Lines and Angles Class 9 MCQ with Solutions Math

Lines and Angles Class 9 MCQ with Solutions

1. In fig. OE is the bisector of $\angle$ AOB and OF is the angle bisector of $\angle$ AOC, then the value of $\angle$ EOF is :

(A) 90°

(B) 180°

(D) None of these

Ans. (A)

Sol. $2\angle EOA + 2\angle AOF$ = 180°

\angle EOA + \angle AOF

= 90°

\angle EOF

= 90°

2. Supplementary and complementary angles need not be :

(A) Equal to 180°, 90°

(B) Adjacent

(D) None of these

Ans. (B)

Sol. They need to be adjacent

3. From the adjoining figure $\angle$ POR, $\angle$ QOR form a linear pair and a – b = 40°. Then a, b are:

(A) 110°, 70°

(B) 70°, 100°

(D) 120°, 80°

Ans. (A)

Sol. a + b = 180°

a – b = 40°

so a = 110°

b = 70°

4. The sum of the two angles in a triangle is 95° and their difference is 25°. Then the angles of the triangle is

(A) 75°, 50°, 55°

(B) 85°, 65°, 30°

(D) 60°, 35°, 85°

Ans. (D)

Sol. Let three angles a, b, c

so a + b = 95°

a – b = 25°

so a = 60°

b = 35°

& c = 180 – (a + b)

c = 85°

5. The value of x in the following fig. is :

(A) 30°

(B) 45°

(D) None of these

Ans. (A)

Sol. $\angle QPS$ = 90° – 60° = 30°

According to law of light

\angle SPR

= 30°

x = $\angle SPR$ = 30° Alternate angle

6. In the figure if BD || EF, then $\angle$ CEF is

(A)100°

(B) 120°

(D) 160°

Ans. (C)

Sol. $\angle CEF = 180^\circ - 40^\circ = 140^\circ$

7. In the figure PQ || ST, then $\angle$ QRS is equal to :

(A) 30°

(B) 40°

(D) 60°

Ans. (A)

Sol.

Draw UV || PQ || ST

so $\angle QRU$ = 180° – 100° = 80°

\angle SRV

= 180° – 110° = 70°

so $\angle R$ = 180 – 80° – 70° = 30°

8. In figure, l || m and transversal n intersects them at P and Q respectively, find the value of x.

(A) 25

(B) 27.5

(D) 17.5

Ans. (A)

Sol. 180 – (4x – 30) = 4x + 10

180 – 4x + 30 = 4x + 10

200 = 8x

x = 25°

9. In figure, for what value of x will line l be parallel to line m ?

(A) 50

(B) 70

(D) 40

Ans. (A)

Sol. 180 – (x – 10) = 2x + 40

150 = 3x

x = 50

10. Number of lines that can be drawn through a given point are :

(A) 1

(B) 2

(D) Infinite

Ans. (D)

Sol. Infinite lines can be drawn

11. From the adjoining figure the value of y is :

(A) 24

(B) 22

(D) 10

Ans. (D)

Sol. 5y° + 5y° + 4y° + 4y° + 9y° + 9y° = 360°

36y° = 360°

y° = 10°

12. In the given figure, if $\angle$ BOC = 7x + 20° and $\angle$ COA = 3x° then the value of x for which AOB becomes a straight line is :

(A) 16°

(B) 48°

(D) 72°

Ans. (A)

Sol. 7x + 20° + 3x = 180°

10 x = 160° ⇒ x = 16°

13. Two angles whose measures are a & b are such that 2a – 3b = 60° then find $\frac{{4a}}{{5b}}$ , if they form a linear pair

(A) 1.6

(B) 2

(D) 4.5

Ans. (A)

Sol. a + b = 180° ….(1)

2a – 3b = 60° ….(2)

i.e. $\frac{{a + b}}{{2a - 3b}}$ = $\frac{{180^\circ }}{{60^\circ }} = \frac{3}{1}$

a + b = 6a – 9b

10b = 5a

\frac{a}{b}

\frac{2}{1}

i.e. $\frac{{4a}}{{5b}}$ = $\frac{8}{5} = 1.6$

14. The sum of internal and external bisectors of an angle is :

(A) 90^o

(B) 180^o

(D) 360^o

Ans. (A)

Sol. Let angle is x, i.e. internal bisector = $\frac{x}{2}$ , external bisector = $90 - \frac{x}{2}$ .

Sum of internal and external bisector = 90°

15. The angle between the bisectors of two adjacent supplementary angles is :

(A) Obtuse angle

(B) Acute angle

(D) Can’t be determined

Ans. (C)

Sol. Let $\angle$ AOC and $\angle$ AOB are two supplementary angles so that angle between the bisectors of two adjacent supplementary angle is right angle.

