Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In ∆ABC, base BC is produced both ways to D and E respectivley forming ∠ABE = 104° and ∠ACD = 136°

Question 2.
In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ∆ABC.

Solution:
In ∆ABC, sides BC, CA and BA are produced to D, E and F respectively.
∠ACD = 105° and ∠EAF = 45°
∠ACD + ∠ACB = 180° (Linear pair)
⇒ 105° + ∠ACB = 180°
⇒ ∠ACB = 180°- 105° = 75°
∠BAC = ∠EAF (Vertically opposite angles)
= 45°
But ∠BAC + ∠ABC + ∠ACB = 180°
⇒ 45° + ∠ABC + 75° = 180°
⇒ 120° +∠ABC = 180°
⇒ ∠ABC = 180°- 120°
∴ ∠ABC = 60°
Hence ∠ABC = 60°, ∠BCA = 75°
and ∠BAC = 45°

Question 3.
Compute the value of x in each of the following figures:

Solution:
(i) In ∆ABC, sides BC and CA are produced to D and E respectively


(ii) In ∆ABC, side BC is produced to either side to D and E respectively
∠ABE = 120° and ∠ACD =110°
∵ ∠ABE + ∠ABC = 180° (Linear pair)

(iii) In the figure, BA || DC

Question 4.
In the figure, AC ⊥ CE and ∠A: ∠B : ∠C = 3:2:1, find the value of ∠ECD.

Solution:
In ∆ABC, ∠A : ∠B : ∠C = 3 : 2 : 1
BC is produced to D and CE ⊥ AC
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangles)
Let∠A = 3x, then ∠B = 2x and ∠C = x
∴ 3x + 2x + x = 180° ⇒ 6x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ ∠A = 3x = 3 x 30° = 90°
∠B = 2x = 2 x 30° = 60°
∠C = x = 30°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ 90° + ∠ECD = 90° + 60° = 150°
∴ ∠ECD = 150°-90° = 60°

Question 5.
In the figure, AB || DE, find ∠ACD.

Solution:
In the figure, AB || DE
AE and BD intersect each other at C ∠BAC = 30° and ∠CDE = 40°
∵ AB || DE
∴ ∠ABC = ∠CDE (Alternate angles)

⇒ ∠ABC = 40°
In ∆ABC, BC is produced
Ext. ∠ACD = Int. ∠A + ∠B
= 30° + 40° = 70°

Question 6.
Which of the following statements are true (T) and which are false (F):
(i) Sum of the three angles of a triangle is 180°.
(ii) A triangle can have two right angles.
(iii) All the angles of a triangle can be less than 60°.
(iv) All the angles of a triangle can be greater than 60°.
(v) All the angles of a triangle can be equal to 60°.
(vi) A triangle can have two obtuse angles.
(vii) A triangle can have at most one obtuse angles.
(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.
(ix) An exterior angle of a triangle is less than either of its interior opposite angles.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
Solution:
(i) True.
(ii) False. A right triangle has only one right angle.
(iii) False. In this, the sum of three angles will be less than 180° which is not true.
(iv) False. In this, the sum of three angles will be more than 180° which is not true.
(v) True. As sum of three angles will be 180° which is true.
(vi) False. A triangle has only one obtuse angle.
(vii) True.
(viii)True.
(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.
(x) True.
(xi) True.

Question 7.
Fill in the blanks to make the following statements true:
(i) Sum of the angles of a triangle is ………
(ii) An exterior angle of a triangle is equal to the two …….. opposite angles.
(iii) An exterior angle of a triangle is always …….. than either of the interior opposite angles.
(iv) A triangle cannot have more than ………. right angles.
(v) A triangles cannot have more than ……… obtuse angles.
Solution:
(i) Sum of the angles of a triangle is 180°.
(ii) An exterior angle of a triangle is equal to the two interior opposite angles.
(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.
(iv) A triangle cannot have more than one right angles.
(v) A triangles cannot have more than one obtuse angles.

Question 8.
In a ∆ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.
Solution:
Given : In ∆ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior ∠B and ∠C meet at P and bisectors of exterior angles B and C meet at Q.

