Class 7 Lines and Angles


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1 ID : in7linesandangles [1] Class 7 Lines and Angles For more such worksheets visit Answer the questions (1) ABCD is a quadrilateral whose diagonals intersect each other at point O such that OA = OB = OD. If OAB = 50, then find the measures of ODA. (2) If lines AB and CD intersect as shown below, find the value of angle x. (3) If lines AC and BD intersects at point O such that AOB: BOC = 4:5, find COD. (4) If AB and PQ are parallel, compute the measure of angle Z.
2 (5) If AB and PQ are parallel, compute the measure of W. ID : in7linesandangles [2] (6) If lines XY and MN intersect as shown below and a:b = 4:5, find c. (7) If OD is perpendicular to AB, and DOC = 25, find BOC  AOC. (8) If AB and CD are parallel, find the measure of angle x.
3 (9) Compute the measure of q. ID : in7linesandangles [3] (10) If AB and CD are parallel, find the value of x. (11) If CD is perpendicular to AB and CE bisects angle ACB, find DCE. (12) AP and BP are bisectors of CAB and CBD, respectively. Find APB. (13) If AD and BD are bisectors of CAB and CBA, respectively. Find the sum of angles x and y.
4 ID : in7linesandangles [4] Choose correct answer(s) from the given choices (14) Which of the following two shapes can be joined together to form a semicircle? (a) (b) (c) (d) a. d and b b. b and c c. d and a d. a and c (15) Which of the following is false for a triangle? a. Each angle is equal to 60 b. Two angles are acute angles c. One angle is an obtuse angle d. Two angles are right angles 2017 Edugain ( All Rights Reserved Many more such worksheets can be generated at
5 Answers ID : in7linesandangles [5] (1) 40 We are given that OA = OB = OD. OAB = 50 In triangle OAB, OA = OB. Therefore, triangle OAB is an isosceles triangle. So, OAB = OBA = 50...(1) Similarly, in triangle ODA, OA = OD. Therefore, triangle ODA too is an isosceles triangle. So, OAD = ODA...(2) In triangle ABD, Sum of all the three angles = 180 ADB + DBA + BAD = 180 ADB + DBA + BAO + OAD = 180 ODA ODA = 180 (Using the facts (1) and (2) we concluded in the previous step) ODA = 40.
6 (2) 115 ID : in7linesandangles [6] In triangle ADE, DAE + ADE + AED = (Since, the sum of all the angles of a triangle is 180 ) AED = AED = 180 AED = AED = 30 Since, BEC and AED are the opposite angles of the triangle BCE and ADE and we know that the opposite angles of a triangle are equal. Therefore, BEC = AED = (1) Now, in triangle BCE, CBE + BCE + BEC = (Since, the sum of all the angles of a triangle is 180 ) 35 + x + 30 = Using (1) 65 + x = 180 x = x = 115 Step 4 Hence, the value of x is 115.
7 (3) 80 ID : in7linesandangles [7] Given, AOB: BOC = 4:5, or AOB BOC 5 AOB = 4 BOC AOB = 4 BOC 5 = 4 5, cross multiplying the fractions we get, Since AC is a straight line, we can say that AOB + BOC = 180.Substituting the value of AOB, in the above equation. We have, 4 BOC + BOC = BOC 5 = 180 BOC = = 100. Since AC and BD are straight lines that intersect, COD = BOC = = 80 Step 4 This means that COD = 80
8 (4) 98 ID : in7linesandangles [8] Draw a line MN that is parallel to lines PQ and AB. We know, Z = Z1 + Z2 Now, AB and NM are two parallel lines cut by a transversal OA. Hence, NOA = OAB (Alternate angles) Z2 = 42 Now, PQ and NM are two parallel lines cut by a transversal OP. Hence, NOP = OPQ (Alternate angles) Z1 = 56 Step 4 We know, Z = Z1 + Z2 = = 98 Hence, Z = 98.
9 (5) 96 ID : in7linesandangles [9] Draw a line MN parallel to both lines PQ and AB. According to the figure, W1 + W2 + W = 360 W = W1  W2 Now, AB and MN are two parallel lines cut by a transversal OA. Hence, MOA = OAB (Alternate angles) W2 = 46 Now, PQ and MN are two parallel lines cut by a transversal OP. Hence, MOP = OPQ (Alternate angles) W1 = 50 Step 4 We know, W = W1 + W2 = = 96
10 (6) 130 ID : in7linesandangles [10] As XY is a straight line, and the sum of the angles on a straight line is equal to 180. We have, a + b + 90 = 180 a + b = 90. We may notice that the angles MOY and XON are vertically opposite angles. So, a + 90 = c (vertically opposite angles are equal). We are given a:b = 4:5, or a b = 4 5. Cross multiplying the fractions, we get, 5a = 4b. Step 4 We put b = 5a 4 in a + b = 90 and get, a + 5a 4 = 90, or 9a 4 = 90, or 9a = 360. Dividing each side by 9, we get, a = 360, or Step 5 Now, since b = 5a 4. We can say that b = , or b = 50. Step 6 From step 2, we have, c = a + 90, or c = , or c = 130.
