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Q1. In the given figure, PQ||RS and EF||QS. If PQS = 60°, then the measure of RFE is:

• 1) 60^o
• 2) 120^o
• 3) 115^o
• 4) 180^o

Solution

RSQ + PQS = 180^o RSQ = 180⁰-60⁰=120^o So,RFE = RSQ = 120⁰(corresponding angles)

Q2. In figure, if AB||CD, FAE = 90^o and AFE = 40^o then find ECD.

Solution

By angle sum property in triangle FAE. ₉₀^o _{+ 40}^o ₊ AEF = 180^o _Or, AEF = 50^o CEB = AEF = 50^o (vertically opp. angles) ECD + CEB = 180^o _Or, ECD + ₅₀^o = 180^o ECD = 130^o

Q3. The angles of a triangle are in the ratio 5:3:7, the triangle is:

• 1) An acute angled triangle
• 2) A right triangle
• 3) An obtuse angled triangle
• 4) An isosceles triangle

Solution

Let the angles be 5x, 3x and 7x respectively. By sum property, 5x + 3x + 7x = 180⁰ 15x = 180^ox = 180^o/15 = 12 So, the angles are 60^o, 36^o and 84^o, i.e. the triangle is acute angled.

Q4. In figure, lines AB and CD intersect at O. If then COB is:

• 1) 90^o
• 2) 80^o
• 3) 70^o
• 4) 100^o

Solution

Here, 4x + 5x = 180^o Or, 9x = 180^o Or, x = 20^o Thus, COB = AOD = 4 x 20^o = 80^o.

Q5. In the given figure , AD is the median, then BAD is:

• 1) 55^o
• 2) 40^O
• 3) 100^o
• 4) 50^o

Solution

_{Since triangle ABC is isosceles and hence median is also the angle bisector.}

Q6. In the given fig., the measure of ABC is:

• 1) 80^o
• 2) 100^o
• 3) 20^o
• 4) 60^o

Solution

Here, x = 20^o as vertically opp. And, y = 180^o - 100^o = 80^o as linear pair. Therefore, B = 180^o - 20^o- 80^o = 80^o

Q7. An exterior angle of triangle is 80^o and the interior opposite angles are in the ratio 1:3. Measure of each interior opposite angle is:

• 1) 30^o, 90^o
• 2) 20^o, 60^o
• 3) 40^o, 120^o
• 4) 30^o, 60^o

Solution

Let interior opposite angles are: x and 3x Therefore, x + 3x = 80^o Or, 4x = 80^o Or, x = 20^o Hence, required angles are: 20^o, 60^o

Q8. Prove that if two lines intersect, the vertically opposite angles are equal.

Solution

Since, 1 + 2 = 180^o … (1) (linear pair) And, 2 + 3 = 180^o … (2) (linear pair) From (1) and (2), we get 1 + 2 = 2 + 3 Or, 1 = 3 Similarly, 2 = 4 Hence proved.

Q9. In fig. if x°, y° and z° are exterior angles of ΔABC, then x° + y° + z° is:

• 1) 180^o
• 2) 360^o
• 3) 270^o
• 4) 90^o

Solution

Sum of exterior angles of a triangle is 360^o

Q10. If the angles of a triangle are in the ratio 1:2:3 then find the measure of angles.

Solution

Let the angles are: x, 2x and 3x Thus by angle sum property, x + 2x + 3x = 180^o Or, 6x = 180^o Or, x = 30^o Hence angles are: 30^o, 60^o, 90^o.

Q11. The sides BA and DC of a quadrilateral ABCD are produced as shown in fig. Show that x + y = a + b.

Solution

_Here, BAD = (180^o - b) (linear pair) BCD = (180^o - a) (linear pair) Now by angle sum property of quadrilateral, we have x + (180 - b) + y + (180 - a) = 360^o x + y - b - a = 0 x + y = a + b x + y = a + b

Q12. One of the angles of a triangle is 75^o. If the difference of the other two angles is 35^o, then the larger angle of the other two angles of the triangle has a measure of:

• 1) 135^o
• 2) 70^o
• 3) 80^o
• 4) 100^o

Solution

Sum of other two angles = 180^o - 75^o = 105^o Difference of other two angles = 35^o Measure of the larger angle =

Q13. In figure, AB||DC, BDC = 30° and BAD = 80°, find x, y and z.

