RD Sharma Solutions Class 9 Maths Chapter 11 – Coordinate Geometry (Updated for 2021-22)

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RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1

Question 1.
In a ∆ABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Solution:
∵ Sum of three angles of a triangle is 180°
∴ In ∆ABC, ∠A = 55°, ∠B = 40°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
∴ ∠C= 180° -95° = 85°

Question 2.
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
∴ x + 2x + 3x = 180° (Sum of angles of a triangle)
⇒6x = 180°
⇒x = 180∘6  = 30°
∴ First angle = x = 30°
Second angle = 2x = 2 x 30° = 60°
and third angle = 3x = 3 x 30° = 90°
∴ Angles are 30°, 60°, 90°

Question 3.
The angles of a triangle are (x – 40)°, (x – 20)° and (12 x – 10)°. Find the value of x.
Solution:
∵ Sum of three angles of a triangle = 180°
∴ (x – 40)° + (x – 20)° + (12x-10)0 = 180°
⇒ x – 40° + x – 20° + 12x – 10° = 180°
⇒ x + x+ 12x – 70° = 180°
⇒ 52x = 180° + 70° = 250°
⇒ x = 250∘x25  = 100°
∴ x = 100°

Question 4.
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.
Solution:
Let each of the two equal angles = x
Then third angle = x + 30°
But sum of the three angles of a triangle is 180°
∴ x + x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒3x = 150° ⇒x = 150∘3 = 50°
∴ Each equal angle = 50°
and third angle = 50° + 30° = 80°
∴ Angles are 50°, 50° and 80°

Question 5.
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
In the triangle ABC,

∠B = ∠A + ∠C
But ∠A + ∠B + ∠C = 180°
⇒∠B + ∠A + ∠C = 180°
⇒∠B + ∠B = 180°
⇒2∠B = 180°
∴ ∠B = 180∘2 = 90°
∵ One angle of the triangle is 90°
∴ ∆ABC is a right triangle.

Question 6.
Can a triangle have:
(i) Two right angles?
(ii) Two obtuse angles?
(iii) Two acute angles?
(iv) All angles more than 60°?
(v) All angles less than 60°?
(vi) All angles equal to 60°?
Justify your answer in each case.
Solution:
(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.
(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are
two obtuse angle, then the third angle will be negative which is not possible.
(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.
(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.
(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.
(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

Question 7.
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10°, find the three angles.
Solution:
Let three angles of a triangle be x°, (x + 10)°, (x + 20)°
But sum of three angles of a triangle is 180°
∴ x + (x+ 10)° + (x + 20) = 180°
⇒ x + x+10°+ x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30° = 150°
∴ x = 180∘2 = 50°
∴ Angle are 50°, 50 + 10, 50 + 20
i.e. 50°, 60°, 70°

Question 8.
ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Solution:
In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O

Now ∠B + ∠C = 180° – 12° = 108°
∵ OB and OC are the bisectors of ∠B and ∠C respectively
∴ ∠OBC + ∠OCB = 12 (B + C)
= 12 x 108° = 54°
But in ∆OBC,
∴ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 54° + ∠BOC = 180°
∠BOC = 180°-54°= 126°
OR
According to corollary,
∠BOC = 90°+ 12 ∠A
= 90+ 12 x 72° = 90° + 36° = 126°

Question 9.
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:
In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively
To prove : ∠BOC cannot be a right angle
Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively

∴ ∠BOC = 90° x 12 ∠A
Let ∠BOC = 90°, then
12 ∠A = O
⇒∠A = O
Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

Question 10.
If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that the triangle is a right triangle.
Solution:
Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°

To prove : ∆ABC is a right angled triangle
Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O
∴ ∠BOC = 90°+ 12∠A
But ∠BOC =135°
∴ 90°+ 12 ∠A = 135°
⇒ 12∠A= 135° -90° = 45°
∴ ∠A = 45° x 2 = 90°
∴ ∆ABC is a right angled triangle

Question 11.
In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB

To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° + 12∠A
But ∠BOC = 120°
∴ 90°+ 12 ∠A = 120°
∴ 12 ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C = 120∘2 = 60°
Hence ∠A = ∠B = ∠C = 60°

Question 12.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
In a ∆ABC,
Let ∠A < ∠B + ∠C

⇒∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)
Similarly, we can prove that
∠B < 90° and ∠C < 90°
∴ Each angle of the triangle are acute angle.

RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2

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 = 180∘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°+ 12 ∠A …(i)
Similarly, QB and QC are the bisectors of exterior angles B and C
∴ ∠BQC = 90° + 12 ∠A …(ii)
Adding (i) and (ii),
∠BPC + ∠BQC = 90° + 12 ∠A + 90° – 12 ∠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 = 12 ∠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

RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry VSAQS

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 = 180∘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 = 180∘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 = 180∘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°+ 12
90°+ 12 ∠A= 120°
⇒ 12 ∠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
⇒ 12 ∠ACD = 12 ∠A + 12 ∠B
⇒ ∠2= 12 ∠A + ∠1 …(i)
But in ∆BCE,
Ext. ∠2 = ∠E + ∠l
⇒ ∠E + ∠l = ∠2 = 12 ∠A + ∠l [From (i)]
⇒ ∠E = 12 ∠A = 68∘2  =34°

Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry MCQS

Mark the correct alternative in each of the following:
Question 1.
If all the three angles of a triangle are equal, then each one of them is equal to
(a) 90°
(b) 45°
(c) 60°
(d) 30°
Solution:
∵ Sum of three angles of a triangle = 180°
∴ Each angle = 180∘3  = 60° (c)

Question 2.
If two acute angles of a right triangle are equal, then each acute is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
In a right triangle, one angle = 90°
∴ Sum of other two acute angles = 180° – 90° = 90°
∵ Both angles are equal
∴ Each angle will be = 90∘2  = 45° (b)

Question 3.
An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to
(a) 75°
(b) 80°
(c) 40°
(d) 50°
Solution:
In a triangle, exterior angles is equal to the sum of its interior opposite angles
∴ Sum of interior opposite angles = 100°
∵ Both angles are equal
∴ Each angle will be = 100∘2  = 50° (d)

Question 4.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C

But ∠A + ∠B + ∠C = 180°
( Sum of angles of a triangle)
∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°
⇒ ∠A = 180∘2  = 90°
∴ ∆ is a right triangle (d)

Question 5.
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = 12∠A, then ∠A is equal to
(a) 80°
(b) 75°
(c) 60°
(d) 90°
Solution:
Side BC of ∆ABC is produced to D, then

Ext. ∠ACB = ∠A + ∠B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

Question 6.
In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
(a) 35°
(b) 90°
(c) 70°
(d) 55°
Solution:
In ∆ABC, ∠B = ∠C
AX is the bisector of ext. ∠CAD
∠DAX = 70°

∴ ∠DAC = 70° x 2 = 140°
But Ext. ∠DAC = ∠B + ∠C
= ∠C + ∠C (∵ ∠B = ∠C)
= 2∠C
∴ 2∠C = 140° ⇒ ∠C = 140∘2 = 70°
∴ ∠ACB = 70° (c)

Question 7.
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is
(a) 55°
(b) 85°
(c) 40°
(d) 9.0°
Solution:
In ∆ABC, BA is produced to D such that ∠CAD = 95°
and let ∠C = 55° and ∠B = x°

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle
∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°
⇒ x = 95° – 55° = 40°
∴ Other interior angle = 40° (c)

Question 8.
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
(a) 90°
(b) 180°
(c) 270°
(d) 360°
Solution:
In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACE are formed

We know an exterior angles of a triangle is equal to the sum of its interior opposite angles
∴ ∠FAB = ∠B + ∠C
∠DBC = ∠C + ∠A and
∠ACE = ∠A + ∠B Adding we get,
∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B
= 2(∠A + ∠B + ∠C)
= 2 x 180° (Sum of angles of a triangle)
= 360° (d)

