RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.1

Exercise 10.1 of RD Sharma Solutions for Class 9 Maths Chapter 1 Congruent Triangles is based on the topic of congruent triangles, and students need to solve problems based on this topic. A triangle is a three-sided polygon having three angles which sum up to 180⁰. When it comes to the congruence of triangles, two triangles are congruent if they superimpose on each other or if their sides and angles have the same lengths and measurements, respectively. Check the detailed RD Sharma Solutions for Class 9 Chapter 10 Congruent Triangles below.

RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.1

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Access Answers to RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.1 Page Number 10.12

Question 1: In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that segment DE ∥ BC.

Solution:

Sides BA and CA have been produced such that BA = AD and CA = AE.

To prove: DE ∥ BC

Consider △ BAC and △DAE,

BA = AD and CA= AE (Given)

∠BAC = ∠DAE (vertically opposite angles)

By the SAS congruence criterion, we have

△ BAC ≃ △ DAE

We know that corresponding parts of congruent triangles are equal

So, BC = DE and ∠DEA = ∠BCA, ∠EDA = ∠CBA

Now, DE and BC are two lines intersected by a transversal DB s.t.

∠DEA=∠BCA (alternate angles are equal)

Therefore, DE ∥ BC. Proved.

Question 2: In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP, respectively. Prove that LN = MN.

Solution:

Draw a figure based on the given instruction,

In △PQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP, respectively (Given)

To prove: LN = MN

As two sides of the triangle are equal, so △ PQR is an isosceles triangle

PQ = QR and ∠QPR = ∠QRP ……. (i)

Also, L and M are midpoints of PQ and QR, respectively

PL = LQ = QM = MR = QR/2

Now, consider Δ LPN and Δ MRN,

LP = MR

∠LPN = ∠MRN [From (i)]

∠QPR = ∠LPN and ∠QRP = ∠MRN

PN = NR [N is the midpoint of PR]

By SAS congruence criterion,

Δ LPN ≃ Δ MRN

We know that the corresponding parts of congruent triangles are equal.

So LN = MN

Proved.

Question 3: In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that

(i) PT = QT (ii) ∠ TQR = 15⁰

Solution:

Given: PQRS is a square, and SRT is an equilateral triangle.

To prove:

(i) PT =QT and (ii) ∠ TQR =15°

Now,

PQRS is a square:

PQ = QR = RS = SP …… (i)

And ∠ SPQ = ∠ PQR = ∠ QRS = ∠ RSP = 90^o

Also, △ SRT is an equilateral triangle:

SR = RT = TS …….(ii)

And ∠ TSR = ∠ SRT = ∠ RTS = 60°

From (i) and (ii)

PQ = QR = SP = SR = RT = TS ……(iii)

From figure,

∠TSP = ∠TSR + ∠ RSP = 60° + 90° = 150° and

∠TRQ = ∠TRS + ∠ SRQ = 60° + 90° = 150°

⇒ ∠ TSP = ∠ TRQ = 150⁰ ………………… (iv)

By SAS congruence criterion, Δ TSP ≃ Δ TRQ

We know that corresponding parts of congruent triangles are equal

So, PT = QT

Proved part (i).

Now, consider Δ TQR.

QR = TR [From (iii)]

Δ TQR is an isosceles triangle.

∠ QTR = ∠ TQR [angles opposite to equal sides]

The sum of angles in a triangle = 180^∘

⇒ ∠QTR + ∠ TQR + ∠TRQ = 180°

⇒ 2 ∠ TQR + 150° = 180° [From (iv)]

⇒ 2 ∠ TQR = 30°

⇒ ∠ TQR = 15⁰

Hence proved part (ii).

Question 4: Prove that the medians of an equilateral triangle are equal.

Solution:

Consider an equilateral △ABC, and Let D, E, and F are midpoints of BC, CA and AB.

Here, AD, BE and CF are medians of △ABC.

Now,

D is the midpoint of BC ⇒ BD = DC

Similarly, CE = EA and AF = FB

Since ΔABC is an equilateral triangle

AB = BC = CA …..(i)

BD = DC = CE = EA = AF = FB …………(ii)

And also, ∠ ABC = ∠ BCA = ∠ CAB = 60° ……….(iii)

Consider Δ ABD and Δ BCE

AB = BC [From (i)]

BD = CE [From (ii)]

∠ ABD = ∠ BCE [From (iii)]

By SAS congruence criterion,

Δ ABD ≃ Δ BCE

⇒ AD = BE ……..(iv)

Now, consider Δ BCE and Δ CAF,

BC = CA [From (i)]

∠ BCE = ∠ CAF [From (iii)]

CE = AF [From (ii)]

By SAS congruence criterion,

Δ BCE ≃ Δ CAF

⇒ BE = CF …………..(v)

From (iv) and (v), we have

AD = BE = CF

Median AD = Median BE = Median CF

The medians of an equilateral triangle are equal.

Hence proved

Question 5: In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.

Solution:

To find: ∠ B and ∠ C.

Here, Δ ABC is an isosceles triangle since AB = AC

∠ B = ∠ C ……… (i)

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

∠ A + ∠ B + ∠ C = 180°

∠ A + ∠ B + ∠ B= 180° (using (i)

120⁰ + 2∠B = 180⁰

2∠B = 180⁰ – 120⁰ = 60⁰

∠ B = 30^o

Therefore, ∠ B = ∠ C = 30^∘

Question 6: In a Δ ABC, if AB = AC and ∠ B = 70°, find ∠ A.

Solution:

Given: In a Δ ABC, AB = AC and ∠B = 70°

∠ B = ∠ C [Angles opposite to equal sides are equal]

Therefore, ∠ B = ∠ C = 70^∘

The sum of angles in a triangle = 180∘

∠ A + ∠ B + ∠ C = 180^o

∠ A + 70^o + 70^o = 180^o

∠ A = 180^o – 140^o

∠ A = 40^o

RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.1

RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.1 is based on the following topics and subtopics:

• Congruence of Line Segments
• Congruence of Angles
• Congruence of Triangles
• Congruence Relation
• Congruence Criteria

-SAS (Side-Angle-Side)

-SSS (Side-Side-Side)

-ASA (Angle-Side-Angle)

– RHS (Right angle-Hypotenuse-Side)

• Congruence Criterion