Solution

Since

Q14. In ABC, BC = AB and B = 80°, then A is equal to:

• 1) 100^o
• 2) 50^o
• 3) 40^o
• 4) 80^o

Solution

Given BC = ABC=A B= 80^o In triangle ABC,A+B+C = 180^o A+B+A = 180^o 2A + 80^o=180^o _{2A = 100} A = 50^o

Q15. In fig., if m||n and a : b = 2 : 3, the measure of h is

• 1) 150^o
• 2) 108^o
• 3) 120^o
• 4) 72^o

Solution

Q16.

Solution

Q17. In fig., in ABC, AB = AC. The value of x is:

• 1) 80^o
• 2) 120^o
• 3) 130^o
• 4) 100^o

Solution

Here, B + C = 180^o - 80^o = 100^o. But, B = C therefore C = 50^o. So, x = 180^o - 50^o = 130^o.

Q18. Given ∠ POR = 3x and ∠ QOR = 2x + 10. If POQ is a straight line, then the value of x is

• 1) 36˚
• 2) 34˚
• 3) 30˚
• 4) None of these

Solution

POQ is a straight line. ∴ ∠ POR + ∠ QOR = 180˚  ∴ 3x + 2x + 10 = 180˚ ∴ 5x = 170˚ ∴ x = 34˚

Q19. Find the measure of the angle which is complement of itself.

• 1) 30^o
• 2) 90^O
• 3) 45^O
• 4) 180^o

Solution

Q20. An exterior angle of a triangle is 80^o and two interior opposite angles are equal. Measure of each of these angle is:

• 1) 40^o
• 2) 100^o
• 3) 60^o
• 4) 120^o

Solution

Since, Sum of two interior opp. angles = Measure of ext. angle Therefore, Sum of two interior opp. angles = 80^o Hence, the measure of each angle = 40^o.

Q21.

Solution

Q22. The number of lines that can pass through a given point is:

• 1) Two
• 2) None
• 3) Only one
• 4) Infinitely many

Solution

Infinitely many

Q23. In figure, value of x is:

• 1) 50^o
• 2) 20^o
• 3) 40^o
• 4) 30^o

Solution

Here, 4x + 2x + x + 150^o = 360^o Or, 7x = 210^o Or, x = 30^o

Q24. Given lines l₁ l₂ and l₃ in fig., are parallel. The value of x is :

• 1) 140^o
• 2) 80^o
• 3) 40^o
• 4) 50^o

Solution

We know corresponding angles are equal. So, 1 = 2 = 40^o Therefore, x = 180^o - 40^o = 140^o

Q25. In fig., if QTPR, TQR = 40° and SPR = 30° find the value of x and y.

Solution

Consider QTR TQR + QTR + R = 180^o (Sum of angle of a triangle in 180^o) 40^o + 90^o + R = 180^o R = 50^o Now ,Consider PRS y = SPR + x (Exterior angle is equal to sum of interior angles) y = 30^o+50^o =80^o Hence, x=50^o, y=80^o

Q26. In a triangle ABC if A = 53^o and C = 44^o then the value of B is:

• 1) 83^o
• 2) 73^o
• 3) 46^o
• 4) 93^o

Solution

B = 180^o - 53^o - 44^o = 83^o

Q27. Measure of an angle which is supplement of itself is :

• 1) 30^o
• 2) 90^o
• 3) 45^o
• 4) 180^o

Solution

Since 90^o + 90^o = 180^o.

Q28. If the bisectors of B and C of triangle of a ABC intersect at O, then prove that BOC = 90^o + .

Solution

In ABC A + B + C =180^o    (angle sum of a ) OBC + BCO + BOC =180^o    (angle sum of BOC) But OBC = B and BCO = C B +C + BOC =180^o (B + C + A) +BOC = 180^o + A         (adding A to both sides) (180^o) + BOC = 180^o + A  90^o + BOC = 180^o + A BOC = 90^o +A Hence proved.

Q29. In figure, if line segment AB intersects CD at O such that OAD = 80⁰, ODA= 50⁰ and OCB= 40⁰, then find OBC.

Solution

_{In AOD ADO+OAD+}AOD = 180^o (Angle sum property of) BOC = 50^o (Vertically opposite angles of) Now, in COB OCB+OBC+BOC = 180^o OBC = 90^o (Angle sum property of)

Q30. In fig., in PQR, right angled at P. PS is perpendicular to QR. Prove that QPS = PRQ.

