# Grade 7 Lines and Angles

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1 ID : ph-7-lines-and-angles [1] Grade 7 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 lines AB and CD intersects as shown below, f ind value of angle x. (3) If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 3:2, then f ind the greater of the two angles. Choose correct answer(s) f rom given choice (4) If OD is perpendicular to AB, and DOC = 50, f ind ( BOC - AOC). a. 95 b. 90 c. 100 d. 110

2 (5) If AB and CD are parallel, f ind the value of angle x. ID : ph-7-lines-and-angles [2] a. 110 b. 120 c. 70 d. 100 (6) Which of f ollowing two shapes can be joined together to f orm a semi-circle? (a) (b) (c) (d) a. d and b b. a and d c. a and c d. c and b (7) If AB and CD are parallel, f ind the value of angle x. a. 115 b. 105 c. 75 d. 95

3 (8) If AB and CD are parallel, f ind the value of x+y. ID : ph-7-lines-and-angles [3] a. 90 b. 80 c. 100 d. 110 (9) If AB and DE are parallel to each other, f ind value of angle BCD. a. 40 b. 55 c. 50 d. 45 (10) What is the angle between hour and minute hands, when clock shows 10:00 o'clock? a. 50 b. 30 c. 90 d. 60 (11) If AP and BP are bisectors of angles CAB and CBD respectively, f ind the angle APB. a. 45 b. 40 c. 50 d. 55 (12) If angles of a triangle are in ratio 4:6:3, the triangle is a. a right triangle b. an obtuse angled triangle c. an acute angled triangle d. an isosceles triangle

4 ID : ph-7-lines-and-angles [4] Fill in the blanks (13) If AD and BD are bisectors of CAB and CBA respectively, the value of ADB =. (14) If AB and CD are parallel, X = Check True/False (15) A triangle must have at least two acute angles. True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

5 Answers ID : ph-7-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 = 98 ACB/2 = 98/2 ACE = 49...[From equation (1)] Now in triangle ADC, CAD + ADC + DCA = DCA = 180 DCA = DCA = 35 Step 4 Now, DCE = ACE - DCA DCE = DCE = 14 Step 5 Hence, the value of angle DCE is 14.

6 (2) 25 ID : ph-7-lines-and-angles [6] In triangle BCE, CBE + BCE + BEC = [Since the sum of all three angles of a triangle is 180 ] BEC = BEC = 180 BEC = BEC = 25 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 Step 4 Hence, the value of x is 25.

7 (3) 108 ID : ph-7-lines-and-angles [7] 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 = x The ratio of the two interior angles on the same side of a transversal intersecting two x parallel lines = x It is given that the ratio of the two interior angles on the same side of a transversal intersecting two parallel lines = 3:2 x Theref ore, = x 2 By cross multiplying both sides 2x = 3(180 - x) 2x = x 2x + 3x = 540 5x = 540 x = x = 108 Step 4 The f irst angle = 108 The second angle = = 72

8 Step 5 ID : ph-7-lines-and-angles [8] Thus, the greater of the two angles is = 108 (4) c. 100 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 = AOC = 40 Now BOC - AOC = BOD + DOC - AOC [Since BOC = BOD + DOC] = [Since BOD = 90, DOC = 50 and AOC = 40 ] = 100 Step 4 Theref ore BOC - AOC = 100

9 (5) a. 110 ID : ph-7-lines-and-angles [9] 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, f = 70 and a = x f + g = g = 180 g = g = 110 As g is equal to a, x is 110. Theref ore, the value of x is 110.

10 ID : ph-7-lines-and-angles [10]

11 (7) b. 105 ID : ph-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, a = 75 and f = x a + d = d = 180 d = d = 105 As d is equal to f, x is 105. Theref ore, the value of x is 105.

12 ID : ph-7-lines-and-angles [12]

13 (9) c. 50 ID : ph-7-lines-and-angles [13] Lets increase the line AB till point F and AF is also parallel to line DE. According to question ABC = 113 and CDE = 117. 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 = 117 [Since CDE = 117 ] Step 4 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 = 113 ] 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 = 67 and BGC = 63 ] BCG = 180 BCG = BCG = 50 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 = 50.

14 (10) d. 60 ID : ph-7-lines-and-angles [14] At 10:00 o'clock, hour hand of the clock will be at 10 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. Step 4 Theref ore the angle between hour and minute hands, when clock shows 10:00 o'clock = 30 (12-10) = 60

15 (11) a. 45 ID : ph-7-lines-and-angles [15] 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 + 90 )/2}...[From equation (1)] APB = CAB/2 - ( ABC + CAB/ ) APB = CAB/2 - ABC - CAB/2-45 APB = ( CAB + ABC) APB = [From equation (2)] APB = 45 Step 5 Hence, the value of angle APB is 45.

16 (12) c. an acute angled triangle ID : ph-7-lines-and-angles [16] According to the question, all angles of the triangle are in ratio 4:6:3. We can assume three angles of the triangle to be 4x, 6x and 3x where x is common f actor. We know that the sum of the three angles of a triangle is 180. Theref ore, 4x + 6x + 3x = x = 180 x = Now, 4x = = 55.38, 6x = = and 3x = = Theref ore, the three angles of the triangle are 55.38, and Since, all angles of the triangle are less than 90, the triangle is an acute angled triangle.

17 (13) 140 ID : ph-7-lines-and-angles [17] 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 = 180 CAB + CBA = CAB + CBA = 80 CAB/2 + CBA/2 = 80/2 = (3) Now, In triangle ABD, BAD + ABD + ADB = 180 CAB/2 + CBA/2 + ADB = Using (1) &(2) 40 + ADB = Using (3) ADB = = 140 Step 4 Hence, ADB = 140

18 (14) 21 ID : ph-7-lines-and-angles [18] 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

19 (15) True ID : ph-7-lines-and-angles [19] 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 must have at least two acute angles" is True.

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