Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-10.3, Congruent Triangles, Class 9, Maths

Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

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.

Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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)

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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.

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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 =  Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠ ABC = Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠ 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 (Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics ∠ 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.

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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)

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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.

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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.

 Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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°

The document Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-10.3, Congruent Triangles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What are congruent triangles?
Ans. Congruent triangles are triangles that have the same shape and size. They have exactly the same angles and sides, and when superimposed, they coincide perfectly.
2. How do you prove two triangles congruent?
Ans. Two triangles can be proved congruent using various methods such as the Side-Side-Side (SSS) congruence criterion, Side-Angle-Side (SAS) congruence criterion, Angle-Side-Angle (ASA) congruence criterion, Angle-Angle-Side (AAS) congruence criterion, and Hypotenuse-Leg (HL) congruence criterion.
3. What is the importance of congruent triangles in mathematics?
Ans. Congruent triangles play a crucial role in various mathematical concepts and constructions. They help in proving the congruence of other geometric figures, solving geometric problems, and establishing theorems in geometry.
4. How can congruent triangles be used to solve problems?
Ans. Congruent triangles can be used to solve problems involving measurements, angles, and sides of triangles. By identifying congruent parts of triangles, we can establish relationships between the corresponding parts of congruent triangles and use them to find unknown values.
5. Can congruent triangles have different orientations?
Ans. No, congruent triangles cannot have different orientations. Congruent triangles have the same shape and size, and their corresponding angles and sides are equal. Therefore, their orientations must also be the same when superimposed.
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