Solution

Given: In PQR, right at P. PS is perpendicular to QR. To prove: QPS = PRQ Proof: In QPS, QPS + PSQ + PQS = 180^o QPS + 90^o + PQS = 180^o QPS = 90^o - PQS QPS = 90^o - Q ...(i) In PQR P + Q + R = 180^o Q + R = 180^o - P = 180^o - 90^o = 90^o _Or, R = 90^o - Q ...(ii) (i) and (ii) QPS = PRQ.

Q31. Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is

• 1) 45˚
• 2) 20˚
• 3) 36˚
• 4) 30˚

Solution

Let the measure of one angle be x. Then, the measure of its complement is 90˚ - x. According to the question, 2x = 3(90˚ - x) ∴ 2x = 270˚ - 3x ∴ 5x = 270˚ ∴ x = 54˚ ∴ 90˚ - x = 90˚ - 54˚ = 36˚ Hence, the measure of smaller angle is 36˚. 

Q32. If two parallel lines are intersected by a transversal then, pair of alternate interior angles are:

• 1) Sum of the two angles is 360^o
• 2) Complementary
• 3) Supplementary
• 4) Equal

Solution

If two parallel lines are intersected by a transversal then pair of alternate interior angles are equal.

Q33. In figure, value of x is:

• 1) 70^o
• 2) 40^o
• 3) 50^o
• 4) 60^o

Solution

Since, x + 10^o + x + x + 20^o = 180^o Or, 3x = 150^o Or, x = 50^o.

Q34. Two supplementary angles are in the ratio 4:5. The angles are :

• 1) 90^o, 90^o
• 2) 30^o, 150^o
• 3) 80^o, 100^o
• 4) 45^o,45^o

Solution

Here, 4x + 5x = 180^o Or, 9x = 180^o Or, x = 20^o 4x = 4 x 20^o = 80^o 5x = 5 x 20^o = 100^o Thus, angles are: 80^o, 100^o

Q35.

Solution

Q36.

• 1) 34^o
• 2) 0^o
• 3) 10^o
• 4) 24^o

Solution

Here, 2x + 10 + 3x = 180^o Or, 5x = 170^o Or, x = 34^o

Q37. In fig., find the value of x.

Solution

In ABC ABC+100⁰= 180⁰ (linear pair) ABC = 80⁰ x+ 80⁰ = 115⁰ (Exterior angle property) x = 35⁰

Q38. In figure, and bisector of meets BC at a point D. Find the value of

Solution

Q39.

Solution

Q40. In figure, the value of (x + y) is:

• 1) 100^o
• 2) 120^o
• 3) 60^o
• 4) 80^o

Solution

By exterior angle theorem, the sum is 100^o.

Q41. In fig., find the value of x.

Solution

Here Since forma straight line 3x + 30^o = 180^o

Q42. In triangle ABC, if A + B = 108^o and B +C = 130^o, find A, B and. C

Solution

A + B = 108^o and B +C = 130^o A + B + B +C = (108^o + 130^o) (A + B +C) + B = 238^o  B = 238^o – 180^o               (A + B +C = 180^o) B = 58^o C = 130^o – 58^o= 72^o         A = 108^o – 58^o = 50^o Hence, A = 50^o, B = 58^o and C = 72^o

Q43. In ABC fig., x + y + z is equal to:

• 1) 120^o
• 2) 360^o
• 3) 240^o
• 4) 180^o

Solution

Sum of exterior angles of a triangle is 360^o.

Q44. Find the value of x.  

• 1) 45°
• 2) 65°
• 3) 35°
• 4) 55°

Solution

From the figure, the interior angles are supplementary. x + 3x - 40° = 180°  ∴ 4x = 180° + 40°  ∴ 4x = 220°  ∴ x = 55° 

Q45. If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

• 1) An isosceles triangle
• 2) A right triangle
• 3) An obtuse angled triangle
• 4) An equilateral triangle

Solution

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right angled triangle.

Q46. In given figure, find the values of x and y.

Solution

Here, x=180^O - 50^O = 130^O (Linear pair) y=130^O (Vertically opp. angles)

Q47. Given four points such that no three of them are collinear, then the number of lines that can be drawn through them are:

• 1) 4 lines
• 2) 8 lines
• 3) 2 lines
• 4) 6 lines

Solution

A line can be drawn with every pair of two points. Thus 6 lines can be drawn.

Q48. From figure, identify the incorrect statement, given that l||m and t is transversal.  

• 1) 2 and 8 are supplementary
• 2) 2 and 1 are supplementary
• 3) 2 and 3 are supplementary
• 4) 2 and 5 are supplementary

Solution

2 and 8 are not supplementary.