To prove : ∠BPC + ∠BQC = 180°
Proof : ∵ PB and PC are the internal bisectors of ∠B and ∠C
∠BPC = 90°+ \(\frac { 1 }{ 2 }\) ∠A …(i)
Similarly, QB and QC are the bisectors of exterior angles B and C
∴ ∠BQC = 90° + \(\frac { 1 }{ 2 }\) ∠A …(ii)
Adding (i) and (ii),
∠BPC + ∠BQC = 90° + \(\frac { 1 }{ 2 }\) ∠A + 90° – \(\frac { 1 }{ 2 }\) ∠A
= 90° + 90° = 180°
Hence ∠BPC + ∠BQC = 180°

Question 9.
In the figure, compute the value of x.

Solution:
In the figure,
∠ABC = 45°, ∠BAD = 35° and ∠BCD = 50° Join BD and produce it E

Question 10.
In the figure, AB divides ∠D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution:
In the figure AB = DB

Question 11.
ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = \(\frac { 1 }{ 2 }\) ∠A.
Solution:
Given : In ∠ABC, CB is produced to E bisectors of ext. ∠ABE and into ∠ACB meet at D.

Question 12.
In the figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Solution:

Question 13.
In a AABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Solution:
Given : In ∆ABC,
∠C > ∠B and AD is the bisector of ∠A

To prove : ∠ADB > ∠ADC
Proof: In ∆ABC, AD is the bisector of ∠A
∴ ∠1 = ∠2
In ∆ADC,
Ext. ∠ADB = ∠l+ ∠C
⇒ ∠C = ∠ADB – ∠1 …(i)
Similarly, in ∆ABD,
Ext. ∠ADC = ∠2 + ∠B
⇒ ∠B = ∠ADC – ∠2 …(ii)
From (i) and (ii)
∵ ∠C > ∠B (Given)
∴ (∠ADB – ∠1) > (∠ADC – ∠2)
But ∠1 = ∠2
∴ ∠ADB > ∠ADC

Question 14.
In ∆ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180°-∠A.
Solution:
Given : In ∆ABC, BD ⊥ AC and CE⊥ AB BD and CE intersect each other at O

To prove : ∠BOC = 180° – ∠A
Proof: In quadrilateral ADOE
∠A + ∠D + ∠DOE + ∠E = 360° (Sum of angles of quadrilateral)
⇒ ∠A + 90° + ∠DOE + 90° = 360°
∠A + ∠DOE = 360° – 90° – 90° = 180°
But ∠BOC = ∠DOE (Vertically opposite angles)
⇒ ∠A + ∠BOC = 180°
∴ ∠BOC = 180° – ∠A

Question 15.
In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.

Solution:
Given : In AABC, BA is produced and AE is the bisector of ∠CAD
∠B = ∠C

To prove : AE || BC
Proof: In ∆ABC, BA is produced
∴ Ext. ∠CAD = ∠B + ∠C
⇒ 2∠EAC = ∠C + ∠C (∵ AE is the bisector of ∠CAE) (∵ ∠B = ∠C)
⇒ 2∠EAC = 2∠C
⇒ ∠EAC = ∠C
But there are alternate angles
∴ AE || BC

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1
• RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS
• RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Question 1.
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 455 }{ 1000 } \)
= \(\frac { 91 }{ 200 } \) = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 545 }{ 1000 } \) = \(\frac { 109 }{ 200 } \) = 0.545

Question 2.
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
Two heads : 95 times
One tail : 290 times
No head: 115 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = \(\frac { m }{ n } \)
= \(\frac { No. of possible events }{ Total number of events } \)
= \(\frac { 95 }{ 500 } \) = \(\frac { 19 }{ 100 } \) = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 500 } \) = \(\frac { 580 }{ 1000 } \) = \(\frac { 58 }{ 100 } \) = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 115 }{ 500 } \) = \(\frac { 23 }{ 100 } \) = 0.23

