NCERT Solutions For Class 7 Maths Chapter 7

NCERT Solutions Class 7 Maths Congruence of Triangles

NCERT Solutions For Class 7 Maths Chapter 7 Congruence of Triangles are given here in a simple and detailed way. These NCERT Solutions for class 7 maths chapter 7 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 class 7 maths chapter 7 Congruence of Triangles solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The NCERT solutions for class 7 maths chapter 7 Congruence of Triangles PDF is also available here that the students can download and study.

NCERT Solutions For Class 7 Maths Chapter 7 Exercises

Excerise 7.1

 

1. In quadrilateral ACBD, AC = AD and AB bisects A . Show that ΔABC ΔABD. What can you say about BC and BD?

1

Sol.  

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

 

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

(i) ΔABD ΔBAC
(ii) BD = AC
(iii) ABD = BAC.

2

Sol.

Given that ,
AD = BC and DAB = CBA

(i) In ΔABD and ΔBAC,
AB = BA (Common)
DAB = CBA (Given)
AD = BC (Given)
Therefore, ΔABD ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ΔBAC
Therefore ABD 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 ΔAOD and ΔBOC,
A = B (Perpendicular)
AOD = BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, ΔAOD ΔBOC 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 ΔABC ΔCDA.

4

Sol. 

Given that,
l m and p q
To prove,
ΔABC ≅ ΔCDA
Then Proof that,
In ΔABC and ΔCDA,
BCA =DAC (Alternate interior angles)
AC = CA (Common)
BAC = DCA (Alternate interior angles)
Therefore, ΔABC ΔCDA by ASA congruence condition.

 

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

(i) ΔAPB ΔAQB

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

Sol.

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

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

 

6. AC = AE, AB = AD and BAD = EAC. Show that BC = DE.               

6 

Sol.

Given that,
AC = AE, AB = AD and BAD = EAC
To show,
BC = DE
then Proof that,
BAD = EAC (Adding DAC both sides)
BAD + DAC = EAC + DAC
BAC = EAD
In ΔABC and ΔADE,
AC = AE (Given)
BAC = EAD
AB = AD (Given)
Therefore, ΔABC  ΔADE 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 BAD = ABE and EPA = DPB . Show that
(i) ΔDAP ΔEBP
(ii) AD = BE

7

Sol.

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

(i) EPA =DPB (Adding DPE both sides)
EPA + DPE = DPB + DPE
DPA = EPB
In ΔDAP ΔEBP,
DPA = EPB
AP = BP (P is mid-point of AB)
BAD =ABE (Given)
Therefore, ΔDAP ΔEBP 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) ΔAMC ΔBMD

(ii) DBC is a right angle.

(iii) ΔDBC ΔACB

(iv) CM = 1/2 AB

8

Sol.

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

(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
CMA = DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ΔBMD by SAS congruence condition.

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

(iii) In ΔDBC and ΔACB,
BC = CB (Common)
ACB = DBC (Right angles)
DB = AC
Therefore, ΔDBC ΔACB by SAS congruence condition.

(iv)  DC = AB (ΔDBC ΔACB)
DM = CM = AM = BM (M is mid-point)
DM + CM = AM + BM
CM + CM = AB
CM = 1/2AB

 

Exercise 7.2

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

(a) OB = OC  (b) AO bisects A

9

Sol.

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

(i) Since ABC is an isosceles with AB = AC,
B = C
⇒ 1/2B = 1/2C
OBC = 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.
BAO = CAO (by CPCT)
Thus, AO bisects 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.