RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Created by: Abhishek Kapoor

Class 9 : RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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Q. (1) In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.

 Solution:

Given that, in two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other.

RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRevRD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have to prove that the triangles are congruent.

Let us consider two right triangles such that

∠ B =  ∠ C = 90                                            …….(i)

AB = DE                                                                …….(ii)

∠ C =  ∠ F       From(iii)

Now observe the two triangles ABC and DEF

∠ C =  ∠ F [From (iii)]

∠ B =  ∠ E [From (i)]

and AB = DE                                                      [From (ii)]

So, by AAS congruence criterion, we have Δ ABC ≅ Δ DEF

Therefore, the two triangles are congruent

Hence proved

 

Q. (2) If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.

Solution:

Given that the bisector of the exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles. Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD ∥ BC

Let ∠ EAD = (i), ∠  DAC = (ii), ∠ ABC = (iii) and ∠ ACB = (iv)

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have,

(i) = (ii)                       [AD is a bisector of ∠  EAC]

(i) = (iii)                      [Corresponding angles]

and (ii) = (iv)             [alternative angle]

(iii) = (iv)

  • AB = AC

Since, in Δ ABC, two sides AB and AC are equal we can say that Δ ABC is isosceles triangle.

 

Q. (3) In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Solution:

Let Δ ABC be isosceles such that AB = AC.

  • ∠ B =  ∠ C

RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

Given that vertex angle A is twice the sum of the base angles B and C. i.e., ∠ A = 2(∠ B + ∠ C)

∠ A = 2(∠ B + ∠ B)

∠ A = 2(2 ∠ B)

∠ A = 4(∠ B)

Now, We know that sum of angles in a triangle = 180°

∠ A + ∠ B + ∠ C = 180°

4 ∠ B + ∠ B + ∠ B = 180°

6 ∠ B = 180°

∠ B = 30°

Since, ∠ B =  ∠ C

  • ∠ B =  ∠ C = 30°

And ∠ A = 4 ∠ B

  • ∠ A = 4 x 30° = 120°

Therefore, angles of the given triangle are 120°, 30° and 30°.

= 428 and LB = LC]

 

Q. (4) PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.

Solution : Given that PQR is a triangle such that PQ = PR ant S is any point on the side PQ and ST ∥ QR.

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

To prove,

PS = PT

Since, PQ = PR

  • PQR is an isosceles triangle.
  • ∠ Q =  ∠ R (or) ∠ PQR =  ∠ PRQ

Now, ∠ PST =  ∠ PQR and ∠ PTS =  ∠ PRQ [Corresponding angles as ST parallel to QR]

Since, ∠ PQR =  ∠ PRQ

  • ∠ PST =  ∠ PTS

Now, In Δ PST, ∠ PST =  ∠ PTS

Δ PST is an isosceles triangle

Therefore, PS = PT

 

Q. (5) In a Δ ABC, it is given that AB = AC and the bisectors of B and C intersect at O. If M is a point on BO produced, prove that ∠ MOC =  ∠ ABC.

Solution:

Given that in Δ ABC,

AB = AC and the bisector of  ∠ B and  ∠ C intersect at O. If M is a point on BO produced

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have to prove ∠ MOC =  ∠ ABC

Since,

AB = AC

  • ABC is isosceles
  • ∠ B =  ∠ C (or)
  • ∠ ABC =  ∠ ACB

Now,

BO and CO are bisectors of ∠ ABC and  ∠ ACB respectively

= > ABO =  ∠ OBC = ∠ ACO =  ∠ OCB =  RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev∠ ABC = RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev∠ ACB                       ……(i)

We have, in Δ OBC

∠ OBC + ∠ OCB + ∠ BOC = 180°                     ……(ii)

And also

∠ BOC + ∠ COM = 180° ……(iii) [Straight angle]

Equating (ii) and (iii)

  • ∠ OBC + ∠ OCB + ∠ BOC =  ∠ BOC + ∠ MOC
  • ∠ OBC + ∠ OCB =  ∠ MOC [From (i)]
  • 2 ∠ OBC =  ∠ MOC [From (i)]
  • 2 (RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev ∠ ABC) =  ∠ MOC [From (i)]
  • ∠ ABC =  ∠ MOC

Therefore, ∠ MOC =  ∠ ABC

 

Q. (6) P is a point on the bisector of an angle ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

Sol:

Given that P is a point on the bisector of an angle ABC, and PQ ∥ AB.

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have to prove that Δ BPQ is isosceles.