Question 3.
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100

(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 36 }{ 100 } \) = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= \(\frac { m }{ n } \) = \(\frac { 12 }{ 100 } \) = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 86 }{ 100 } \) = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 48 }{ 100 } \) = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 52 }{ 100 } \) = 0.52

Question 4.
1500 families with 2 children were selected randomly and the following data were recorded:

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution:
Total number of families (n) = 1500

(i) Probability of a family having no girls (m) = 211
∴Probability P(A)= \(\frac { m }{ n } \) = \(\frac { 211 }{ 1500 } \) = 0.1406
(ii) Probability of a family having one girl (in) = 814
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 814 }{ 1500 } \) = 0.5426
(iii) Probability of a family having 2 girls (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166
(iv) Probability of a family having at the most one girls
∴m = 814 + 211 = 1025
∴Probability P(A) =\(\frac { m }{ n } \) = \(\frac { 1025 }{ 1500 } \) = 0.6833
(v) Probability of a family having more girls than boys (m) = 475
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 475 }{ 1500 } \) = 0.3166

Question 5.
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:
(i) he hits boundary
(ii) he does not hit a boundary.
Solution:
Total balls played (n) 30
No. of boundaries = 6
(i) When the batsman hits the boundary = 6
∴m = 6
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \) = 0.2
(ii) When the batsman does not hit the boundary (m) = 30 – 6 = 24
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \) = 0.8

Question 6.
The percentage of marks obtained by a student in monthly unit tests are given below:

Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Solution:
Percentage of marks obtain in

(i) Probability of getting more than 70% marks (m) = In unit test II, III, V = 3
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 3 }{ 5 } \) = 0.6
(ii) Getting less then 70% marks = units test I and IV
∴m = 2
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 5 } \) = 0.4
(iii) Getting a distinction = In test V (76 of marks)
∴m = 1
Total unit test (n) = 5
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 1 }{ 5 } \) = 0.2

Question 7.
To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:

Find the probability that a student chosen at random
(i) likes Mathematics
(ii) does not like it.
Solution:
Total number of students (n) = 200

(i) Probability of students who like mathematics (m) = 135
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 135 }{ 200 } \) = 0.675
(ii) Probability of students who dislike mathematics (m) = 65
∴Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 65 }{ 200 } \) = 0.325

Question 8.
The blood groups of 30 students of class IX are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O,
A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:
(i) A (ii) B (iii) AB (iv) O
Solution:
Total number of students of IX class = 30
No. of students of different blood groups
A AB B O
9 3 6 12

Question 9.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour
(in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Number of total bags (n) = 11
No. of bags having weight more than 5 kg (m) = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 11 } \)

Question 10.
Following table shows the birth month of 40 students of class IX.

Find the probability that a student was born in August.
Solution:
Total number of students (n) = 40

Number of students who born in Aug. (m) = 6
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 6 }{ 40 } \) = \(\frac { 3 }{ 20 } \)

Question 11.
Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Find the probability of concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
Total number of days (n) = 30

Probability of cone, of S02 of the interval 0.12-0.16 (m) = 2
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2 }{ 30 } \) = \(\frac { 1 }{ 15 } \)

Question 12.
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

If a family is chosen, find the probability that the family is:
(i)earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii)earning Rs 16000 or more per month and owning exactly I vehicle.
(iii)earning less than Rs 7000 per month and does not own any vehicle.
(iv)earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v)owning not more than 1 vehicle.
(vi)owning at least one vehicle.
Solution:
Total number of families (n) = 2400

(i) Number of families earning income Rs 10000-13000 and owning exactly 2 vehicles (m) = 29
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 29 }{ 2400 } \)
(ii) Number of families earning income Rs 16000 or more having one vehicle (m) = 579
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 579 }{ 2400 } \)
(iii) Number of families earning income less than Rs 7000 having no own vehicle (m) = 10
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 10 }{ 2400 } \) = \(\frac { 1 }{ 240 } \)
(iv) Number of families having X13000 to X16000 having more than two vehicles (m) = 25
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 25 }{ 2400 } \) = \(\frac { 1 }{ 96 } \)
(v) Number of families owning not more than one vehicle (m)
= 10 + 1 + 2 + 1 + 160 + 305 + 533 + 469 + 579 = 2062
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2062 }{ 2400 } \) = \(\frac { 1031 }{ 1200 } \)
(vi) Number of families owning at least one vechile (m) = 2048 + 192 + 110 = 2356
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 2356 }{ 2400 } \) = \(\frac { 589 }{ 600 } \)

