Prasanna

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS

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

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.
Define a triangle.
Solution:
A figure bounded by three lines segments in a plane is called a triangle.

Question 2.
Write the sum of the angles of an obtuse triangle.
Solution:
The sum of angles of an obtuse triangle is 180°.

Question 3.
In ∆ABC, if ∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.
Solution:
In ∆ABC, ∠B = 60°, ∠C = 80°
OB and OC are the bisectors of ∠B and ∠C
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 60° + 80° = 180°
⇒ ∠A + 140° = 180°
∴ ∠A = 180°- 140° = 40°

= 90° + – x 40° = 90° + 20° = 110°

Question 4.
If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.
Solution:
Sum of angles of a triangle = 180°
Ratio in the angles = 2 : 1 : 3
Let first angle = 2x
Second angle = x
and third angle = 3x
∴ 2x + x + 3x = 180° ⇒ 6x = 180°
∴ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ First angle = 2x = 2 x 30° = 60°
Second angle = x = 30°
and third angle = 3x = 3 x 30° = 90°
Hence angles are 60°, 30°, 90°

Question 5.
State exterior angle theorem.
Solution:
Given : In ∆ABC, side BC is produced to D

To prove : ∠ACD = ∠A + ∠B
Proof: In ∆ABC,
∠A + ∠B + ∠ACB = 180° …(i) (Sum of angles of a triangle)
and ∠ACD + ∠ACB = 180° …(ii) (Linear pair)
From (i) and (ii)
∠ACD + ∠ACB = ∠A + ∠B + ∠ACB
∠ACD = ∠A + ∠B
Hence proved.

Question 6.
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Solution:
In ∆ABC,
∠A + ∠C = ∠B

But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠B + ∠A + ∠C = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ Third angle = 90°

Question 7.
In the figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

Solution:
Given : In figure, AB || CD, EF || BC ∠BAC = 65°, ∠DHF = 35°

∵ EF || BC
∴ ∠A = ∠ACH (Alternate angle)
∴ ∠ACH = 65°
∵∠GHC = ∠DHF
(Vertically opposite angles)
∴ ∠GHC = 35°
Now in ∆GCH,
Ext. ∠AGH = ∠GCH + ∠GHC
= 65° + 35° = 100°

Question 8.
In the figure, if AB || DE and BD || FG such that ∠FGH = 125° and ∠B = 55°, find x and y.

Solution:
In the figure, AB || DF, BD || FG

∠FGH = 125° and ∠B = 55°
∠FGH + FGE = 180° (Linear pair)
⇒ 125° + y – 180°
⇒ y= 180°- 125° = 55°
∵ BA || FD and BD || FG
∠B = ∠F = 55°
Now in ∆EFG,
∠F + ∠FEG + ∠FGE = 180°
(Angles of a triangle)
⇒ 55° + x + 55° = 180°
⇒ x+ 110°= 180°
∴ x= 180°- 110° = 70°
Hence x = 70, y = 55°

Question 9.
If the angles A, B and C of ∆ABC satisfy the relation B – A = C – B, then find the measure of ∠B.
Solution:
In ∆ABC,
∠A + ∠B + ∠C= 180° …(i)
and B – A = C – B

⇒ B + B = A + C ⇒ 2B = A + C
From (i),
B + 2B = 180° ⇒ 3B = 180°
∠B = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 60°
Hence ∠B = 60°

Question 10.
In ∆ABC, if bisectors of ∠ABC and ∠ACB intersect at O at angle of 120°, then find the measure of ∠A.
Solution:
In ∆ABC, bisectors of ∠B and ∠C intersect at O and ∠BOC = 120°

But ∠BOC = 90°+ \(\frac { 1 }{ 2 }\)
90°+ \(\frac { 1 }{ 2 }\) ∠A= 120°
⇒ \(\frac { 1 }{ 2 }\) ∠A= 120°-90° = 30°
∴ ∠A = 2 x 30° = 60°

Question 11.
If the side BC of ∆ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.
Solution:
In ∆ABC, side BC is produced on both sides forming exterior ∠ABE and ∠ACD
Ext. ∠ABE = ∠A + ∠ACB
and Ext. ∠ACD = ∠ABC + ∠A

Adding we get,
∠ABE + ∠ACD = ∠A + ∠ACB + ∠A + ∠ABC
⇒ ∠ABE + ∠ACD – ∠A = ∠A 4- ∠ACB + ∠A + ∠ABC – ∠A (Subtracting ∠A from both sides)
= ∠A + ∠ABC + ∠ACB = ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

Question 12.
In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find ∠ACD: ∠ADC.
Solution:
In ∆ABC, AB = AC
AB is produced to D such that BD = BC
DC are joined
In ∆ABC, AB = AC
∴ ∠ABC = ∠ACB
In ∆ BCD, BD = BC
∴ ∠BDC = ∠BCD
and Ext. ∠ABC = ∠BDC + ∠BCD = 2∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 2∠BCD (∵ ∠ABC = ∠ACB)
Adding ∠BDC to both sides
⇒ ∠ACB + ∠BDC = 2∠BDC + ∠BDC
⇒ ∠ACB + ∠BCD = 3 ∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 3∠BDC

Question 13.
In the figure, side BC of AABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.