Q49. If E is a point on side QR of then:

• 1) ER>RP
• 2) QP>QE
• 3) QE>QP
• 4) QE= ER

Solution

Q50. Show that in the given figure, AB || EF.

Solution

Q51. Lines PQ and RS intersect each other at point O. If POR: ROQ= 5:7, find all the remaining angles.

Solution

But POR : ROQ = 5 : 7 (Vertically opposite angles)

Q52. The angles of a triangle are in the ratio 2 : 3 : 4. The angles, in order, are :

• 1) 80^o, 40^o, 60^o
• 2) 40^o, 60^o, 80^o
• 3) 60^o, 40^o, 80^o
• 4) 20^o, 60^o, 80^o

Solution

Let the angles are: 2x, 3x and 4x. Thus, 2x + 3x + 4x = 180^o Or, 9x = 180^o Or, x = 20^o Hence angles are: 40^o, 60^o, 80^o.

Q53. In figure, if AB||DE, ABC = 75^o and CDE = 145^o, then find BCD.

Solution

Construction: Draw PCQ || to AB and DE. DCQ + CDE = 180^o (DE || CQ, co-interior angles are supplementary) QCD = 180^o -145^o = 35^o BCP + CBA = 180^o ( AB||PQ, co-interior angles are supplementary) BCP =180^o - 75^o = 105^o BCP + BCD + DCQ = 180^o (Angles in straight line) BCD = 180^o - 105^o - 35^o = 40^o

Q54. In given figure, OA, OB are opposite rays and

Solution

Q55. In figure, BP AC, PBC = 25^o and DAC = 30^o. Find the value of x.

Solution

OPA + PAO + AOP = 180^o Or, 90^o + 30^o + AOP = 180^o Or, AOP = 180^o - 120^o = 60^o AOP = BOD = 60^o (vertically opp. angles) 25^o + x+ 60^o = 180 ^o (Angle sum property) x = 180^o - 85^o = 95^o

Q56. In figure, the side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = QPR.

Solution

(Angle sum property of triangle) (TR bisects PRS) Hence Proved

Q57. In figure, POQ is a line, ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = .

Solution

From the figure ROS = QOS - ROQ ROS = QOS - 90^o ...(i) ( OR QO) ROS = ROP - POS ROS = 90^o - POS ...(2) ( OR PQ) (1) + (2) ROS + ROS = QOS - 90^o + 90^o - POS 2ROS = QOS - POS ROS = (QOS - POS)

Q58. In figure, PQ||RS QPO = 70^o ROT = 20^o. Find the value of x.

• 1) 50^o
• 2) 20^o
• 3) 110^o
• 4) 70^o

Solution

Q59. A pair of angles is called linear pair if sum of two adjacent angles is:

• 1) 270^o
• 2) 180^o
• 3) 90^o
• 4) 360^o

Solution

A pair of angles is called linear pair if sum of two adjacent angles is 180^o

Q60. l, m and n are parallel lines intersected by transversal 't' at A, B and C respectively. Find the measure of 1, 2 and 3. Give reasons.

Solution

Q61. If the measure of an angle is twice the measure of its supplementary angle then the measure of the angle is:

• 1) 60°
• 2) 90°
• 3) 130°
• 4) 120°

Solution

Let x be the angle.   x=2(180^o -x)   x = 360^o - 2x   Or, 3x=360^o Or, x=120^o

Q62. In figure, prove that .

Solution

Q63.

Solution

Q64. In figure, the value of x is:

• 1) 10^o
• 2) 40^o
• 3) 30^o
• 4) 20^o

Solution

2x + 5x + 40^o = 180^o Or, 7x = 140^o Or, x = 20^o

Q65.

Solution

Q66. In figure find the value of QRP when QP||TR.

Solution

Here, PRT = 64^o (vertically opp. angles as PQ||TR) Therefore, QRP = 180^o - (64^o + 67^o) (linear pair) Or, QRP = 180^o - 131^o = 49^o.

Q67.

Solution

Q68. If x + y = s + w, prove that AOB is straight line.

Solution

x + y + w + s = 360^o Let x + y = s + w = t Therefore, 2t = 360^o Or, t = 180^o Or, x + y = 180^o Thus, AOB is a straight line.

Q69. RS || PQ. Find the value of x.

