ID : ww-7-lines-and-angles [1] Grade 7 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) WXYZ is a quadrilateral whose diagonals intersect each other at the point O such that OW = OX = OZ. If OWX = 25, then f ind the measures of OZW. (2) If AB and PQ are parallel, compute the angle Z. Choose correct answer(s) f rom given choice (3) If lines AC and BD intersects at point O such that AOB: BOC = 1:2, f ind AOD. a. 128 b. 113 c. 121 d. 120
(4) If two horizontal lines are parallel, f ind the value of angle x. ID : ww-7-lines-and-angles [2] a. 75 b. 85 c. 105 d. 65 (5) Which of f ollowing two shapes can be joined together to f orm a semi-circle? (a) (b) (c) (d) a. b and a b. d and a c. c and d d. c and b (6) If AD and BD are bisectors of CAB and CBA respectively, f ind value of ADB. a. 145 b. 136 c. 155 d. 150
(7) If AB and CD are parallel, f ind the value of x+y. ID : ww-7-lines-and-angles [3] a. 70 b. 90 c. 60 d. 80 (8) If AB and CD are parallel, f ind the value of angle x. a. 120 b. 60 c. 110 d. 130 (9) If AB and DE are parallel to each other, f ind value of angle BCD. a. 38 b. 53 c. 48 d. 43 (10) If AP and BP are bisectors of angles CAB and CBD respectively, f ind the angle APB. a. 35 b. 40 c. 45 d. 55
(11) Find value of angle x ID : ww-7-lines-and-angles [4] a. 50 b. 60 c. 80 d. 70 Fill in the blanks (12) If AB and DE are parallel, f ind the value of ACB (13) If AC and EF are parallel, ADB = Check True/False (14) A triangle can have two obtuse angles. True False (15) A triangle must have at least two acute angles. True False 2016 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : ww-7-lines-and-angles [5] (1) 65 (2) 88 Construction: Draw a line MN which is parallel to the line PQ and the line AB. If you look at the given f igure caref ully, you will notice that, the addition of the angle Z1, angle Z2 and angle Z makes f ull angle and f ull angle is exactly 360. Theref ore, angle Z1 + angle Z2 + angle Z = 360, angle Z = 360 - angle Z1 - angle Z2. Angle P = 40, angle Z1 and angle P are alternate angles, angle Z1 = angle P, angle Z1 = 40. Similarly, angle A = 48, angle Z2 and angle A are alternate angles, angle Z2 = angle A angle Z2 = 48. Step 4 Angle Z = angle Z1 + angle Z2 = 40 + 48 = 88. Step 5 Theref ore, the angle Z is 88. (3) d. 120
(4) a. 75 ID : ww-7-lines-and-angles [6] When a straight line cuts any two parallel lines, its Alternate Angles are equal. Since angles x and 75 are Alternate angles. Theref ore, x = 75...[Alternate angles of two parallel lines are equal] (5) d. c and b The angle f ormed in a semi-circle is 180. If you look at the f ollowing shapes caref ully, you will notice that shapes c and b makes an angle 180 by joining and hence the shapes c and b can be joined together to f orm a semi-circle. (a) (b) (c) (d)
(6) a. 145 ID : ww-7-lines-and-angles [7] It is given that AD and BD are bisectors of CAB and CBA respectively. Theref ore, BAD = CAB/2------(1) ABD = CBA/2-------(2) In triangle ABC, CAB + CBA + ACB = 180...[The sum of all three angles of a triangle is 180 ] CAB + CBA + 110 = 180 CAB + CBA = 180-110 CAB + CBA = 70 CAB/2 + CBA/2 = 70/2 = 35 -------(3) Now, In triangle ABD, BAD + ABD + ADB = 180 CAB/2 + CBA/2 + ADB = 180...Using (1) &(2) 35 + ADB = 180...Using (3) ADB = 180-35 = 145 Step 4 Hence, ADB = 145 (7) a. 70 According to question AB and CD are parallel, theref ore x = 30 [Alternate interior angles] y = 40 [Alternate interior angles] Now the value of x+y = 30 + 40 = 70
(8) a. 120 ID : ww-7-lines-and-angles [8] 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, h = 60 and a = x and h + g = 180 60 + g = 180 g = 180-60 g = 120 As g is equal to a So, x is 120. Theref ore, the value of x is 120.
(9) c. 48 ID : ww-7-lines-and-angles [9] Lets increase the line AB till point F and AF is also parallel to line DE. According to question ABC = 120 and CDE = 108. 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 = 108 [Since CDE = 108 ] Step 4 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 = 180-120 [Since ABC = 120 ] CBG = 60 -----(1) and CGB + CGF = 180 CGB = 180 - CGF CGB = 180-108 CGB = 72 -----(2) Step 5 The sum of all three angles of a triangle is 180. Now in triangle BCG, CBG + BGC + BCG = 180 60 + 72 + BCG = 180 [Since CBG = 60 and BGC = 72 ] 132 + BCG = 180 BCG = 180-132 BCG = 48 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 = 48.
(10) c. 45 ID : ww-7-lines-and-angles [10] 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 + 90 ----(1) In triangle ABC, CAB + ABC + ACB = 180 CAB + ABC = 180-90 CAB + ABC = 90 -----(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) Step 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 + 90 )/2}...[From equation (1)] APB = 180 - CAB/2 - ( ABC + CAB/2 + 45 ) APB = 180 - CAB/2 - ABC - CAB/2-45 APB = 135 - ( CAB + ABC) APB = 135-90...[From equation (2)] APB = 45 Step 5 Hence, the value of angle APB is 45.
(11) b. 60 ID : ww-7-lines-and-angles [11] If you look at the f igure caref ully, you will notice that line AB is a straight line. The angles of straight line add up to 180. Line AB is a straight line, theref ore x + 70 + 50 = 180 120 + x = 180 x = 180-120 x = 60. Theref ore the value of angle x is 60.
(12) 124 ID : ww-7-lines-and-angles [12] If you look at the f igure caref ully, you will notice that the angle BAC = 21, 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 + 21 + ACB = 180 [Since ABC = 35 and BAC = 21 ] 56 + ACB = 180 ACB = 180-56 ACB = 124 Step 4 Theref ore the value of ACB is 124.
(13) 70 ID : ww-7-lines-and-angles [13] If you look at the f igure caref ully, you will notice that ADE = 55 and DBC = 125. 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 = 125 - ADE [Since ADB = 125 ] ADB = 125-55 [Since ADE = 55 ] ADB = 70 Step 4 Theref ore ADB = 70
(14) False ID : ww-7-lines-and-angles [14] 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.
(15) True ID : ww-7-lines-and-angles [15] 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 must have at least two acute angles" is True.