16. In given figure. OD is angle bisector of $\angle$ AOB then which of the following is not true :

(A) $\angle$ AOD = $\angle$ DOB

(B) Point P on OD remains equidistant from OA and OB

(D) None of these

Ans. (C)

Sol. Here $\angle$ AOD and $\angle$ BOD are adjacent angles i.e. option (C) is false.

17. The complementary angles are such that two times the measure of one is equal to three times the measure of the other, then the measure of the larger angle is

(A) 36°

(B) 54°

(D) 46°

Ans. (B)

Sol. Let angles are x°, (90° – x°)

⇒ 2 x° = 3 (90 – x°)

2x = 270° – 3x°

5x° = 270°

⇒ x° = 54°

18. In figure, find x :

(A) 120°

(B) 130°

(D) 140°

Ans. (B)

Sol. 2x° + 100° = 360°

2x° = 260°

x° = 130°

19. In the given figure, AB is a mirror; PQ is the incident ray and QR, the reflected ray. If $\angle$ PQR = 112°, then $\angle$ PQA equals

(A) 68°

(B) 112°

(D) 54°

Ans. (C)

Sol. Let $\angle$ PQA = x

i.e. $\angle$ RQB = x

2x + 112° = 180° ⇒ x = 34°

20. In the adjoining figure, AOB is a straight line. Then $\angle$ BOD equals

(A) 45°

(B) 110°

(D) 90°

Ans. (C)

Sol. x° + 65° + 2x° – 20° = 180°

3 x° = 135°

x° = 45°

⇒ $\angle$ BOD = 2 × 45° – 20° = 70°

21. In given figure find x :

(A) 141°

(B) 70°

(D) 45°

Ans. (A)

Sol. x + 25° + 104° + 90° = 360°

x = 141°

22. What value of x will make AOB a straight line ?

(A) 30^o

(B) 50^o

(D) Can’t be determined

Ans. (B)

Sol. 2x + 30° + 2x – 50° = 180°

4x = 200

x = 50°

23. If two parallel lines are intersected by a transversal line, then the bisectors of the interior angles form a :

(A) Square

(B) Rectangle

(D) Trapezium

Ans. (B)

Sol. Let AB, CB, AD and CD are angle bisector of interior angles

i.e. $\angle$ BAD = $\angle$ BCD = $\angle$ ABC = $\angle$ ADC = 90°

i.e. ABCD is a rectangle.

24. The measure of angle, if 4 times its supplement is 104^o more than 8 times its complement is :

(A) 26^o

(B) 52^o

(D) 65^o

Ans. (A)

Sol. Let angle be x°.

So, 4 (180° – x) = 8 (90° – x) + 104°

180° – x = 180° – 2x + 26°

x = 26°

25. The measure of an angle is four times the measure of its supplementary angle. Then the angles are :

(A) 36°, 144°

(B) 40°, 160°

(D) 50°, 200°

Ans. (A)

Sol. Let angles are x°, (180° – x°)

i.e. x° = 4 (180° – x°)

x° = 720° – 4x°

5 x° = 720°

⇒ x° = 144°

i.e. angles 144° and 36°.

26. The supplement of an angle is one third of itself. Then the angle and its supplement are :

(A) 135°, 45°

(B) 60°, 180°

(D) 60°, 120°

Ans. (A)

Sol. Let angles are x, 180° – x

i.e. 180° – x = $\frac{1}{3}x$

⇒ 540° – 3x = x

4x = 540°

x = 135°

i.e. angles are 135°, 45°.

27. An angle is 14° more than its complementary angle then angle is :

(A) 38°

(B) 52°

(D) 48°

Ans. (B)

Sol. Let angles are x°, (90 – x°)

So, x° = 14° + (90° – x)

2x° = 104°

x° = 52°

28. The angle which is twice its supplement is

(A) 120°

(B) 90°

(D) 30°

Ans. (A)

Sol. Let angle is x, then its supplement is 180° – x

x° = 2(180° – x°)

3 x° = 360°

x° = 120°

29. The angle which exceeds its complement by 20° is :

(A) 45°

(B)55°

(D) 110°

Ans. (B)

Sol. Let angle is x, then its complement is 90° – x

x° = (90° – x) + 20°

2 x° = 110°

⇒ x = 55°

30. In the adjoining figure AB || CD and BC || ED then the value of x is :

(A) 95^o

(B) 90^o

(D) 80^o

Ans. (A)

Sol. x° = 180° – 85° = 95°

31. In the adjacent figure the value of x if AB || CD and EF || CD, is:

(A) 45°

(B) 55°

(D) 70°

Ans. (B)

Sol. $\angle$ ECD = 180° – 150° = 30°

Now $\angle$ ABC = $\angle$ BCE + $\angle$ ECD

= 25° + 30° = 55°

In the adjoining figure AB || CD,

32. $\angle 1\,\,:\,\angle 2 = 3:2$ Then $\angle 6$ is :

(A) 72^o

(B) 36^o

(D) 144^o

Ans. (A)

