NCERT Solutions For Class 7 Maths Chapter 7

NCERT Solutions Class 7 Maths Congruence of Triangles

Ncert Solutions For Class 7 Maths Chapter 7 PDF Download

NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles are given here in a simple and detailed way. While solving the exercise questions from the NCERT Book For Class 7 Maths, students often face difficulty and eventually pile up their doubts. The NCERT solutions can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way.

NCERT chapter 7 class 7 maths solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The solutions given here are in easy steps to help the students understand the respective solutions properly. These solutions can also help the students to have an in-depth understanding of the concepts and develop a deeper interest in the concepts. The NCERT solutions for class 7 maths Congruence of Triangles PDF is also available here that the students can download and study anytime they want.

NCERT Solutions For Class 7 Maths Chapter 7 Exercises

Excerise 7.1

 

1. In quadrilateral ACBD, AC = AD and AB bisects \(\angle\)A . Show that \(\Delta\)ABC \(\cong\) \(\Delta\)ABD. What can you say about BC and BD?

1

Sol.  

Given that,
AC = AD and AB bisects \(\angle\)A
To prove that,
\(\Delta\)ABC \(\cong\) \(\Delta\)ABD
Proof,
In \(\Delta\)ABC and \(\Delta\)ABD,
AB = AB
AC = AD
\(\angle\)CAB =\(\angle\)DAB (AB is bisector)
Therefore, \(\Delta\)ABC \(\cong\)  \(\Delta\)ABD by SAS congruence condition.
BC and BD are of equal length.

 

2. ABCD is a quadrilateral in which AD = BC and \(\angle\)DAB = \(\angle\)CBA . Prove that

(i) \(\Delta\)ABD \(\cong\) \(\Delta\)BAC
(ii) BD = AC
(iii) \(\angle\)ABD = \(\angle\)BAC.

2

Sol.

Given that ,
AD = BC and \(\angle\) DAB = \(\angle\)CBA

(i) In \(\Delta\)ABD and \(\Delta\)BAC,
AB = BA (Common)
\(\angle\) DAB = \(\angle\)CBA (Given)
AD = BC (Given)
Therefore, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC by SAS congruence condition.
(ii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC
Therefore BD = AC by CPCT
(iii) Since, \(\Delta\)ABD \(\cong\) \(\Delta\)BAC
Therefore \(\angle\) ABD \(\cong\) \(\angle\)BAC by CPCT

 

3. AD and BC are equal perpendiculars to a line segment AB . Show that CD bisects AB.       

3    

Sol. 

Given that,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In \(\Delta\)AOD and \(\Delta\)BOC,
\(\angle\)A = \(\angle\)B (Perpendicular)
\(\angle\)AOD = \(\angle\)BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, \(\Delta\)AOD \(\cong\) \(\Delta\)BOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

 

4. l and m are two parallel lines intersected by another pair of parallel lines p and q . Show that \(\Delta\)ABC \(\cong\) \(\Delta\)CDA.

4

Sol. 

Given that,
l \(\parallel\) m and p \(\parallel\) q
To prove,
\(\Delta\)ABC ≅ \(\Delta\)CDA
Then Proof that,
In \(\Delta\)ABC and \(\Delta\)CDA,
\(\angle\)BCA =\(\angle\)DAC (Alternate interior angles)
AC = CA (Common)
\(\angle\)BAC = \(\angle\)DCA (Alternate interior angles)
Therefore, \(\Delta\)ABC \(\cong\) \(\Delta\)CDA by ASA congruence condition.

 

5. Line l is the bisector of an angle \(\angle\)A and B is any point on l. BP and BQ are perpendiculars from B to the arms of \(\angle\)A . Show that:

(i) \(\Delta\)APB \(\cong\) \(\Delta\)AQB

(ii) BP = BQ or B is equidistant from the arms of \(\angle\)A.              5 

Sol.

Given,
l is the bisector of an angle \(\angle\)A.
BP and BQ are perpendiculars.

(i) In \(\Delta\)APB and \(\Delta\)AQB,
\(\angle\)P = \(\angle\)Q (Right angles)
\(\angle\)BAP =\(\angle\)BAQ (l is bisector)
AB = AB (Common)
Therefore, \(\Delta\)APB \(\cong\) \(\Delta\)AQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of \(\angle\)A.

 

6. AC = AE, AB = AD and \(\angle\)BAD = \(\angle\)EAC. Show that BC = DE.               

6 

Sol.

Given that,
AC = AE, AB = AD and \(\angle\)BAD = \(\angle\)EAC
To show,
BC = DE
then Proof that,
\(\angle\)BAD = \(\angle\)EAC (Adding \(\angle\)DAC both sides)
\(\angle\)BAD + \(\angle\)DAC = \(\angle\)EAC + \(\angle\)DAC
\(\Rightarrow\) \(\angle\)BAC = \(\angle\)EAD
In \(\Delta\)ABC and \(\Delta\)ADE,
AC = AE (Given)
\(\angle\)BAC = \(\angle\)EAD
AB = AD (Given)
Therefore, \(\Delta\)ABC \(\cong\)  \(\cong\)\(\Delta\)ADE by SAS congruence condition.
BC = DE by CPCT.

