RS Aggarwal Solutions Class 7 Maths Chapter 16 Ex 16.1 (Updated For 2024)

RS Aggarwal Solutions Class 7 Maths Chapter 16 Ex 16.1: Be it your Class 7 Maths exams or assignments, RS Aggarwal Solutions Class 7 Maths has got your back. All the solutions of RS Aggarwal Solutions Class 7 Maths Chapter 16 Ex 16.1 are prepared by subject matter experts which makes them very credible and they are even as per the current CBSE Syllabus.

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RS Aggarwal Solutions Class 7 Maths Chapter 16 Ex 16.1

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Question 1.
Solution:
(i) ∆ABC ≅ ∆EFD, Then
A ↔ E, B ↔ F and C ↔ D
AB = EF, BC = FD and CA = DE
∠A = ∠E, ∠B = ∠F and ∠C = ∠D
(ii) ∆CAB ≅ ∆QRP
C ↔ Q, A ↔ R and B ↔ P
CA = QR, AB = RP and BC = PQ
∠C = ∠Q, ∠A = ∠R and ∠B = ∠P
(iii) ∆XZY ≅ ∆QPR
X ↔ Q, Z ↔ P, Y ↔ R
XZ = QP, ZY = PR and YX = RQ
∠X = ∠Q, ∠Z = ∠P and ∠Y = ∠R
(iv) ∆MPN ≅ ∆SQR
M ↔ S, P ↔ Q and N ↔ R
MP = SQ, PN = QR and NM = RS
∠M = ∠S, ∠P = ∠Q and ∠N = ∠R.

Question 2.
Solution:
(i) In fig (i)
In ∆ABC and ∆DEF
∠C = ∠E
CA = ED
CB = EF
∆ACB ≅ ∆DEF (SAS condition)
(ii) In fig (ii)
In ∆RPQ and ∆LNM
Side PQ = NM
Hyp. RQ = LM
∆RPQ ≅ ∆LNM (RHS condition)
(iii) In ∆YXZ and ∆TRS
XY = RT
∠X = SR and YZ = TS
∆YXZ ≅ ∆TRS (SSS condition)
(iv) In ∆DEF and ∆PNM
∠E = ∠N
∠F = ∠M
EF = NM
∆DEF ≅ ∆PNM (ASA condition)
(v) In ∆ABC and ∆ADC
AC = AC (common)
∠ CAB = ∠ CAD (each 50°)
∠ ACB = ∠ DCA (each 60°)
∆ABC ≅ ∆ADC (ASA condition)

Question 3.
Solution:
In fig,
PL ⊥ OA and PM ⊥ OB and PL = PM
Now in right ∆PLO and ∆PMO,
Side PL = PM (given)
Hypotenuse OP = OP (common)
∆PLO ≅ ∆PMO (RHS condition)
Yes ∆PLO ≅ ∆PMO
Hence proved.

Question 4.
Solution:
In the figure,
AD = BC and AD || BC.
In ∆ABC and ∆ADC,
AC = AC (common)

BC = AB (given)
∠ACB = ∠CAD (Alternate angles)
∆ABC ≅ ∆ADC (SAS condition)
AB = DC (c.p.c.t)
Hence proved.

Question 5.
Solution:
In ∆ABD and ∆ACD,
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD ≅ ∆ADC (SSS condition)
∠BAD = ∠CAD (c.p.c.t.)
and ∠ADB = ∠ADC (c.p.c.t.)
But ∠ADB + ∠ADC = 180° (Linear pair)
∠ADB = ∠ADC = 90°
Hence proved.

Question 6.
Solution:
given : In ∆ABC, AD is the bisector of ∠A i.e. ∠BAD = ∠CAD
AD ⊥ BC.
To prove : ∆ABC is an isosceles
Proof : In ∆ADB and ∆ADC.
AD = AD (common)
∠ BAD = ∠ CAD (AD is the bisector of ∠A)
∠ ADB = ∠ ADC (each = 90°, AD ⊥ BC)
∆ADM ≅ ∆ADC (ASA condition)
AB = AC (c.p.c.t)
Hence ∆ABC is an isosceles triangle.
Hence proved.

Question 7.
Solution:
In the figure,
AB = AD, CB = CD
To prove : ∆ABC ≅ ∆ADC
Proof : In ∆ABC and ∆ADC
AC = AC (common)
AB = AD (given)
CB = CD (given)
∆ABC ≅ ∆ADC (SSS condition)
Hence proved.

RS Aggarwal Solutions Class 7 Maths Chapter 16 Ex 16.1 – Overview

Basic Concepts Of Congruence 

• Two figures can be called congruent if they have the same shape and size.
• We represent the congruence sign by ‘=’.
• Whenever the one-plane figure gets superposed on another plane figure covering it from all sides thoroughly as if it seems to be the one figure only, then we can that those two figures are congruent.
• The congruence of triangles doesn’t only happen when the corresponding angle measures or the length of the corresponding side are equivalent but also the vertices play an important role equally. It should be equivalent as well to be said as congruent. 
• Two figures namely, Fig-1 & Fig-2 are said to be congruent only if Fig-1 overlaps Fig-2 completely. Hence, we can write it as F1 = F2.
• Also, the two line segments, let us say WX and YZ will be called congruent only if they possess an equal length. 
• If two angles, let us say angle ABC and Angle DEF, will be called congruent only if they possess equal measures. Henceforth, in this case, we can write them as ∠ABC = ∠DEF.

Congruence of two triangles

1. SSS Congruence property of two triangles – The two triangles are said to be congruent, only in the case if the three sides of the first triangle are equivalent to the three corresponding sides of another triangle.
2. SAS Congruence property of two triangles – The two triangles are said to be congruent only in the case if the two sides of the first triangle and the corresponding sides of the second triangle are equivalent to each other. Also, if the corresponding angle formed in the first triangle and the second triangle is equivalent then the SAS congruence property is said to be true.
3. ASA congruence property of two triangles – In this case, the two triangles are said to be congruent only if the two angles and the one side of the first triangle is equivalent to the corresponding two angles and the corresponding side of the second triangle.
4. RHS Congruence property of two right-angled triangles – In this case, the two right-angled triangles are said to be congruent only if the hypotenuse side along with one other side of the same triangle is equivalent to the corresponding hypotenuse side along with the other corresponding side of another triangle.

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