Class 7 Lines and Angles

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1 ID : in-7-lines-and-angles [1] Class 7 Lines and Angles For more such worksheets visit Answer t he quest ions (1) If AD and BD are bisectors of CAB and CBA respectively, f ind sum of angle x and y. Choose correct answer(s) f rom given choice (2) If angles of a triangle are in ratio 5:3:8, the triangle is a. an acute angled triangle b. a right triangle c. an isosceles triangle d. an obtuse angled triangle (3) If AD and BD are bisectors of CAB and CBA respectively, f ind value of ADB. a. 143 b. 128 c. 135 d. 138 (4) The value of the supplement of the complement of 77 is : a. 13 b. 86 c. 167 d. none of these

2 ID : in-7-lines-and-angles [2] Fill in the blanks (5) If CD is perpendicular to AB, and CE bisect angle ACB, the angle DCE =. (6) If AP and BP are bisectors of angles CAB and CBD respectively, the angle APB.. (7) Find the angle between A) South-West and South : B) South and North : C) West and East : D) North and West : (8) If AB and CD are parallel, value of angle x is.

3 (9) If lines AB and CD intersects as shown below, the value of angle x =. ID : in-7-lines-and-angles [3] (10) If OD is perpendicular to AB, and DOC = 50, ( BOC - AOC). =. (11) If AB and CD are parallel, X = (12) If AB and DE are parallel, f ind the value of ACB

4 ID : in-7-lines-and-angles [4] (13) If two horizontal lines are parallel, value of angle x is. (14) Value of angle X is Check True/False (15) A triangle can have two obtuse angles. True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

5 Answers ID : in-7-lines-and-angles [5] (1) 35 It is given that AD and BD are bisectors of CAB and CBA respectively. Theref ore, x = CAB/ (1) y = CBA/ (2) In triangle ABC, CAB + CBA + ACB = [The sum of all three angles of a triangle is 180 ] CAB + CBA = 180 CAB + CBA = CAB + CBA = 70 CAB/2 + CBA/2 = 70/2 x + y = 35...[From equation (1) and (2)] Hence, the sum of the angles x and y is 35. (2) b. a right triangle According to the question, all angles of the triangle are in ratio 5:3:8. We can assume three angles of the triangle to be 5x, 3x and 8x where x is common f actor. We know that the sum of the three angles of a triangle is 180. Theref ore, 5x + 3x + 8x = x = 180 x = Now, 5x = = 56.25, 3x = = and 8x = = Theref ore, the three angles of the triangle are 56.25, and 90. Since, one of the angle of the triangle is 90, the triangle is a right triangle.

6 (3) c. 135 ID : in-7-lines-and-angles [6] It is given that AD and BD are bisectors of CAB and CBA respectively. Theref ore, BAD = CAB/ (1) ABD = CBA/ (2) In triangle ABC, CAB + CBA + ACB = [The sum of all three angles of a triangle is 180 ] CAB + CBA + 90 = 180 CAB + CBA = CAB + CBA = 90 CAB/2 + CBA/2 = 90/2 = (3) Now, In triangle ABD, BAD + ABD + ADB = 180 CAB/2 + CBA/2 + ADB = Using (1) &(2) 45 + ADB = Using (3) ADB = = 135 Step 4 Hence, ADB = 135 (4) c. 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 = = 13 T he sum of supplementary angles is 180. Theref ore, the supplement of 13 = = 167 Step 4 Theref ore the value of the supplement of the complement of 77 is 167.

7 (5) 11 ID : in-7-lines-and-angles [7] 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 = 102 ACB/2 = 102/2 ACE = 51...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = DCA = 180 DCA = DCA = 40 Step 4 Now, DCE = ACE - DCA DCE = DCE = 11 Step 5 Hence, the value of angle DCE is 11.

8 (6) 55 ID : in-7-lines-and-angles [8] 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 (1) In triangle ABC, CAB + ABC + ACB = 180 CAB + ABC = CAB + ABC = (2) It is given that AP and BP are bisectors of angles CAB and CBD respectively. Theref ore, PAB = CAB/ (3) CBP = CBD/ (4) Step 4 Now, in triangle ABP, PAB + ABP + APB = [Since the sum of all three angles of a triangle is 180 ] APB = PAB - ABP APB = CAB/2 - ABP..[From equation (3), PAB = CAB/2] APB = CAB/2 - ( ABC + CBP) APB = CAB/2 - ( ABC + CBD/2)...[From equation (4), CBP = CBD/2] APB = CAB/2 - { ABC + ( CAB )/2}...[From equation (1)] APB = CAB/2 - ( ABC + CAB/ ) APB = CAB/2 - ABC - CAB/2-55 APB = ( CAB + ABC) APB = [From equation (2)] APB = 55 Step 5 Hence, the value of angle APB is 55.

9 (7) A) 45 ID : in-7-lines-and-angles [9] Let us look at the directions as shown below: We can see that the angle between the South-West and South directions is 45. B) 180 Let us look at the directions as shown below: We can see that the angle between the South and East directions is 90. Also, the angle between the East and North directions is 90. Theref ore, the angle between the South and North directions is = = 180

10 C) 180 ID : in-7-lines-and-angles [10] Let us look at the directions as shown below: We can see that the angle between the West and South directions is 90. Also, the angle between the South and East directions is 90. Theref ore, the angle between the West and East directions is = = 180 D) 90 Let us look at the directions as shown below: We can see that the angle between the North and West directions is 90.

11 (8) 50 ID : in-7-lines-and-angles [11] 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, e = 130 and d = x e + h = h = 180 h = h = 50 As h is equal to d, x is 50. Theref ore, the value of x is 50.

12 ID : in-7-lines-and-angles [12] (9) 120 In triangle BCE, CBE + BCE + BEC = [Since the sum of all three angles of a triangle is 180 ] BEC = BEC = 180 BEC = BEC = 30 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 ] 30 + x + 30 = Using (1) 60 + x = 180 x = x = 120 Step 4 Hence, the value of x is 120.

13 (11) 21 ID : in-7-lines-and-angles [13] Parallel line AB and CD are intersected by a transversal as shown below, Here angle P and Q are corresponding angles. i.e. P = Q On comparing given angles with P and Q, 3x + 2x = 105 5x = 105 x = 21

14 (12) 81 ID : in-7-lines-and-angles [14] If you look at the f igure caref ully, you will notice that the angle BAC = 49, BDE = 50. 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 = 50 The sum of all three angles of a triangle is 180. Now in triangle ABC, ABC + BAC + ACB = ACB = 180 [Since ABC = 50 and BAC = 49 ] 99 + ACB = 180 ACB = ACB = 81 Step 4 Theref ore the value of ACB is 81.

15 (13) 55 ID : in-7-lines-and-angles [15] We know that the angle made by a straight line is 180. Theref ore, we can write, y = 180 or y = = 55 When a straight line cuts any two parallel lines, its Corresponding Angles are equal. Since angles x and y are Corresponding angles. Theref ore, x = y = 55...[Corresponding angles of two parallel lines are equal] (14) 54 If you look at the angles 99 and X + 45, these are opposite angles We know that opposite angles are equal. Theref ore X + 45 = 99 X = X = 54

16 (15) False ID : in-7-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 9 Lines and Angles

ID : cn-9-lines-and-angles [1] Grade 9 Lines and Angles For more such worksheets visit www.edugain.com Answer the questions (1) If AB and CD are parallel, find the value of x. (2) Lines AB and CD intersect