Question 13.
The following table gives the life time of 400 neon lamps:

A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.
Solution:
Total number of neon lamps (n) = 400

A bulb is chosen:
(i)No. of bulbs having life time less than 400 hours (m) = 14
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 14 }{ 400 } \) = \(\frac { 7 }{ 200 } \)
(ii)No. of bulbs having life time between 300 to 800 hours (m) = 14 + 56 + 60 + 86 + 74 = 290
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 290 }{ 400 } \) = \(\frac { 29 }{ 40 } \)
(iii)No. of bulbs having life time at least 700 hours (m) = 74 + 62 + 48 = 184
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 184 }{ 400 } \) = \(\frac { 23 }{ 50 } \)

Question 14.
Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.
Solution:
Number of total workers (n) = 30

A worker is selected.
(i)No. of workers having less than Rs 150 (m) = 3 + 4 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(ii)No. of workers having at least Rs 210 (m) = 4 + 3 = 7
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 7 }{ 30 } \)
(iii)No. of workers having more than or equal to Rs 150 but less than Rs 210 = 5 + 6 + 5 = 16
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 16 }{ 30 } \) = \(\frac { 8 }{ 15 } \)

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4

Other Exercises

• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)  = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°

Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.

Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°

Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.

Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°

Question 4.
In the figure, state which lines are parallel and why.

Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE

Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.

Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°

Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:

Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,

∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)  = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°

Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l

∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD

Question 9.
In the figure, ∠1 = 60° and ∠2 = (\(\frac { 2 }{ 3 }\))3 a right angle. Prove that l || m.

Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = \(\frac { 2 }{ 3 }\) rd of a right angle 2
= \(\frac { 2 }{ 3 }\) x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m

Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.

Solution:
In the figure,

l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°

Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l

Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°

∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)

∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°

Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.

Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n

Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?

Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n

Question 16.
Which pair of lines in the figure are parallel? Give reasons.

Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
∴ AD || BC
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC

Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l

To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m

Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.

Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.

Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.

Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH

∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°

Question 21.
In the figure, show that AB || EF.

Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF

Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.

Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D

∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°

Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.


Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary

Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD

Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
Adding (i) and (ii)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB

Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°

Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD

Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
Adding (i) and (ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°

Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.

Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)

BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF

Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.

Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G

Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 25 Probability Ex 25.1
• RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS
• RD Sharma Class 9 Solutions Chapter 25 Probability MCQS

Question 1.
Define a trial.
Solution:
When we perform an experiment, it is called a trial of the experiment.

Question 2.
Define an elementary event.
Solution:
An outcome of a trial of an experiment is called an elementary event.

Question 3.
Define an event.
Solution:
An event association to a random experiment is said to occur in a trial.

Question 4.
Define probability of an event.
Solution:
In n trials of a random experiment if an event A happens m times, then probability of happening
of A is given by P(A) = \(\frac { m }{ n } \)

Question 5.
A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is \(\frac { 2 }{ 5 } \), find the number of red balls in the bag
Solution:
No. of white balls = 4
Let number of red balls = x
Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question 6.
A die is thrown 100 times. If the probability of getting an even number is \(\frac { 2 }{ 5 } \). How many times an odd number is obtained?
Solution:
Total number of a die is thrown = 100
Let an even number comes x times, then probability of an even number = \(\frac { x }{ 100 } \)

Question 7.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Find the probability of getting at most two heads.
Solution:
Total number of three coins are tossed (n) = 200
Getting at the most 2 heads (m) = 72 + 77 + 28 = 177
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 177 }{ 200 } \)