Question 9.
In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =
(a) 50°
(b) 90°
(c) 40°
(d) 100°
Solution:
In ∆ABC, ∠A = 100°
AD is bisector of ∠A and AD ⊥ BC

Now, ∠BAD = 100∘2 = 50°
In ∆ABD,
∠BAD + ∠B + ∠D= 180°
(Sum of angles of a triangle)
⇒ ∠50° + ∠B + 90° = 180°
∠B + 140° = 180°
⇒ ∠B = 180° – 140° ∠B = 40° (c)

Question 10.
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
(a) 48°, 60°, 72°
(b) 50°, 60°, 70°
(c) 52°, 56°, 72°
(d) 42°, 60°, 76°
Solution:
In ∆ABC, BC is produced to D and ∠ACD = 108°

Ratio in ∠A : ∠B = 4:5
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles
∴ ∠ACD = ∠A + ∠B = 108°
Ratio in ∠A : ∠B = 4:5

Question 11.
In a ∆ABC, if ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC =
(a) 60°
(b) 120°
(c) 150°
(d) 30°
Solution:
In ∆ABC, ∠A = 60°, ∠B = 80°
∴ ∠C = 180° – (∠A + ∠B)
= 180° – (60° + 80°)
= 180° – 140° = 40°
Bisectors of ∠B and ∠C meet at O

Question 12.
Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB = 55°, then ∠BOD =
(a) 100°
(b) 80°
(c) 90°
(d) 135°
Solution:
In the figure,

AB and CD intersect at O
and AC || DB, ∠CAB = 45°
and ∠CDB = 55°
∵ AC || DB
∴ ∠CAB = ∠ABD (Alternate angles)
In ∆OBD,
∠BOD = 180° – (∠CDB + ∠ABD)
= 180° – (55° + 45°)
= 180° – 100° = 80° (b)

Question 13.
In the figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is

(a) 20°
(b) 50°
(c) 60°
(d) 70°
Solution:
In the figure, EC || AB
∠ECD = 70°, ∠BDO = 20°

∵ EC || AB
∠AOD = ∠ECD (Corresponding angles)
⇒ ∠AOD = 70°
In ∆OBD,
Ext. ∠AOD = ∠OBD + ∠BDO
70° = ∠OBD + 20°
⇒ ∠OBD = 70° – 20° = 50° (b)

Question 14.
In the figure, x + y =
(a) 270
(b) 230
(c) 210
(d) 190°

Solution:
In the figure

Ext. ∠OAE = ∠AOC + ∠ACO
⇒ x = 40° + 80° = 120°
Similarly,
Ext. ∠DBF = ∠ODB + ∠DOB
y = 70° + ∠DOB
[(∵ ∠AOC = ∠DOB) (vertically opp. angles)]
= 70° + 40° = 110°
∴ x+y= 120°+ 110° = 230° (b)

Question 15.
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
(a) 25°
(b) 30°
(c) 45°
(d) 60°
Solution:
Ratio in the measures of the triangle =3:4:5
Sum of angles of a triangle = 180°
Let angles be 3x, 4x, 5x
Sum of angles = 3x + 4x + 5x = 12x
∴ Smallest angle = 180x3x12x = 45° (c)

Question 16.
In the figure, if AB ⊥ BC, then x =
(a) 18
(b) 22
(c) 25
(d) 32

Solution:
In the figure, AB ⊥ BC
∠AGF = 32°
∴ ∠CGB = ∠AGF (Vertically opposite angles)
= 32°

In ∆GCB, ∠B = 90°
∴ ∠CGB + ∠GCB = 90°
⇒ 32° + ∠GCB = 90°
⇒ ∠GCB = 90° – 32° = 58°
Now in ∆GDC,
Ext. ∠GCB = ∠CDG + ∠DGC
⇒ 58° = x + 14° + x
⇒ 2x + 14° = 58°
⇒ 2x = 58 – 14° = 44
⇒ x = 442 = 22°
∴ x = 22° (b)

Question 17.
In the figure, what is ∠ in terms of x and y?
(a) x + y + 180
(b) x + y – 180
(c) 180° -(x+y)
(d) x+y + 360°

