## 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.

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