Question 8.
In the Q. No. 7, what is the probability of getting at least two heads?
Solution:
Total number of possible events = 200
No. of events getting at the least = 2 heads (m) = 23 + 72 = 95
Probability P(A) = \(\frac { m }{ n } \) = \(\frac { 95 }{ 200 } \) = \(\frac { 19 }{ 40 } \)

Hope given RD Sharma Class 9 Solutions Chapter 25 Probability VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

Mark the correct alternative in each of the following:
Question 1.
One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution:
Let required angle = x
Then its supplement = (180° – x)
x = 3(180° – x) = 540° – 3x
⇒ x + 3x = 540°
⇒ 4x = 540°
⇒ x = \(\frac { { 540 }^{ \circ } }{ 4 }\)  = 135°
∴ Required angle = 135° (b)

Question 2.
Two straight lines AB and CD intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(a) 86°
(b) 90°
(c) 94°
(d) 137°
Solution:
Sum of angles at a point O = 360°
Sum of three angles ∠AOC + ∠COB + ∠BOD = 274°

∴ Fourth angle ∠AOD = 360° – 274°
= 86° (a)

Question 3.
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution:
CD is a line
∴ ∠BOD + ∠BOC = 180° (Linear pair)

⇒ 63° + ∠BOC = 180°
⇒ ∠BOC = 180° – 63°
∴ ∠BOC =117° (b)

Question 4.
Consider the following statements:
When two straight lines intersect:
(i) adjacent angles are complementary
(ii) adjacent angles are supplementary
(iii) opposite angles are equal
(iv) opposite angles are supplementary Of these statements
(a) (i) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution:
Only (ii) and (iii) arc true. (b)

Question 5.
Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a striaght line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution:
∵ POQ is a straight line
∴ ∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180 – 10° = 170°
∴ x = \(\frac { { 170 }^{ \circ } }{ 5 }\)  = 34° (b)

Question 6.
In the figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°

Solution:
AOB is a straight line,
OC and OD are rays on it
and ∠AOC + ∠BOD = 85°
But ∠AOC + ∠BOD + ∠COD = 180°
⇒ 85° + ∠COD = 180°
∠COD = 180° – 85° = 95° (c)

Question 7.
In the figure, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°

Solution:
In the figure,

y = x (Vertically opposite angles)
∠1 = 3x
∠2 = 3x
∴ 2(x + 3x + 2x) = 360° (Angles at a point)
2x + 6x + 4x = 360°
12x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 12 }\)  = 30°
∴ y = x = 30° (b)

Question 8.
In the figure, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30

Solution:
∠1 = 3x+ 10 (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°

⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = \(\frac { { 80 }^{ \circ } }{ 4 }\) = 20   (c)

Question 9.
In the figure, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f= c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only

Solution:
In the figure,
(i) a + b = d + c
a° = d°
b° = e°
c°= f°
(ii) a + b + e = 180°
a + e + c = 180°
⇒ a + c + e = 180°
(iii) b + f= e + c
∴ (ii) and (iii) are true statements (d)

Question 10.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
(a) 54°
(b) 120°
(c) 108°
(d) 136°
Solution:
In figure, l || m and p is transversal

= \(\frac { 3 }{ 5 }\) x 180° = 108° (c)

Question 11.
In the figure, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°

Solution:
In the figure, AB || CD,
and / is transversal
∠1 = x (Vertically opposite angles)
and 120° + x + ∠1 = 180° (Co-interior angles)

Question 12.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, then ∠AOC =
(a) 70°
(b) 80°
(c) 90°
(d) 180°
Solution:
Two lines AB and CD intersect at O

∠AOC + ∠COB + ∠BOD = 270° …(i)
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° …(ii)
Subtracting (i) from (ii),
∠DOA = 360° – 270° = 90°
But ∠DOA + ∠AOC = 180°
∴ ∠AOC = 180° – 90° = 90° (c)

Question 13.
In the figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°

Solution:
In the figure,

PQ || RS, ∠AEF = 95°
∠BHS = 110°, ∠ABC = x
∵ PQ || RS,
∴ ∠AEF = ∠1 = 95° (Corresponding anlges)
But ∠1 + ∠2 = 180° (Linear pair)
⇒ ∠2 = 180° – ∠1 = 180° – 95° = 85°
In ∆AGH,
Ext. ∠BHS = ∠2 +x
⇒ 110° = 85° + x
⇒ x= 110°-85° = 25° (b)