Sol. $\angle$ 1 + $\angle$ 2 = 3x + 2x = 5x

So, 5 x = 180°

⇒ x = 36°

\angle

2 = 2x = 72°

⇒ $\angle$ 6 = 72° (corresponding to $\angle$ 2)

33. From the adjoning figure AB || DE, then the value of x° is :

(A) 25°

(B) 35°

(D) 55°

Ans. (B)

Sol. Draw a line through C parallel to AB and DE.

x° = 180° – (85° + 60°) = 180° – 145° = 35°

34. In figure, if l || m, then the value of x is :

(A) 60^o

(B) 65^o

(D) 35^o

Ans. (D)

Sol.

x° = 60° – 25° = 35°

35. P and Q are two plane mirrors placed parallel to each other. An incident ray AB to the first mirror is reflected twice in the direction CD. Then

(A) AB || CD

(B) AB || PQ

(D) Can’t say

Ans. (A)

Sol. $\angle$ 1 = $\angle$ 2, $\angle$ 2 = $\angle$ 3, $\angle$ 3 = $\angle$ 4

i.e. $\angle$ 1 + $\angle$ 2 = $\angle$ 3 + $\angle$ 4

i.e. AB || CD.

36. In figure, l || m and transversal n intersects them at P and Q respectively, find the value of x.

(A) 25

(B) 27.5

(D) 17.5

Ans. (A)

Sol. 4x – 30° + 4x + 10° = 180°

8 x° = 200°

x° = 25°

37. Which one of the following is not correct :

(A) Two lines which are both parallel to the same line are parallel to each other.

(B) Two distinct lines cannot have more than one point in common.

(D) A line contains infinite number of points

Ans. (C)

Sol. Two intersecting lines can not be parallel to the same line.

38. Which one of the following is correct :

(A) A line segment has no definite length

(B) Two lines always intersect at a point

(D) A linesegment AB is same as the linesegment BA

Ans. (D)

Sol. A linesegment AB is same as the linesegment BA.

39. If one angle of triangle is equal to the sum of the other two then triangle is :

(A) acute triangle

(B) obtuse triangle

(D) Can’t say

Ans. (C)

Sol. Let angles are $\angle$ A, $\angle$ B, $\angle$ C.

\angle

A =

\angle

B +

\angle

i.e., $\angle$ A = 90º

i.e., triangle is right angle triangle.

40. If two angles are supplementary, then the sum of the two angles is :

(A) 90°

(B) 360°

(D) 270º

Ans. (C)

Sol. Sum of two supplementary angles is 180º.

41. In figure, AB || CD then x = :

(A) 64°

(B) 62°

(D) 58°

Ans. (A)

Sol.

From figure x + y = 112° and y = 48°

i.e. x + 48° = 112°

x = 64°

42. In figure, AB || DC and DE || BF then y :

(A) 100°

(B) 105°

(D) 115°

Ans. (B)

Sol. $\angle$ DCF = 180° – (40° + 65°) = 75°

i.e. y = 180° – 75° = 105°

43. Two parallel lines have :

(A) a common point

(B) two common points

(D) infinite common points

Ans. (C)

Sol. Two parallel lines have no common point but two co-incident lines have infinite common points.

44. In the given figure find y, if x + y = 80° :

(A) 20°

(B) 60°

(D) 30°

Ans. (A)

Sol. x + y = 80° ….(1)

4x + 6y = 360° ….(2)

From eqn. (2)

4(x + y) + 2y = 360°

2y = 360° – 320° = 40°

y = 20°

45. In the given figure find x :

(A) 120°

(B) 40°

(D) 30°

Ans. (C)

Sol. 2x + 60° = 180°

⇒ 2x = 120°

⇒ x = 60°

46. Two intersecting lines have :

(A) a common point

(B) two common point

(D) infinite common points

Ans. (A)

Sol. Two intersecting lines have only one common point.

47. If two parallel lines intersected by a transversal, then each pair of consecutive interior angles so formed is :

(A) Equal

(B) Complementary

(D) None of these

Ans. (C)

Sol. Sum of these two angles is 180°, i.e. they are supplementary pair.

48. From the adjoining figure x = 30°. The value of y° is :

(A) 25°

(B) 24°

(D) 45°

Ans. (B)

Sol. 2x° = 2 × 30° = 60°

5y = 120°

⇒ y = 24°

49. From the adjoining figure the value of y is :

(A) 24°

(B) 22°

(D) 10°

Ans. (C)

Sol. 7y + 3y + 10y + 10y + 3y + 7y = 360°

40y = 360°

⇒ y = 9°

50. In figure, if AB || CD and CD || EF, then $\angle$ FAC = :

(A) 30°

(B) 40°

(D) 50°

Ans. (B)

Sol. $\angle$ CAB = 180° – 140° = 40°

\angle

FAC =

\angle

FAB –

\angle

CAB

= 80° – 40° = 40°

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