11 (7) 50 ID : in7linesandangles [11] According to the question, DOC = 25 and OD is perpendicular to AB. Therefore, AOD = 90 and BOD = 90. Also, DOC + AOC = AOD 25 + AOC = 90 (As, AOD = 90 and DOC = 25 ) AOC = AOC = 65 Now, BOC  AOC = BOD + DOC  AOC (As, BOC = BOD + DOC) = (As, BOD = 90, DOC = 25 and AOC = 65 ) = 50 Step 4 Therefore, BOC  AOC = 50 (8) 21 The parallel lines AB and CD are intersected by a transversal as shown below, Here, P and Q are corresponding angles, i.e., P = Q On comparing the given angles with P and Q, 3x + 2x = 105 5x = 105 Hence, x = 21.
12 (9) 47 ID : in7linesandangles [12] If we look at the angles 93 and (q + 46), we find that they are vertically opposite angles. We know that the vertically opposite angles are equal. Therefore, (q + 46) = 93 q = q = 47
13 (10) 65 ID : in7linesandangles [13] It is given that line AB and CD are parallel lines and the third line (say EF) cuts them as shown in the figure. a = c (Vertically opposite angles) c = e (Alternate interior angles) Therefore, we can write, a = c = e = g Also, b = d (Vertically opposite angles) d = f (Alternate interior angles) Therefore, we can write, b = d = f = h We know that the sum of two adjacent angles is equal to 180. Therefore, we can write, a + b = 180, b + c = 180, c + d = 180, d + a = 180 Given, h = 115 and a = x and h + g = g = 180 g = g = 65 As, g is equal to a. So, x is 65. Hence, the value of x is 65.
14 (11) 12 ID : in7linesandangles [14] It is given that, CE bisects angle ACB. Therefore, ACE = ACB/ (1) In triangle ABC, CAB + ABC + ACB = (Since, the sum of all the angles of a triangle is 180 ) ACB = ACB = 180 ACB = ACB = 84 ACB/2 = 84/2 ACE = 42...(From equation (1)) Now, in triangle ADC, CAD + ADC + DCA = DCA = 180 DCA = DCA = 30 Step 4 Now, DCE = ACE  DCA DCE = DCE = 12 Step 5 Hence, the value of DCE is 12.
15 (12) 55 ID : in7linesandangles [15] According to the question, CBD is the exterior angle of triangle ABC and we know that an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles. Therefore, 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 CAB and CBD, respectively. Therefore, PAB = CAB/ (3) CBP = CBD/ (4) Step 4 Now, in triangle ABP, PAB + ABP + APB = (Since, the sum of all the angles of a triangle is 180 ) APB = PAB  ABP APB = CAB/2  ABP..(From equation (3), PAB = CAB/2) APB = 180 CAB/2  ( ABC + CBP) APB = CAB/2  ( ABC + CBD/2)...(From equation (4), CBP = CBD/2) APB = 180 CAB/2  { ABC + ( CAB )/2}...(From equation (1)) APB = 180 CAB/2  ( ABC + CAB/ ) APB = 180 CAB/2  ABC  CAB/255 APB = ( CAB + ABC) APB = (From equation (2)) APB = 55 Step 5 Hence, the value of APB is 55.
16 (13) 45 ID : in7linesandangles [16] It is given that AD and BD are bisectors of CAB and CBA, respectively. Therefore, x = CAB/ (1) y = CBA/ (2) In triangle ABC, CAB + CBA + ACB = (The sum of all the angles of a triangle is 180 ) CAB + CBA + 90 = 180 CAB + CBA = CAB + CBA = 90 CAB/2 + CBA/2 = 90/2 x + y = 45...(From equations (1) and (2)) Hence, the sum of the angles x and y is 45. (14) c. d and a We know that the angle formed in a semicircle is 180. If we look at the following shapes carefully, we notice that shapes 'd and a' can be joined together to form a semicircle as their sum adds up to 180. Hence, d and a is the correct answer. (15) d. Two angles are right angles The sum of all the angles of a triangle must be 180. If we look at the all of the options carefully, we notice that the statement "Two angles are right angles" does not satisfy the condition of a triangle. Hence, the statement is false.
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