Question 14.
In the figure, if l1 || l2, what is the value of x?
(a) 90°
(b) 85°
(c) 75°
(d) 70°

Solution:
In the figure,

∠1 = 58° (Vertically opposite angles)
Similarly, ∠2 = 37°
∵ l₁ || l₂, EF is transversal
∠GEF + EFD = 180° (Co-interior angles)
⇒ ∠2 + ∠l +x = 180°
⇒ 37° + 58° + x = 180°
⇒ 95° + x= 180°
x = 180°-95° = 85° (b)

Question 15.
In the figure, if l₁ || l₂, what is x + y in terms of w and z?
(a) 180-w + z
(b) 180° + w- z
(c) 180 -w- z
(d) 180 + w + z

Solution:
In the figure, l₁ || l₂

p and q are transversals
∴ w + x = 180° ⇒ x = 180° – w (Co-interior angle)
z = y (Alternate angles)
∴ x + y = 180° – w + z (a)

Question 16.
In the figure, if l₁ || l₂, what is the value of y?
(a) 100
(b) 120
(c) 135
(d) 150

Solution:
In the figure, l₁ || l₂ and l₃ is the transversal

Question 17.
In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?
(a) 90 + x
(b) 90 + 2x
(c) 90 – \(\frac { x }{ 2 }\)
(d) 90 – 2x

Solution:
In the figure,

l₁ || l₂ and l₃ || l₄ and m is the angle bisector
∴ ∠2 = ∠3 = y
∵ l₁ || l₂
∠1 = x (Corresponding angles)
∵ l₃ || l₄
∴ ∠1 + (∠2 + ∠3) = 180° (Co-interior angles)
⇒ x + 2y= 180°
⇒ 2y= 180°-x
⇒ y = \(\frac { { 540 }^{ \circ }-x }{ 4 }\)
= 90° – \(\frac { x }{ 2 }\) (c)

Question 18.
In the figure, if 11| m, what is the value of x?
(a) 60
(b) 50
(c) 45
d) 30

Solution:
In the figure, l || m and n is the transversal

⇒ y = 25°
But 2y + 25° = x+ 15°
(Vertically opposite angles) ⇒ x = 2y + 25° – 15° = 2y+ 10°
= 2 x 25°+10° = 50°+10° = 60° (a)

Question 19.
In the figure, if AB || HF and DE || FG, then the measure of ∠FDE is
(a) 108°
(b) 80°
(c) 100°
(d) 90°

Solution:
In the figure,
AB || HF, DE || FG

∴ HF || AB
∠1 =28° (Corresponding angles)
But ∠1 + ∠FDE + 72° – 180° (Angles of a straight line)
⇒ 28° + ∠FDE + 72° = 180°
⇒ ∠FDE + 100° = 180°
⇒ ∠FDE = 180° – 100 = 80° (b)

Question 20.
In the figure, if lines l and m are parallel, then x =
(a) 20°
(b) 45°
(c) 65°
(d) 85°

Solution:
In the figure, l || m

∴ ∠1 =65° (Corresponding angles)
In ∆BCD,
Ext. ∠1 = x + 20°
⇒ 65° = x + 20°
⇒ x = 65° – 20°
⇒ x = 45° (b)

Question 21.
In the figure, if AB || CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°

Solution:
In the figure, AB || CD
Through P, draw PQ || AB or CD

∠A + ∠1 = 180° (Co-interior angles)
⇒ 132° + ∠1 = 180°
⇒ ∠1 = 180°- 132° = 48°
∴ ∠2 = 148° – ∠1 = 148° – 48° = 100°
∵ DQ || CP
∴ ∠2 = x (Corresponding angles)
∴ x = 100° (a)

Question 22.
In tlie figure, if lines l and in are parallel lines, then x =
(a) 70°
(b) 100°
(c) 40°
(d) 30°