Solution:
In the figure, BC is produced both sides CA and BA are also produced
In ∆ABC,
∠B = 180° -y
and ∠C 180° – x

∴ z = ∠A = 180° – (B + C)
= 180° – (180 – y + 180 -x)
= 180° – (360° – x – y)
= 180° – 360° + x + y = x + y – 180° (b)

Question 18.
In the figure, for which value of x is l₁ || l₂?
(a) 37
(b) 43
(c) 45
(d) 47

Solution:
In the figure, l₁ || l₂
∴ ∠EBA = ∠BAH (Alternate angles)
∴ ∠BAH = 78°

⇒ ∠BAC + ∠CAH = 78°
⇒ ∠BAC + 35° = 78°
⇒ ∠BAC = 78° – 35° = 43°
In ∆ABC, ∠C = 90°
∴ ∠ABC + ∠BAC = 90°
⇒ x + 43° = 90° ⇒ x = 90° – 43°
∴ x = 47° (d)

Question 19.
In the figure, what is y in terms of x?

Solution:
In ∆ABC,
∠ACB = 180° – (x + 2x)
= 180° – 3x …(i)
and in ∆BDG,
∠BED = 180° – (2x + y) …(ii)
∠EGC = ∠AGD (Vertically opposite angles)
= 3y

In quad. BCGE,
∠B + ∠ACB + ∠CGE + ∠BED = 360° (Sum of angles of a quadrilateral)
⇒ 2x+ 180° – 3x + 3y + 180°- 2x-y = 360°
⇒ -3x + 2y = 0
⇒ 3x = 2y ⇒ y = 32x (a)

Question 20.
In the figure, what is the value of x?
(a) 35
(b) 45
(c) 50
(d) 60

Solution:
In the figure, side AB is produced to D
∴ ∠CBA + ∠CBD = 180° (Linear pair)
⇒ 7y + 5y = 180°
⇒ 12y = 180°
⇒ y = 18012 = 15
and Ext. ∠CBD = ∠A + ∠C
⇒ 7y = 3y + x
⇒ 7y -3y = x
⇒ 4y = x
∴ x = 4 x 15 = 60 (d)

Question 21.
In the figure, the value of x is
(a) 65°
(b) 80°
(c) 95°
(d) 120°

Solution:
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°

Now in ∆ABD,
Ext. ∠DBC = ∠A + ∠D
= 55° + 25° = 80°
Similarly, in ∆BCE,
Ext. ∠DEC = ∠EBC + ∠ECB
= 80° + 40° = 120° (d)

Question 22.
In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20°
(b) 25°
(c) 30°
(d) 35°

Solution:
In the figure, AC = BC, BP || CQ

∵ BP || CQ
∴ ∠PBC – ∠QCD
⇒ 20° + ∠ABC = 70°
⇒ ∠ABC = 70° – 20° = 50°
∵ BC = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides)
= 50°
Now in ∆ABC,
Ext. ∠ACD = ∠B + ∠A
⇒ x + 70° = 50° + 50°
⇒ x + 70° = 100°
∴ x = 100° – 70° = 30° (c)

Question 23.
In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
(a) x = 55°, y = 40°
(b) x = 50°, y = 45°
(c) x = 60°, y = 35°
(d) x = 35°, y = 60°

Solution:
In the figure,
∵ AB || CD, EF intersects them at P and Q respectively,
∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°
∵ AB || CD
∴ ∠APQ = ∠CQF (Corresponding anlges)
⇒ y + 25° = 65°
⇒ y = 65° – 25° = 40°
and APQ + PQC = 180° (Co-interior angles)

y + 25° + ∠1 +30°= 180°
40° + 25° + ∠1 + 30° = 180°
⇒ ∠1 + 95° = 180°
∴ ∠1 = 180° – 95° = 85°
Now, ∆PQR,
∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)
⇒ 40° + x + 85° = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° = 55°
∴ x = 55°, y = 40° (a)

