 - ADE [Since ADB = 150 ] ADB = 150-50 [Since ADE = 50 ] ADB = 100 Theref ore ADB = 100 (12) b. Two angles are obtuse angles The sum of all three angles of a triangle must be 180. Now if you look at the all of the options caref ully, you will notice that the statement "Two angles are obtuse angles", can't satisf y the condition of a triangle and hence the statement "Two angles are obtuse angles" is f alse.
(13) b. 15 ID : ae-7-lines-and-angles [13] 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 ] 55 + 25 + ACB = 180 80 + ACB = 180 ACB = 180-80 ACB = 100 ACB/2 = 100/2 ACE = 50...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = 180 55 + 90 + DCA = 180 DCA = 180-90 - 55 DCA = 35 Now, DCE = ACE - DCA DCE = 50-35 DCE = 15 Step 5 Hence, the value of angle DCE is 15.
(14) 60 ID : ae-7-lines-and-angles [14] Let us assume that x be the f irst interior angle on the same side of a transversal intersecting two parallel lines. We know that the sum of two interior angles on the same side of a transversal intersecting two parallel lines is 180. T hus, the second interior angle on the same side of a transversal intersecting two parallel lines = 180 - x The ratio of the two interior angles on the same side of a transversal intersecting two x parallel lines = 180 - x It is given that the ratio of the two interior angles on the same side of a transversal intersecting two parallel lines = 1:2 x Theref ore, = 1 180 - x 2 By cross multiplying both sides 2x = 1(180 - x) 2x = 1 180-1x 2x + 1x = 180 3x = 180 x = 180 3 x = 60 The f irst angle = 60 The second angle = 180-60 = 120
Step 5 ID : ae-7-lines-and-angles [15] Thus, the smaller of the two angles is = 60