Solution:
In the figure, l || m

∠l =70° (Corresponding angles)
In ∆DEF,
Ext. ∠l = x + 30°
⇒ 70° = x + 30°
⇒ x = 70° – 30° = 40° (c)

Question 23.
In the figure, if l || m, then x =
(a) 105°
(b) 65°
(c) 40°
(d) 25°

Solution:
In the figure,
l || m and n is the transversal

∠1 = 65° (Alternate angles)
In ∆GHF,
Ext. x = ∠1 + 40° = 65° + 40°
⇒ x = 105°
∴ x = 105° (a)

Question 24.
In the figure, if lines l and m are parallel, then the value of x is
(a) 35°
(b) 55°
(c) 65°
(d) 75°

Solution:
In the figure, l || m
and PQ is the transversal

∠1 = 90°
In ∆EFG,
Ext. ∠G = ∠E + ∠F
⇒ 125° = x + ∠1 = x + 90°
⇒ x = 125° – 90° = 35° (a)

Question 25.
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
(a) 45°
(b) 30°
(c) 36°
(d) none of these
Solution:
Let first angle = x
Then its complementary angle = 90° – x
∴ 2x = 3(90° – x)
⇒ 2x = 270° – 3x
⇒ 2x + 3x = 270°
⇒ 5x = 270°
⇒ x = \(\frac { { 270 }^{ \circ } }{ 5 }\)  = 54°
∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)

Question 26.

Solution:

Question 27.
In the figure, AB || CD || EF and GH || KL.
The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°

Solution:
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.

∵ AB || CD
∴ ∠1 = 25° (Alternate angle)
and GH || KL
∴ ∠4 = 60° (Corresponding angles)
∠5 = ∠4 = 60° (Vertically opposite angle)
∠5 + ∠2 = 180° (Co-interior anlges)
∴ ⇒ 60° + ∠2 = 180°
∠2 = 180° – 60° = 120°
Now ∠HKL = ∠1 + ∠2 = 25° + 120°
= 145° (c)

Question 28.
AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB = 35°, then ∠CFQ will be
(a) 55°
(b) 70°
(c) 110°
(d) 130°
Solution:
AB || CD and PQ is the transversal EL is the bisector of ∠FEB and ∠LEB = 35°

∴ ∠FEB = 2 x 35° = 70°
∵ AB || CD
∴ ∠FEB + ∠EFD = 180°
(Co-interior angles)
70° + ∠EFD = 180°
∴ ∠EFD = 180°-70°= 110°
But ∠CFQ = ∠EFD
(Vertically opposite angles)
∴ ∠CFQ =110° (c)

Question 29.
In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?
(a) 12
(b) 15
(c) 18
(d) 20

Solution:
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y = \(\frac { { 180 }^{ \circ } }{ 9 }\)  = 20° (d)

Question 30.
In the figure, if CP || DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°

Solution:
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D

∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
In ∆ADC,
Ext. ∠ACP = ∠CDA + ∠DAC
⇒ x = ∠1 + 25°
= 105° + 25° = 130° (a)

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Find the median of the following data (1-8)

Question 1.
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
We know that median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
(When n is even)
= \(\frac { n+1 }{ 2 } th\quad term\)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Arranging in ascending order, 29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here n = 10 which an even
Median = \(\frac { 1 }{ 2 } \)[5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (45+54) = \(\frac { 99 }{ 2 } \) = 49.5

Question 2.
133, 73, 89, 108, 94, 104, 94, 85, 100, 120
Solution:
133, 73, 89, 108, 94, 104, 94, 85, 100, 120
Arranging in ascending order, 73, 85, 89, 94, 94, 100, 104, 108, 120, 133
Here n = 10 which is an even
Median = \(\frac { 1 }{ 2 } \)[5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (94+100) = \(\frac { 1 }{ 2 } \) x 194 = 97