Since,

BP is bisector of ∠ ABC =  ∠ ABP =  ∠ PBC                 ……(i)

Now,

PQ ∥ AB

  • ∠ BPQ =  ∠ ABP ……(ii) [alternative angles]

From (i) and (ii), we get

∠ BPQ =  ∠ PBC (or) ∠ BPQ =  ∠ PBQ

Now,

In BPQ,

∠ BPQ =  ∠ PBQ

Δ BPQ is an isosceles triangle.

Hence proved

 

Q. (7) Prove that each angle of an equilateral triangle is 60°.

Sol:

Given to prove each angle of an equilateral triangle is 60°.

Let us consider an equilateral triangle ABC.

Such that AB = BC = CA

Now, AB = BC

  • ∠ A =  ∠ C ……(i) [Opposite angles to equal sides are equal]

And BC = AC

  • ∠ B =  ∠ A ……(ii)

From (i) and (ii), we get

∠ A =  ∠ B =  ∠ C                                        ……(iii)

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We know that

Sum of angles in a triangle = 180

  • ∠ A + ∠ B + ∠ C = 180
  • ∠ A + ∠ A + ∠ A = 180
  • 3 ∠ A = 180
  • ∠ A = 60

∠ A =  ∠ B =  ∠ C = 60

Hence, each angle of an equilateral triangle is 60°.

 

Q. (8) Angles Δ A, B, C of a triangle ABC are equal to each other. Prove that ABC is equilateral.

Sol:

Given that angles A, B, C of a triangle ABC equal to each other.

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have to prove that Δ ABC is equilateral

We have, ∠ A =  ∠ B =  ∠ C

Now,

∠ A =  ∠ B

  • BC = AC [opposite sides to equal angles are equal]

And ∠ B =  ∠ C

  • AC = AB

From the above we get

AB = BC = AC

  • Δ ABC is equilateral.

 

Q. (9) ABC is a triangle in which ∠ B = 2 ∠ C. D is a point on BC such that AD bisects ∠ BAC and AB = CD. Prove that [∠ BAC = 72 .

Solution: Given that in ABC, ∠ B = 2 ∠ C and D is a point on BC such that AD bisectors ∠ BAC and

AB = CD.

 RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths Class 9 Notes | EduRev

We have to prove that ∠ BAC = 72°

Now, draw the angular bisector of ∠ ABC, which meets AC in P.

Join PD

Let C =  ∠ ACB =  y

  • ∠ B =  ∠ ABC = 2 ∠ C = 2y and also
  • Let ∠ BAD =  ∠ DAC
  • ∠ BAC = 2x [ AD is the bisector of ∠ BAC ]

Now, in Δ BPC,

  • ∠ CBP = y [BP is the bisector of ∠ ABC ]
  • ∠ PCB = y
  • ∠ CBP =  ∠ PCB = y [PC = BP]

Consider, Δ ABP and Δ DCP, we have

Δ ABP =  Δ DCP = y

AB = DC                            [Given]

And PC = BP                     [From above]

So, by SAS congruence criterion, we have ΔABP≅ΔDCP

Now,

∠ BAP =  ∠ CDF and AP = DP [Corresponding  parts of congruent triangles are equal]

  • ∠ BAP =  ∠ CDP = 2

Consider, Δ APD,

We have AP = DP

=  ∠ ADP =  ∠ DAP

But ∠ DAP = x

  • ∠ ADP =  ∠ DAP = x

Now

In Δ ABD.

∠ ABD + ∠ BAD + ∠ ADB = 180

And also ∠ ADB + ∠ ADC = 180° [Straight angle]

From the above two equations, we get

∠ ABD + ∠ BAD + ∠ ADB =  ∠ ADB + ∠ ADC

  • 2y + x =  ∠ ADP + ∠ PDC
  • 2y + x = x + 2x
  • 2y = 2x
  • y = x (or) x = y

We know,

Sum of angles in a triangle = 180°

So, In Δ ABC,

  • ∠ A + ∠ B + ∠ C = 180°
  • 2x + 2y + y = 180° [∠ A = 2x, ∠ B = 2y, ∠ C = y]
  • 2(y)+3y = 180° [x = y]
  • 5y = 180°
  • y = 36°

Now, ∠ A =  ∠ BAC = 2 x 36° = 72°

 

Q. (10) ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Solution: Given that ABC is a right angled triangle such that ∠ A = 90° and AB = AC

Since, AB = AC

  • Δ ABC is also isosceles.

Therefore, we can say that Δ ABC is right angled isosceles triangle.

  • ∠ C =  ∠ B and ∠ A = 90° ……(i)

Now, we have sum of angled in a triangle = 180°

  • ∠ A + ∠ B + ∠ C = 180°
  • 90° + ∠ B + ∠ B = 180° [From(i)]
  • 2∠ B = 180° – 90°
  • ∠ B = 45°

Therefore, ∠ B =  ∠ C = 45°

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