Class 6 Geometry. Answer the questions. For more such worksheets visit (1) If AB and DE are parallel, find the value of ACB.

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1 ID : in-6-geometry [1] Class 6 Geometry For more such worksheets visit Answer the questions (1) If AB and DE are parallel, find the value of ACB. (2) If AB and DE are parallel to each other, find the measure of BCD. (3) If lines AC and BD intersects at point O such that AOB: BOC = 2:3, find COD. (4) If AB and PQ are parallel, compute the measure of X.

2 ID : in-6-geometry [2] (5) Find the measure of a+b. (6) If AB and CD are parallel, find the value of angle x. (7) What is the value of the supplement of the complement of 15? (8) If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 1:2, then find the greater of the two angles. (9) Compute the measure of x. (10) If lines XY and MN intersect as shown below and a:b = 2:3, find c.

3 (11) Lines AB and CD intersect at point O. If AOC + BOE = 120 and BOD = 50, find BOE. ID : in-6-geometry [3] (12) If AB and CD are parallel, find the value of x. (13) Find the measure of angle x. (14) If AB and PQ are parallel, compute the measure of W. Check True/False (15) The sum of all the angles around a point is 360. True False

4 ID : in-6-geometry [4] 2017 Edugain ( All Rights Reserved Many more such worksheets can be generated at

5 Answers ID : in-6-geometry [5] (1) 115 In the given figure, BAC = 25 and BDE = 40. According to the question, AB and DE are parallel. Therefore, ABC and BDE are alternate interior angles: ABC = BDE (Alternate interior angles) ABC = 40 The sum of all the three angles of a triangle is 180. Now, in ABC, ABC + BAC + ACB = ACB = 180 (As, ABC = 40 and BAC = 25 ) 65 + ACB = 180 ACB = ACB = 115 Step 4 Therefore, the value of ACB is 115.

6 (2) 48 ID : in-6-geometry [6] Let us extend the line AB to point F such that AF is parallel DE. According to the question, ABC = 114 and CDE = 114. If we look at the given figure carefully, we notice that CGF and CDE are corresponding angles. Therefore, CGF = CDE (Corresponding angles) CGF = 114 (As, CDE = 114 ) Step 4 The angles on a straight line add up to 180. AF is a straight line. So, ABC + CBG = 180 CBG = ABC CBG = (Since, ABC = 114 ) CBG = (1) and also, CGB + CGF = 180 CGB = CGF CGB = CGB = (2) Step 5 The sum of all the three angles of a triangle is 180. Now, in BCG, CBG + BGC + BCG = BCG = 180 (Since, CBG = 66 and BGC = 66 ) BCG = 180 BCG = BCG = 48 Step 6 If we look at the given figure carefully, we notice that BCD = BCG. Therefore, the value of BCD = 48.

7 (3) 72 ID : in-6-geometry [7] Given, AOB: BOC = 2:3, or AOB BOC 3 AOB = 2 BOC AOB = 2 BOC 3 = 2 3, 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, 2 BOC + BOC = BOC 3 = 180 BOC = = 108. Since AC and BD are straight lines that intersect, COD = BOC = = 72 Step 4 This means that COD = 72

8 (4) 265 ID : in-6-geometry [8] Draw a line MN parallel to both lines AB and PQ. According to the given figure, X1 + X2 + X makes a full angle, i.e., 360. Therefore, X1 + X2 + X = 360 X = X1 - X2 Now, AB and MN are two parallel lines cut by a transversal OB. Hence, NOB = OBA (Alternate angles) X2 = 54 Now, PQ and MN are two parallel lines cut by a transversal OQ. Hence, NOQ = OQP (Alternate angles) X1 = 41 Step 4 We know, X = X1 - X2 = = = 265 Hence, X = 265. (5) 200 If we look at the figure carefully, we notice that the angles b, 160, and a are made around a point. The sum of all the angles around a point is 360. Therefore, b a = 360 a + b = a + b = 200 Hence, the measure of a + b is 200.