 

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB . Show that
(i) \(\Delta\)DAP \(\cong\) \(\Delta\)EBP
(ii) AD = BE

7

Sol.

Given,
P is mid-point of AB.
\(\angle\)BAD = \(\angle\)ABE and \(\angle\)EPA = \(\angle\)DPB

(i) \(\angle\)EPA =\(\angle\)DPB (Adding \(\angle\)DPE both sides)
\(\angle\)EPA + \(\angle\)DPE = \(\angle\)DPB + \(\angle\)DPE
\(\Rightarrow\) \(\angle\)DPA = \(\angle\)EPB
In \(\Delta\)DAP \(\cong\) \(\Delta\)EBP,
\(\angle\)DPA = \(\angle\)EPB
AP = BP (P is mid-point of AB)
\(\angle\)BAD =\(\angle\)ABE (Given)
Therefore, \(\Delta\)DAP \(\cong\) \(\Delta\)EBP by ASA congruence condition.
(ii) AD = BE by CPCT.

 

8. A right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:

(i) \(\Delta\)AMC \(\cong\) \(\Delta\)BMD

(ii) \(\angle\)DBC is a right angle.

(iii) \(\Delta\)DBC \(\cong\) \(\Delta\)ACB

(iv) CM = 1/2 AB

8

Sol.

Given,
\(\angle\)C = 90°, M is the mid-point of AB and DM = CM

(i) In \(\Delta\)AMC and \(\Delta\)BMD,
AM = BM (M is the mid-point)
\(\angle\)CMA = \(\angle\)DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, \(\Delta\)AMC \(\cong\) \(\Delta\)BMD by SAS congruence condition.

(ii) \(\angle\)ACM = \(\angle\)BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
\(\angle\)ACB + \(\angle\)DBC = 180° (co-interiors angles)
\(\Rightarrow\) 90° + \(\angle\)B = 180°
\(\Rightarrow\) \(\angle\)DBC = 90°

(iii) In \(\Delta\)DBC and \(\Delta\)ACB,
BC = CB (Common)
\(\angle\)ACB = \(\angle\)DBC (Right angles)
DB = AC
Therefore, \(\Delta\)DBC \(\cong\) \(\Delta\)ACB by SAS congruence condition.

(iv)  DC = AB (\(\Delta\)DBC \(\cong\) \(\Delta\)ACB)
\(\Rightarrow\) DM = CM = AM = BM (M is mid-point)
\(\Rightarrow\) DM + CM = AM + BM
\(\Rightarrow\) CM + CM = AB
\(\Rightarrow\) CM = 1/2AB

 

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of \(\angle\)B and \(\angle\)C intersect each other at O. Join A to O. Show that :

(a) OB = OC  (b) AO bisects \(\angle\)A

9

Sol.

Sol. Given that,
AB = AC, the bisectors of \(\angle\)B and \(\angle\)C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,
∴ \(\angle\)B = \(\angle\)C
⇒ 1/2\(\angle\)B = 1/2\(\angle\)C
⇒ \(\angle\)OBC = \(\angle\)OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, ΔAOB ≅ ΔAOC by SSS congruence condition.
\(\angle\)BAO = \(\angle\)CAO (by CPCT)
Thus, AO bisects \(\angle\)A.

 

2. In ΔABC, AD is the perpendicular bisector of BC . Show that ΔABC is an isosceles triangle in which AB = AC.

10

Sol.

Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
In ΔADB and ΔADC,
AD = AD (Common)
∠ADB = ∠ADC
BD = CD (AD is the perpendicular bisector)
Therefore, ΔADB ≅ ΔADC by SAS congruence condition.
AB = AC

 

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

11

Sol.

Given that,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In ΔAEB and ΔAFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.
Thus, BE = CF by CPCT.

 

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal . Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

12

Sol.

Given,

BE = CF

(i) In ΔABE and ΔACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, ΔABE ≅ ΔACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

 

5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

13

Sol.

Sol. Given that,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

 

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

14

Sol.

Sol. Given that,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

 

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

15

Sol.

Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

 

8. Show that the angles of an equilateral triangle are 60° each.

16

Sol.

Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

The congruence of triangles is among the most important topics in the entire CBSE class 7 maths book. In this chapter, students learn various important topics. At first, students are introduced to what congruence is an then the congruence of plane figures, line segment, and angles are discussed. Then the congruence of triangles are introduced along with the criteria for congruence of triangles (SAS, ASA, etc.) and at the end, the congruence among right-angled triangles are discussed.

In this chapter, several exercise questions are included which should be solved to analyze the understanding of the concepts taught. Students should solve all these questions and can check the class 7 maths NCERT solutions for chapter 7 here in case of any doubts. By solving these questions, students can easily have an in-depth understanding of the concepts.

This chapter is one of the most crucial topics in the entire geometry section and these concepts are also important for the students to be able to comprehend several higher level topics in the later grades.

Stay tuned with BYJU’S to get the complete NCERT Solutions For Class 7 Maths for all the chapters. These solutions can help the students to clear all their doubts instantly and help them to learn the maths concepts in an in-depth and in a more effective way.