Solution

Q70. In fig., l||m, show that 1 + 2 - 3 = 180^o

Solution

_{Given: l||m To prove:} 1 + 2 - 3 = 180^o _{Construction: Draw XC||AB and extend AB to Y. Proof:} BCX = 180^_o - 1            (sum of int. _{s on same side of transversal is 180^o}) XCD = 3                     (Alternate int. _s) _Now, 2 = BCX + XCD _Or, 2 = (180^o - 1) + 3 1 + 2 - 3 = 180^o

Q71. An angle is of its complement. Find the angles.

Solution

Let Angle ₌ x, So complement ₌ 5x Therefore, x + 5x = 90^o So, x = 15^o And, 5x = 5 15^o = 75^o Angles are: 15^o and 75^o

Q72. The arms of an angle are respectively parallel to the arms of another angle as shown in the figure. Show that the two angles are equal.

Solution

Q73. In figure the side QR of PQR is produce to a point S. If the bisectors of PQR and PRS meet at T, then prove that QTR =.

Solution

QPR = PRS - PQR (by exterior angle property) … (1) QTR = TRS - RQT (by exterior angle property) … (2) PQR = 2TQR (as QT is bisector) … (3) PRS = 2TRS (as RT is bisector) … (4) _{From (1), (3) and (4) we get} QPR = 2TRS - 2TQR Or, QPR = 2(TRS - TQR) Or, QPR = 2(QTR) Using (2) Or, _{Hence proved.}

Q74. In fig., BO and CO are bisectors of DBC and ECB respectively. If BAC = 50^o and ABC = 60^o then find the measure of BOC.

Solution

By angle sum property in triangle ABC 50^o + 60^o + _{ACB = 180}^o _{This gives, ACB = 180}^o _{- 110}^o _{= 70}^o _{Now by exterior angle theorem,} DBC = A + ACB = 50^o + 70^o = 120^o. Therefore, OBC = DBC = 60^o. Again, ECB = A + ABC = 50^o + 60^o = 110^o. Therefore, OCB = DBC = 55^o. By angle sum property in triangle OBC, we get OBC + OCB + BOC = 180^o This gives, 60^o + 55^o + BOC = 180^o Or, BOC = 65^o

Q75. In fig., PQ||RS and T is any point as shown in figure then show that PQT + QTS + RST = 360^o

Solution

Construction: Draw a line AB parallel to PQ through T. Since, AB||PQ and PQ||RS therefore PQ||AB||RS. If the lines are parallel then the sum of interior angles on the same side of transversal is 180^o, therefore 1 + 2 = 180^o … (i) And, 3 + 4 = 180^o … (ii) Adding (i) and (ii), we get 1 + 2 + 3 + 4= 180^o + 180^o Or, PQT + QTS + RST = 360^o Hence proved.

Q76. An angle is equal to five times of its complement. Find the measure of the angle.

Solution

Let the angle be a. Its complement = 90^o - a Therefore, a = 5(90^o - a) _{a = 450o - 5a 6a = 450o} a = 75^o

Q77. In figure, if AB||CD then find PQR.

Solution

RQS = 35^o (alternate op. angles as TS||CD) SQP = 180^o - 125^o = 55^o (angles on same side as AB||TS) PQR = RQS + SQP = 35^o + 55^o = 90^o

Q78. In the figure ACCE and A:B :C = 3: 2: 1, find the value of ECD

Solution

Let A, B and C be 3x, 2x and x respectively. 3x +2x +x = 180^o     (angle sum of ) 6x = 180^o x = 30        A = 90^o, B = 60^o and ACB =30^o Now, ACB + ACE + ECD = 180^o (angles on a straight line)  30^o + 90^o + ECD = 180^o  ECD = 180^o – 120^o = 60^o Hence, ECD = 60^o

Q79. The exterior angles obtained on producing the base of a triangle both ways are 100° and 120° respectively. Find all the angles.

Solution

2 + 3 = 100^o (exterior angle property) 1 + 100^o = 180^o (Linear pair) 1 = 80^o 1 + 2 = 120^o (exterior angle property) 80^o + 2 = 120^o 2 = 40^o _{1 + 2 + 3 = 180}^o (Angle sum property of ) ₈₀^o _{+ 40}^o _{+ 3 = 180}^o _{3 = 60}^o Hence, 1 = 80^o, 2= 40^o, 3 = 60^o

Q80.

Solution

Q81. In fig., OP bisects AOC, OQ bisects BOC and OPOQ. Show that points A, O and B are collinear.

Solution

OR Since OP and OQ bisects angles AOC and BOC respectively. …(I) And COB = 2COQ … (II) Add (I) and (II) AOC + COB = 2POC + 2COQ is a straight line A, O and B are collinear points.

