Number Line Prime Class 7 Solutions Chapter 18 Congruence of Triangles
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 18, Congruence of Triangles. Here students can easily find step by step solutions of all the problems for Congruence of Triangles, Exercise 18A and 18B Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 18 solutions. Here in this post all the solutions are based on latest Syllabus.
Congruence of Triangles Exercise 18A Solution :
Question no – (1)
Solution :
∴ AB = CD
∴ Let, the line segment A and superimpose it on the segment it on the line segment BD. If B coincides with a and C with D. Then Ac and BD have the same length thus,
∴ AC ≅ BD
Question no – (2)
Solution :
∴ AC = BD
∴ The line segment AB and superimpose it on the line segment CD. IF ecoincides with A and B with D. then A B and C D have the same length thus,
∴ AB ≅ CD
Question no – (3)
Solution :
Yes, they are congruent.
Question no – (4)
Solution :
∴ ∠BAC ≅ ∠IGH
And ∠QPR ≅ ∠XYZ
Question no – (5)
Solution :
∴ C > WXYZ ≅ MNOP
Question no – (6)
Solution :
AB = MN and BC = NO
∴ ∠ABC = ∠NMO = 90°, ∠OPM = ∠CDA = 90°
∴ Two quadrilaterals are congruent
∴ ABCD ≅ MNOP
Congruence of Triangles Exercise 18B Solution :
Question no – (1)
Solution :
(a) Given , △PQR ≅ △ABC
P ↔ A, Q ↔ B and R ↔ C
Also, PQ = AB, QR = BC and RD = CA
Similarly, ∠P = ∠B, ∠Q = ∠C, ∠R = A
(b) Given, △PQR ≅ △BCA
P ↔ B, Q ↔ C and R ↔ A
Also, PQ = BC, QR = CD, and RP = AB
Similarly, ∠P = ∠B, ∠Q = ∠C, ∠R = ∠A
Question no – (2)
Solution :
(a) △ABC = △ADC
Here, ∠ABC = ∠ADC = 90°
∴ BC = AD and AB = DC
And AC = AC [common]
∴ △ABC ≅ △ADC [by SAS Congruence condition]
(b) △ AOB and △COD
∴ AB = CD and AB||CD
And ∠AOB = ∠COD
∴ △AOB ≅ △COD [by ASA congruence condition]
Question no – (3)
Solution :
△ABC and △MINO
∴ BC = 6 cm = MN
∴ AB = 5 cm = ON
∴ ∠ABC = ∠ONM = 45°
∴ △ABC = △MNO [by SAS congruence condition]
Question no – (4)
Solution :
△ ABC and △BAD
AD = BC
∴ ∠BAC = ∠ABD
And
∴ ∠CAD = ∠CBD
∴ △ABC ≅ △BAC [by ASA congruence condition]
Question no – (5)
Solution :
△ ABC and △DBE
∴ ∠BAC = ∠BDE
∴ AC = DE
∴ AB = BD
∴ △ABC ≅ △DBE [by SAS congruence condition]
Question no – (6)
Solution :
△CDA and △ABF
∴ AB = CD
∴ ∠CDE = ∠ABF
And
∴ ∠BAF = ∠DCE
∴ △CDE ≅ △ABF [by ASA congruence condition]
∴ EC = AF [as △CDE ≅ △ABF]
Next Chapter Solution :
👉 Chapter 21