9 (6) 120 ID : in-6-geometry [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 figure above. Let us redraw the figure as below: a = c (vertically opposite angles) c = e (alternate interior angles) Therefore we can write, a = c = e = g Again, b = d (vertically opposite angles) d = f (alternate interior angles) Therefore we can write, b = d = f = h We know that sum of two adjacent angle is equal to 180. Therefore, from the diagram, you can write, a + b = 180, b + c = 180, c + d = 180, d + a = 180 Here, a = 120 and g = x As a is equal to g, x is 120. Therefore, the value of x is 120.

10 (7) 105 ID : in-6-geometry [10] If we look at the question carefully, we notice that we have to find the complement of 15 and then find the supplement of the complement of 15. We know, the sum of the complementary angles is 90. Therefore, the complement of 15, = = 75 The sum of the supplementary angles is 180. Therefore, the supplement of 75, = = 105 Step 4 Hence, the value of the supplement of the complement of 15 is 105. (8) 120 Let x be the first 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. Thus, 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 parallel lines

11 = x x ID : in-6-geometry [11] 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 Therefore, = x 2 By cross multiplying, we get: 2x = 1(180 - x) 2x = x 2x + 1x = 180 3x = 180 x = x = 60 Step 4 The first angle = 60 The second angle = = 120 Step 5 Hence, the greater of the two angles is 120. (9) 69 If we look at the angles 113 and (x + 44), we find that they are vertically opposite angles. We know that the vertically opposite angles are equal. Therefore, (x + 44) = 113 x = x = 69

12 (10) 126 ID : in-6-geometry [12] 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 = 2:3, or a b = 2 3. Cross multiplying the fractions, we get, 3a = 2b. Step 4 We put b = 3a 2 in a + b = 90 and get, a + 3a 2 = 90, or 5a 2 = 90, or 5a = 180. Dividing each side by 5, we get, a = 180, or Step 5 Now, since b = 3a 2. We can say that b = , or b = 54. Step 6 From step 2, we have, c = a + 90, or c = , or c = 126.

13 (11) 70 ID : in-6-geometry [13] We are given that BOD = 50. Now, look at the figure. BOD and AOC are vertically opposite angles. So, they are equal. This means, AOC = 50. Since, AOC + BOE = 120. From step 1, we have AOC = 50. We can say that 50 + BOE = 120. Now, subtracting 50 from both the sides, we get, BOE = , or BOE = 70

14 (12) 115 ID : in-6-geometry [14] It is given that lines AB and CD are parallel and the third line (say EF) cuts the lines AB and CD at a certain angle as shown in the figure above. Let us redraw the figure as below: a = c (Vertically opposite angles) c = e (Alternate interior angles) Therefore, we can write, a = c = e = g Again, b = d (Vertically opposite angles) d = f (Alternate interior angles) Therefore, we can write, b = d = f = h We know that 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 Here, h = 115 and d = x As d is equal to h, x is 115. Hence, the value of x is 115.

15 (13) 10 ID : in-6-geometry [15] If we look at the figure carefully, we notice that the angles 160, 13x, and 7x are made around a point. The sum of all the angles around a point is 360. Therefore, x + 7x = x = x = x = 200 x = x = 10 Hence, the measure of angle x is 10.

16 (14) 249 ID : in-6-geometry [16] Draw a line MN parallel to both the line PQ and AB. According to the figure, W = W1 + W2 Now, AB and MN are two parallel lines cut by a transversal OA. Hence, MOA = OAB (Alternate angles) W2 = 54 Now, PQ and MN are two parallel lines cut by a transversal OP. Hence, MOP = OPQ (Alternate angles) W1 = 57 Step 4 We know, W = W1 - W2 = = = 249 (15) True Since, we know that the sum of all the angles around a point must be 360. Therefore, we can say that the statement " The sum of all the angles at a point is 360 " is True.

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