Q82. In given figure, if a ray OC stands on line AB such that then show that

Solution

Q83.

Solution

Q84. In fig., AB||CD and CD||EF. Also EA AB. If BEF= 40^o, then find x, y, z.

Solution

Since sum of the interior angles on the same side of transversal is 180^o, we have y = 180^o - 40^o = 140^o x = y = 140^o (corresponding angles) z = 90^o - 40^o = 50^o

Q85. In fig., l₁||l₂ and m₁||m₂. Prove that 1 + 2 = 180^o.

Solution

1 = 3 (corresponding angle ) 3 = 4 (corresponding angle) 1 = 4 _But, 4 + 2 = 180^o (linear pair) 1 + 2 = 180^o

Q86.

Solution

Q87.

Solution

Q88. Lines AB and CD are parallel and P is any point between the two lines as shown in figure. Show that DPB = x + y.

Solution

Construction :- Draw LM||AB and CD ABP = BPM = x ...(1) (Alternate interior angles) DPM = PDC = y ...(2) (Alternate interior angles) DPB = BPM + DPM From (1) and (2), DPB = x + y

Q89. If fig., if AB||CD, EFCD and GED = 126°, find AGE, GEF and FGE.

Solution

GED = AGE [alternate interior angles] AGE = 126⁰ GED = 126⁰ GEF+FED = 126⁰ GEF+90⁰= 126⁰ [FED = 90⁰] GEF = 36⁰ GEC+GED = 180⁰ (linear pair) GEC+126⁰= 180⁰ GEC = 54⁰ GEC= FGE= 54⁰ [Alternate interior angles]

Q90. If two interior angles on the same side of a traversal intersecting two parallel lines are in the ratio 2: 3, then the smaller of the two angles is:

• 1) 108^o
• 2) 36^o
• 3) 72^o
• 4) 54^o

Solution

Q91. If the bisectors of the angles ABC and ACB of a triangle ABC meet at point O then prove that BOC = 90^o + .

Solution

Given : In ABC, BO and CO are bisectors of ABC and ACB To prove : BOC = A Proof: In _BOC, OBC + BCO + BOC = 180^o OBC + BCO = 180^o - BOC ...(i) In ABC, ABC + BCA + A = 180^o From (i) and (ii) _{Hence proved.}

Q92. In figure, if PQ||RS, then find SRO.

Solution

Since PQ || RS, then OSR = 50^o as alternate interior angles are equal. Now in triangle RSO, SRO + 50^o + 20^o = 180^o Or, SRO = 180^o - 70^o = 110^o

Q93. In figure, PQ PS, PQ||SR, SQR = 25^o and QRT = 65^o. Find the value of x and y.

Solution

QSR + SQR = QRT (exterior angle property) QSR = 65^o - 25^o = 40^o PSR + SPQ = 180^o (PQ||SR, Co-interior angles are supplementary) PSR = 180^o - 90^o = 90^o PSR = y + QSR y = 90 ^o - 40 ^o = 50^o Also x = QSR (PQ||SR, alternate interior angles are equal) x = 40^o

Q94. In figure, PQR is:

• 1) 40^o
• 2) 30^o
• 3) 105^o
• 4) 50^o

Solution

_SincePQR + 75^o = 105^o Since, 105^o is exterior angle. Therefore, PQR = 30^o.

Q95. In fig., if AB||CD, APQ = 50^o and PRD = 127^o, find x and y.

Solution

Since AB||CD, x = APQ = 50^o (alternate interior angles) _{Also, y + 50o =} 127^o (alternate interior angles) Or, y = 77^o.

Q96. In figure prove that l||m.

Solution

By exterior angle theorem, 1 = 45^o + 35^o = 80^o Thus 1 is equal to given angle of 80^o, Since they are corresponding angles so l || m.

Q97. In the given figure, Prove that x = a + b + c.

Solution

Construction: We join B and D and produce to E Proof: 1 +2 =b and 3+4=x … (i) In ABD,3 = 1+ a [external angle property] In BDC, 4 =2 + c [external angle property] … (ii) Adding (i) and (ii), we get 3+4 = (1+2) +a +c x= b + a + c using (i)

Q98. In fig., ray OC stands on the line AB, ray OL and ray OM are angle bisector of AOC and BOC respectively. Prove that LOM = 90^o.

Solution

Let AOL = LOC = x ( OL is the bisector) Let COM = MOB = y ( OM is the bisector) Now, AOL + LOC + COM + MOB = 180^o (Linear pair) x + x + y + y = 180 x + y = 90^o LOC + COM = 90^o LOM = 90^o

Q99. In figure, if PQ||RS, ACS = 117^o then find (x - y).

