Lines and Angles - Test Papers

 CBSE Test Paper 01

CH-6 Lines and Angles

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1. In the adjoining figure, AB ∥ CD and AB ∥ EF. If EA ⊥ BA and ∠BEF = 55°, then the values of x, y and z :-
   
   1. 125°, 125°, 35°
   2. 60°, 60°, 60°
   3. 120°, 130°, 25°
   4. 35°, 125°, 120°
2. 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. an equilateral triangle
   3. a right triangle
   4. an obtuse angled triangle
3. In the adjoining figure, if QP ∥ RT, then x is equal to –
   
   1. 55°
   2. 75°
   3. 65°
   4. 70°
4. The number of lines that can pass through a given point is:
   1. only one
   2. two
   3. one
   4. Infinity
5. In figure, PQ ∥ RS, ∠ QPR = 70°, ∠ ROT = 20° find the value of x. 
   1. 20°
   2. 70°
   3. 50°
   4. 110°
6. Fill in the blanks:

   An equation of the type ________ represents a straight line passing through the origin.

7. Fill in the blanks:

   The common between the three angles of a triangle and a linear pair is ________.

8. In Fig., find the value of x for which the lines l and m are parallel.
   

9. Find the measure of the complementary angle of 60o.

10. In the given figure △ABC is right angled at A. AD is drawn perpendicular to BC. Prove that ∠BAD=∠ACB
    

11. Prove that if one angle of a triangle is equal to the sum of other two angles, then the triangle is right angled.

12. The exterior angle of a triangle is 110° and one of the interior opposite angle is 35°. Find the other two angles of the triangle.

13. AB, CD and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD. If  ∠BOF = 35o. Find ∠BOC and ∠AOD.
    

14. In Fig., if AB∥CD,  ∠APQ=50∘ and ∠PRD=127∘, find x and y.
    

15. In △ABC in given figure, the sides AB and AC of  △ABC are produced to points E and D respectively. If bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90° - 12 ∠ A .
    

CBSE Test Paper 01
CH-6 Lines and Angles

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Solution

1. (a) 125°, 125°, 35°
   Explanation: x + 55 = 180° (Sum of supplementary angles or co-interior angles)
   x = 125°
   x = y = 125° (Corresponding angles)
   z + ∠EAB = y (Exterior angle property)
   z = 125° - 90° = 35°
2. (c) a right triangle
   Explanation: The sum of the angles of triangle  is 180 degrees.
   let the angles of triangle be a ,  b, c
   we have given that one angle of a triangle is equal to the sum of the other two angles
   so we have
   c=a + b
   a + b + c = 180
   Substitute c for a +b
   c + c = 180
   2c = 180
   c = 90
   therefore the triangle is a right triangle.
3. (b) 75°
   Explanation:
   ∠QPR = ∠PRT = 40° (Alternate interior angles)
   In △QPR
   ∠PQR + ∠QPR + ∠PRQ = 180° (Angle sum property)
   65° + 40° + x° = 180°  
   x° = 180° - 40° - 65°
   x° = 75
4. (d) Infinity
   Explanation:

   
   As seen from the above image, any number of lines can be drawn through a given point. 
   Hence the answer may be given as "Infinity".

5. (c) 50°
   Explanation:

   PQ ∥ RS
   ∠QPR = ∠SRO = 70° (Corresponding, Angle)
   NOW IN △RTO
   x + 20° = 70° (exterior angle)
   x = 70° - 20°
   x = 50°

6. y = mx

7. 180o

8. Two lines are parallel when angles on the same side of the transversal are supplementary i.e.,
   x+55∘=180∘⇒x=180∘−55∘ ⇒x=125∘