Solution:
In the figure,

side BC of ∆ABC is produced to D such that bisectors of ∠ABC and ∠ACD meet at E
∠BAC = 68°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ \(\frac { 1 }{ 2 }\) ∠ACD = \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B
⇒ ∠2= \(\frac { 1 }{ 2 }\) ∠A + ∠1 …(i)
But in ∆BCE,
Ext. ∠2 = ∠E + ∠l
⇒ ∠E + ∠l = ∠2 = \(\frac { 1 }{ 2 }\) ∠A + ∠l [From (i)]
⇒ ∠E = \(\frac { 1 }{ 2 }\) ∠A = \(\frac { { 68 }^{ \circ } }{ 2 }\)  =34°

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry 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 24 Measures of Central Tendency MCQS

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 MCQS

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

Mark the correct alternative in each of the following:

Question 1.
Which one of the following is not a measure of central value?
(a) Mean
(b) Range
(c) Median
(d) Mode
Solution:
Range (b)

Question 2.
The mean of n observations is \(\overline { X } \) . If k is added to each observation, then the new mean is
(a) \(\overline { X } \)
(b) \(\overline { X } \) + k
(c) \(\overline { X } \) – k
(d) k\(\overline { X } \)
Solution:
Mean of n observation = \(\overline { X } \)
By adding k to each observation the new mean will be \(\overline { X } \) + k (b)

Question 3.
The mean of n observations is \(\overline { X } \) . If each observation is multiplied by k, the mean of new observations is
(a) k\(\overline { X } \)
(b) \(\frac { \overline { X } }{ k } \)
(c) \(\overline { X } \) + k
(d) \(\overline { X } \) – k
Solution:
Mean of n observations = \(\overline { X } \)
By multiplying each observation by k,
the new mean = k\(\overline { X } \) (a)

Question 4.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is
(a) 98
(b) 99
(c) 100
(d) 101
Solution:
Mean of 7 numbers = 81
Total = 7 x 81 = 567
By discarding one number, the mean of the remaining 7 – 1 = 6 numbers = 78
Total = 6 x 78 = 468
Discarded number = 567 – 468 = 99 (b)

Question 5.
For which set of numbers do the mean, median and mode all have the same value?
(a) 2, 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 1, 2, 5
Solution:
a) In set 2, 2, 2, 2, 4

Mode = 3 as it come in maximum times
This set has mean, median and mode same (b)

Question 6.
For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?
(a) Mean = Median
(b) Mean > Mode
(c) Mean < Mode
(d) Mode = Median
Solution:
The given set is 2, 2, 4, 5, 12

Question 7.
If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is
(a) 10
(b) 12
(c) 14
(d) 16
Solution:
Arithmetic mean of 7, 5, 13, x, 9 is 10

Question 8.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is
(a) 9
(b) 11
(c) 13
(d) none of these
Solution:
Mean = 11
But mean of x, x + 2, x + 4, x+ 6, x + 8

Question 9.
Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) none of these
Solution:
Mode is most frequent value (c)

Question 10.
The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is
(a) 7
(b) 5
(c) 8
(d) 10
Solution:
Wages per day
5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8
=> 4, 5, 5, 5, 5, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10
Here 8 comes in maximum times
Mode = 8 (c)

Question 11.
The median of the following data :
is ,
(a) 0
(b) -1.5
(c) 2
(d) 3.5
Solution:
Arranging in ascending order,
-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6

Question 12.
The algebraic sum of the deviations of a set of n values from their mean is
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
The algebraic sum of deviation of a set of n values from that mean

Question 13.
A, B, C are three sets of values of X:
A : 2, 3, 7, 1, 3, 2, 3
B: 7, 5, 9, 12, 5, 3, 8
C: 4, 4, 11, 7 ,2, 3, 4
Which one of the following statements is
correct?
(a) Mean of A = Mode of C
(b) Mean of C = Median of B
(c) Median of B = Mode of A
(d) Mean, Median and Mode of A are equal.
Solution:
Arranging the sets in ascending order
A{2, 3, 7, 1,3,2,3)
= {1, 2, 2, 3, 3, 3, 7)
B = {7, 5, 9, 12, 5, 3, 8)
= {3, 5, 5, 7, 8, 9, 12)
C = {4, 4, 11,7,2,3,4)
= {2, 3, 4, 4, 4, 7, 11)

Mode = 5 {as it comes max times}
(c) Mean of set C = \(\\ \frac { 2+3+4+4+4+7+11 }{ 7 } \)
= \(\\ \frac { 35 }{ 7 } \) = 5
Median = \(\\ \frac { 7+1 }{ 2 } \) th =\(\\ \frac { 8 }{ 2 } \) =4th term = 4
Mode =4 {as it comes max times}
In set A,mean = median = mode = 3 (d)

Question 14.
The empirical relation between mean, mode and median is
(a) Mode = 3 Median — 2 Mean
(b) Mode 2 Median — 3 Mean
(c) Median 3 Mode — 2 Mean
(d) Mean = 3 Median —2 Mode
Solution:
The empirical relations between mean, mode
and median is
Mode = 3 Median — 2 Mean (a)

Question 15.
The mean of a, b, c, d and e is 28. If the mean of a, c, and e is 24, what is the mean of b and d?
(a) 31
(b) 32
(c) 33
(d) 34
Solution:
Mean of a, b, c, d and e = 28
Total of a, b, c, d and e = 28 x 5 = 140
Mean of a, c and e is = 24
Total of a, c, e = 24 x 3 = 72
Total of b and d = 140 – 72 = 68
Mean = \(\\ \frac { 68 }{ 2 } \) = 34 (d)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency 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 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.