Solution

60^o + y = 117^o (alternate interior angle as PQ||RS) y = 117^o - 60^o = 57^o Again, x + y = 117^o (exterior angle theorem) x = 117^o - 57^o x = 60^o x - y = 3^o

Q100. AB and CD are the bisectors of the two alternate interior angles formed by the intersection of a transversal 't' with parallel lines l and m (Shown in Figure) Show that AB||CD.

Solution

Since lm, t is transversal line EAC= ACH (alt. interior angles)

Q101. In figure,  _{is an isosceles triangle with LM = LN, and LP bisects} NLQ, prove that LP|| MN.

Solution

Here in LMN we have LM= LN _{But they are alternate interior angles. Hence, LP || MN.}

Q102. Prove that the sum of the angles of a triangle is 180⁰.

Solution

Stat: Sum of the angle of a triangle is 180º. Given: A triangle ABC in which the angles are a degree, b degree and c degree. Construction Draw the line PAQ parallel to BC. Take the replica of ÐABC and put it in ÐPAB and ÐACB in ÐQAC Proof: Now angle PAB= b degree and angle QAC= c degree since they are the pairs of alternate interior angles. ÐPAB = b º [ Alternate interior angle] ÐQAC = c º [ Alternate interior angle] Now Angle PAB plus angle BAC plus angle QAC is equal to 180 degree ÐPAB+ÐBAC+ÐQAC= 180º º [ Angle sum of straight line] b º+ c º+ a º = 180º Hence the angle sum of triangle is 180 º Hence we have proved the sum of angles of a triangle to be 180 degree.

Q103. In the figure, POR and QOR form a linear pair. If b - a = 60°, find the value of a and b.

Solution

POQ is a straight line b-a = 60^O (given) …..(i) a+b = 180⁰ (linear pair axiom) …..(ii) Adding (i) and (ii) 2b = 240⁰ b= 120⁰ Equation (i) gives a = b - 60^o = 120^o-60^o= 60⁰ Hence, a = 60⁰ and b = 120⁰

Q104. If l, m, n are three lines such that l||m and n l. Prove that n m.

Solution

n l (given) Therefore, 1 = 90^o Also, l||m, gives 2 = 90^o n is perpendicular to m.

Q105. In figure, if x : y = 3 : 7, then find the value of x,y,z and w.

Solution

x : y = 3 : 7 3a + 7a = 180 10a = 180 a = 18^o _Thus, x = 54, y = 126 _{Since vertically opp. Angles are equal, therefore} y = w = 126^o x = z = 54^o

Q106. In ΔXYZ, YO and ZO are the bisectors of ÐXYZ and ÐXZY respectively of ΔXYZ. If X = 62°, XYZ = 54°, then find ÐOZY.

Solution

In XYZ X+Y+Z= 180⁰ 62⁰+54⁰+Z= 180⁰ Z = 64⁰. OZY = = OZY = 32⁰

Q107. In fig., if y = 20°, prove that the line AOB is a straight line.

Solution

We know angle at a point is 360^o. y+3x-15+y+5+2y+4y+10+x=360^o 8y + 4x = 360^o 160^o + 4x = 360^o (Since, y=20^o) 4x = 200 x= 50^o Now, we have: y + 5 + 3x - 15 + y = 40 + 150 - 10 = 180^o AOB is a straight line.

Q108.

Solution

Q109.

Solution

Q110. In fig., the side BC of ABC is produced to D, the bisector of BAC intersects the side BC at E. Prove that ABC + ACD = 2AEC.

Solution

Given: The side BC of ABC is produced to D, the bisector of BAC intersects the side BC at E. To prove: ABC + ACD = 2AEC Proof: We know that ACD = ABC + BAC … (1) (ext. angle theorem) AE is bisector of BAC, thus i.e., BAC = 2BAE Thus equation (1) reduce to ACD = ABC + 2BAE Adding ABC on both sides, Now ABC + ACD = ABC + ABC + 2BAE = 2 (ABC + BAE) = 2 (AEC) [Ext angle of ABE] Hence proved.

Q111. Find x and y in the given figure.

Solution

Using angle sum property in AOD, x - 6 + x = 90^o 2x = 96^o x = 48^o Using angle sum property in BOC, 4y + 3y + 6 = 90^o 7y = 84 y = 12^o

Q112. If two lines are perpendicular to the same line, prove that they are parallel to each other.