9. The measure of the complementary angle x = (90o – ro)
   Where ro = given measurement
   ∴ x = (90o – 60o) = 30o

10. ∵ AD⊥BC
    ∴∠ADB=∠ADC=900
    from △ABD
    ∠ABD + ∠BAD + ∠ADB = 1800
    ∠ABD + ∠BAD + 900=1800
    ∠ABD + ∠BAD = 900
    ∠BAD = 90° - ∠ABD ..(i)
    But ∠A + ∠B + ∠C = 180° △ABC
    ∠B + ∠C = 900 ∵ ∠A = 900
    ∠C = 90° - ∠B ...(ii)
    From (i) and (ii)
    ∠BAD = ∠C
    ∠ BAD = ∠ACB Hence proved

11. Given in ΔABC ∠B = ∠A + ∠C

    Proof: ∠A+∠B+∠C=180o….. (1) [Sum of three angles of a ΔABC is 180°]
    ∠A+∠C=∠B….. (2)
    From (1) and (2)
    ∠B+∠B=180o
    2∠B = 180o
    ∠B = 90o

    Hence, the triangle is a right angled triangle.

12. Since the exterior angle of a triangle is equal to the sum of interior opposite angles.
    
    ∴ ∠ACD = ∠A + ∠B
    110 = ∠A + 35o
    ∠A = 110o−35o
    ∠A = 75o 
    Now, ∠A+∠B+∠C=180∘(Angle sum property)
    ∠C = 180−(∠A + ∠B)
    ∠C = 180−(75o+35o)
    ∠C = 70o

13. 
    OF bisects ∠BOD . . . [Given]
    ∠BOF = ∠DOF = 35o
    ∠COE = ∠DOF = 35o
    ∠EOF = 180o . . . . [A straight angle = 180o]
    ∴ ∠EOC + ∠BOC + ∠BOF = 180o
    ∴ 35o + ∠BOC + 35o = 180o
    ∴ ∠BOC = 180o – 70o = 110o
    ∠AOD = ∠BOC . . . [Vertically opposite angles]
    = 110o

14. As AB∥CD and PQ is a transversal.
    ∴ ∠APQ=∠PQR (Alternate interior angles)
    ⇒50∘=x ...........(1)
    Also, ∠APR=∠PRD (Alternate interior angles) .
    ⇒∠APQ+∠QPR=127°
    ⇒x+y=127∘
    ⇒50∘+y=127∘. [ From (1) ]
    or ​​​, y = 127° - 50° = 77°
    Hence, x = 50° & y = 77°

15. 

    As ∠ABC and ∠CBE form a linear pair
    ∴∠ABC + ∠CBE = 180°..........(1)
    Given, BO is the bisector of ∠CBE. Hence, 
    ∠CBE = 2∠OBC. 
    ⇒ ∠CBE = 2∠1...............(2)

    Therefore, ∠ABC+ 2∠1 =180° [ from (1) & (2) ]
    ⇒ 2∠1 = 180° - ∠ABC
    ⇒ ∠1 = 90° - 12∠ABC........(3)
    Again, ∠ACB and ∠BCD form a linear pair
    ∴∠ACB + ∠BCD =180°.........(4)
    Given, CO is the bisector of ∠BCD.  Hence,
      ∠BCD = 2∠2............(5)
    So, ∠ACB + 2∠2 = 180° [ from (4) & (5) ]
    ⇒ 2∠ 2 = 180°- ∠ACB
    ⇒ ∠2 = 90° - 12∠ACB ...(6)

    Now in △OBC, we have
    ∠1+ ∠2 + ∠BOC = 180° (Angle sum property of triangle) ...(7)

    From (3), (6) and (7), we have
    90°- 12∠ABC + 90°- 12∠ACB + ∠BOC = 180° .
      ⇒ ∠BOC = 12(∠ABC + ∠ACB )...........(8)
    Now, in △ABC, we have
    ∠BAC + ∠ABC + ∠ACB = 180°
    or, ∠ABC + ∠ACB = 180° - ∠BAC............(9)

    From (8) and (9), we have:- 
    ⇒ ∠BOC = 12(180° - ∠BAC)
    Hence, ∠BOC = 90° - 12 ∠A Proved.