Question 3.
31, 38, 27, 28, 36, 25, 35, 40
Solution:
31, 38, 27, 28, 36, 25, 35, 40
Arranging in ascending order, 25, 27, 28, 31, 35, 36, 38, 40
Here n = 8 which is even
Median = \(\frac { 1 }{ 2 } \)[4th term + 5th term]
= \(\frac { 1 }{ 2 } \) (31+35) = \(\frac { 1 }{ 2 } \) x 66 = 33

Question 4.
15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
15, 6, 16, 8, 22, 21, 9, 18, 25
Arranging in ascending order = 6, 8, 9, 15, 16, 18, 21, 22, 25
Here n = 9 which is odd
Median \(\frac { n+1 }{ 2 } th\quad term\) = \(\frac { 9+1 }{ 2 } th\quad term\) = \(\frac { 10 }{ 2 } th\quad \)
= 5th term = 16

Question 5.
41, 43, 127, 99, 71, 92, 71, 58, 57
Solution:
41, 43, 127, 99, 71, 92, 71, 58, 57
Arranging in ascending order = 41, 43, 57, 58, 71, 71, 92, 99, 127
Here n = 9 which is an odd
Median \(\frac { n+1 }{ 2 } th\quad term\) = \(\frac { 9+1 }{ 2 } th\quad term\) = \(\frac { 10 }{ 2 } th\quad\)
= 5th term = 71

Question 6.
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Arranging in ascending order = 20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Here n = 10 which is even
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (26 + 29) = \(\frac { 1 }{ 2 } \) x 55 = \(\frac { 55 }{ 2 } \) = 27.5

Question 7.
12, 17, 3, 14, 5, 8, 7, 15
Solution:
12, 17, 3, 14, 5, 8, 7, 15
Arranging in ascending order = 3, 5, 7, 8, 12, 14, 15, 17
Here n = 8 which is odd
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [4th term + 5th term]
= \(\frac { 1 }{ 2 } \) (8+12) = \(\frac { 1 }{ 2 } \) x 20 = 10

Question 8.
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
92, 35, 67, 85, 72, 81, 56, 51, 42. 69
Arranging in ascending order = 35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Here n = 10 which is even
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (67+69) = \(\frac { 1 }{ 2 } \) x 136 = 68

Question 9.
Numbers 50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are written in descending order and their median is 25, find x.
Solution:
50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are in descending order
Here n = 8 which is even
Now Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [4th term + 5th term] = \(\frac { 1 }{ 2 } \)[2x + 10 + 2x – 8]
= \(\frac { 1 }{ 2 } \) [4x + 2] = 2x + 1
But median = 25
2x + 1 = 25
=> 2x = 25 – 1 = 24
=> \(\frac { 24 }{ 2 } \) = 12
Hence x = 12

Question 10.
Find the median of the following observations 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution:
46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33
Writing in ascending order = 33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here n = 11 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term
= \(\frac { 11+1 }{ 2 } \) = \(\frac { 12 }{ 2 } \)
= 6th term = 58
By replacing 92 by 93 and 41 by 43, then new order will be
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
Median = 6th term = 58

Question 11.
Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median.
Solution:
41, 43, 127, 99, 61, 92, 71, 58, 57
Arranging in ascending order = 41, 43, 57, 58, 61, 71, 92, 99, 127
Here n = 9 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term
= \(\frac { 10 }{ 2 } \) = 5th term = 61
By change 58 by 92, we get new order = 41, 43, 57, 61, 71, 92, 92, 99, 127
Median = 5th term = 71

Question 12.
The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Solution:
Weights of 15 students are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30
Writing in ascending order = 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45
here n = 15 which is odd
n+1 15+1
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 15+1 }{ 2 } \)
= \(\frac { 16 }{ 2 } \)th term = 8th term = 35 kg
By replacing 44 kg by 46 kg and 27 kg by 25 kg we get new order,
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46
Median = 8th term = 35 kg

Question 13.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Median = 63
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Here n = 10 which is even
median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 16 }{ 2 } \) [x+x+2] = \(\frac { 2x + 2 }{ 2 } \) = x + 1
x + 1 = 63 = x = 63 – 1 = 62
Hence x = 62

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.