Solution

Let l n and m n. 1 = 2 = (each is 90^o) But these are corresponding angles. Hence l || m.

Q113. In fig., if PQ||RS and PXM = 50° and MYS = 120°, find the value of x.

Solution

Q114. In fig., lines l and m intersect each other at O. If x = 40^o then, find the value of y, z and w.

Solution

x = 40^o z = x (vertically opposite angles) z = 40^o x^o + y^o = 180^o 40^o + y^o = 180^o y = 140^o Again x ^o + w ^o = 180^o 40^o + w ^o = 180^o w ^o = 140^o Thus y = w = 140^o, z = 40^o

Q115. In fig, if POR and QOR form a linear pair and a - b = 80° then find the value of a and b.

Solution

a + b = 180^o (linear pair) … (i) a - b = 80^o (given) …(ii) Add (i) and (ii) 2a = 260^o a = 130^o Put value of a in (i) we get, b = 50^o Hence a = 130^o, b = 50^o

Q116. In figure, PQ||SR, SQR = 25^o, QRT = 65 ^o, find x and y:

Solution

Since x + 25 ^o = 65 ^o (Alternate interior angles) x = 65 ^o - 25 ^o = 40^o In PQS, by angle sum property, we have: 40 ^o + 90 ^o + y = 180^o y = 50^o

Q117. In figure, TQ and TR are the bisectors of Q and R respectively, QPR = 80°, PRT = 30°.Determine TQR and QTR.

Solution

Q118. In figure, lines AB and CD intersect at O. If AOD : DOC = 4 : 5, then find COB

Solution

_Let AOD = 4x And, DOC = 5x Therefore, 4x + 5x = 180^o Or, 9x = 180^o Or, x = 20^o Therefore, AOD = 4 x 20^o = 80^o Now, COB = AOD (Vertically opp. angles) COB = 80^o.

Q119.

Solution

Q120. In fig., two straight lines AB and CD intersect each other at O. If COE = 70^o, find the values of a, b and c.

Solution

CO and EO stand on AB, therefore 4b + 70^o + b = 180^o 5b = 110^o b = 22^o Also, a = 4b (vertically app s) a = 4 22^o = 88^o Now a + 2c = 180^o 88 + 2c = 180^o 2c = 92^o c = 46^o Hence a = 88^o, b = 22^o, c = 46^o

Q121. In figure, find the value of x and y and then show that AB||CD.

Solution

60+x = 180⁰ (linear pair axiom) x = 120⁰ z = y = 120⁰ [vertically opposite s] y = x = 120⁰ But x = y are alternate angles So AB CD

Q122. In fig., find the value of x.

Solution

Produce AD to meet BC at point E In AEB We have, A + B + AEB = 180^o (Sum of 3 angles of a triangle is 180^o) Thus, Now, AEB + AEC = 108^o (Linear pair axiom) 100^o + AEC = 180^o … (i) Now in DEC DEC + 50^o = x^o (Exterior angle is equal to sum of the opp. Interior angles) Hence, x = 130^o (By using (i))

Q123. Prove that each angle of an equilateral triangle is 60^o.

Solution

_Since ABC is an equilateral triangle, therefore AB = BC = CA BA = BC ...(1) Similarly, A = B ...(2) From (1) and (2) A = B = C So A + B + C = 180^o _Gives, A = B = C = 60^o

Q124. In the given figure, l||m, then find the value of x?

Solution

1 = 180-60= 120⁰ (sum of angle on the same side of transversal is 180⁰) 1= 2 (vertically opposite angles) 2= 120⁰ x+ 40 +2= 180⁰ (Angle sum property of a triangle) x = 180- (40+120) x = 20⁰

Q125. Prove that sum of the exterior angle of a triangle is 360^o_.

Solution

Let x, y and z be the exterior angles of a triangle. Then the corresponding interior angles are: 180^o - x, 180^o - y, 180^o - z. Since the sum of interior angles is 180^o, therefore 180^o - x + 180^o - y + 180^o - z = 180^o Or, 540^o - (x + y + z) = 180^o Or, x + y + z = 540^o - 180^o = 360^o. Hence the sum of exterior angles is 360^o.

Q126. Prove that an equilateral triangle can be constructed on any given line segment.

Solution

Let x be the length of the sides of an equilateral triangle. Now in a triangle, sum of any two sides is always greater than the third side. In this case, Sum of any two sides = x + x = 2x which is more than x. Hence the construction of an equilateral triangle on any given line segment is possible.