Question 24.
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠BAC = ?
(a) 94°
(b) 54°
(c) 40°
(d) 44°
Solution:
In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°

Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 94° = ∠BAC + ∠ABC
Similarly, ∠ABE = ∠BAC + ∠ACB
⇒ 126° = ∠BAC + ∠ACB
Adding,
94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC
220° = 180° + ∠BAC
∴ ∠BAC = 220° -180° = 40° (c)

Question 25.
If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is
(a) 45°
(b) 95°
(c) 135°
(d) 90°
Solution:
In right ∆ABC, ∠A = 90°

Bisectors of ∠B and ∠C meet at O, then 1
∠BOC = 90° + 12 ∠A
= 90°+ 12 x 90° = 90° + 45°= 135° (c)

Question 26.
The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=


Solution:
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.

Question 27.
In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
(a) 25°
(b) 50°
(c) 100°
(d) 75°
Solution:
In ∆ABC, ∠A = 50°
BC is produced

Bisectors of ∠ABC and ∠ACD meet at ∠E
∴ ∠E = 12 ∠A = 12 x 50° = 25° (a)

Question 28.
The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD =115°,then ∠ALC =
(a) 85°
(b) 7212 °
(c) 145°
(d) none of these
Solution:
In ∆ABC, BC is produced to D
∠B = 30°, ∠ACD = 115°

Question 29.
In the figure , if l1 || l2, the value of x is
(a) 22 12
(b) 30
(c) 45
(d) 60

Solution:
In the figure, l₁ || l₂
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB

∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively
∴ ∠CEB = 90°
∴ a + b = 90° ,
and ∠GEB = 90° (∵ ∠CEB = 90°)
2x = 90° ⇒ x = 902 = 45 (c)

Question 30.
In ∆RST (in the figure), what is the value of x?
(a) 40°
(b) 90°
(c) 80°
(d) 100°

Solution:

Detailed Exercise-wise Explanation with Listing of Important Topics 

A student must remember to check all the significant details of Class 9 Chapter 11 RD Sharma Solutions while getting ready for the final Class 9 Maths exam. In Chapter 11 Solutions RD Sharma the students will cover from all kinds of tricky questions.

A basic idea of the parts falling under Coordinate Geometry can play an important role in securing a good position. You will get all of that in the RD Sharma Class 9 Solutions Chapter 11. Practicing on a regular basis holds the key and here we have tried to throw light on the exercise wise explanations. Some solutions from the various exercises have been presented here take a look-

Question 1: Plot the following points on the graph paper:

(i) (2,5) (ii) (4,-3) (iii) (-5,-7) (iv) (7,-4) (v) (-3,2)

(vi) (7,0) (vii) (-4,0) (viii) (0,7) (ix) (0,-4) (x) (0,0)

Solution:

Question 2: Write the coordinates of each of the following points marked in the graph paper.

Solution:

Point	Distance from the y-axis (units)	Distance from x-axis (units)	Coordinates of Point
A	3	1	(3,1)
B	6	0	(6, 0)
C	0	6	(0, 6)
D	-3	0	(-3,0)
E	-4	3	(-4,3)
F	-2	-4	(-2,-4)
G	0	-5	(0,-5)
H	3	-6	(3,-6)
P	7	-3	(7,-3)
Q	7	6	(7,6)

Explanation for Coordinates of A:

The distance of point A from y-axis is 3 units and from x-axis is 1 unit.

A lies in the first quadrant, so both the coordinates are positive.

This implies coordinates of A are (3,1).

Similarly, other points are,

A (3,1), B (6,0), C (0,6), D (-3,0), E (-4,3), F (-2,-4), G (0,-5), H (3,-6), P (7,-3), Q (7,6)

Important concepts

• Rectangular or cartesian coordinates of a point
• Cartesian co-ordinates axes
• Quadrants
• Cartesian coordinates of a point
• Plotting of points

This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 11. To know more about the CBSE Class 9 Maths exam, ask in the comments.

FAQs on RD Sharma Solutions Class 9 Maths Chapter 11 Coordinate Geometry