11. Chapter 11 . Congruence of Triangles NIOS Class 10 Mathematics Textbook Solutions
Chapter 11 . Congruence of Triangles NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 11. CONGRUENCE OF TRIANGLES
CHECK YOUR PROGRESS 11.1
1. In Δ ABC (Fig. 11.19) if ∠B = ∠C and AD ⊥ BC, then Δ ABD ≅ Δ ACD by the criterion.
Photo
(a) RHS (b) ASA (c) SAS (d) SSS
Answer: (a) RHS
[ In ∆ADB 
and ∆ADC 
,we have
AB=AC 
(∠B=∠C 
)
∠ADB=∠ADC=90° 
AD=AD 
[ Common side]
∆ADB≅∆ADC 
[RHS] ]
2. In Fig. 11.20, ΔABC≅ΔPQR
. This congruence may also be written as
Photo
(a) Δ BAC ≅ Δ RPQ (b) Δ BAC ≅ Δ QPR (c) Δ BAC ≅ Δ RQP (d) Δ BAC ≅ Δ PRQ.
Answer: (b) ΔBAC≅ΔQPR
[ Since, ΔABC≅ΔPQR 
AB=PQ , BC=QR , AC=PR 
S0, ΔBAC≅ΔQPR
(SSS) ]
3. In order that two given triangles are congruent, along with equality of two corresponding angles we must know the equality of :
(a) No corresponding side
(b) Minimum one corresponding side
(c) Minimum two corresponding sides
(d) All the three corresponding sides
Answer: (b) Minimum one corresponding side.
[ For two triangles to be congruent, if two corresponding angles are equal, the third angle is also equal (Angle-Angle-Angle similarity). ]
4. Two triangles are congruent if ....
(a) All three corresponding angles are equal
(b) Two angles and a side of one are equal to two angles and a side of the other.
(c) Two angles and a side of one are equal to two angles and the corresponding side of the other.
(d) One angle and two sides of one are equal to one angle and two sides of the other.
Answer: (c) Two angles and a side of one are equal to two angles and the corresponding side of the other.
[ Two angles and the corresponding side must be equal (ASA or AAS rule). ]
5. In Fig. 11.21, ∠B = ∠C and ΑB = AC. Prove that Δ ABE ≅ Δ ACD. Hence show that CD = BE .
Photo
Solution: Given , ∠B = ∠C and ΑB = AC.
To prove : Δ ABE ≅ Δ ACD and also CD = BE .
Proof: In ΔABE 
and ΔACD 
, we have
∠B=∠C [given]
AB=AC 
[given]
∠A=∠A 
[common angle]
ΔABE≅ΔACD 
[ASA]
BE=CD 
[CPCT]
⇒CD=BE 
Hence proved .
6. In Fig. 11.22, AB is parallel to CD. If O is the mid-point of BC, show that it is also the mid-point of AD.
Photo
Solution: Given, AB is parallel to CD. If O is the mid-point of BC .
We show that O is the mid-point of AD. (i.e., OA = OD )
Proof : ∆AOB 
and ∆COD 
, we have
∠AOB=∠COD 
[Vertically opposite angles]
OB=OC 
[O is the mid-point]
∠OBA=∠OCD 
∆AOB≅∆COD 
[ASA]
OA=OD 
[CPCT]
Thus, O is the mid-point AD .
Hence proved .
7. In Δ ABC (Fig. 11.23), AD is ⊥ BC, BE is ⊥ AC and AD = BE. Prove that AE = BD.
Photo
Solution: Given, AD⊥BC , BE⊥AC and AD = BE.
To prove : AE = BD.
Proof: In ∆ABE and ∆ABD 
, we have
AD=BE 
∠AEB=∠ADB=90° 
AB=AB 
[Common side] 
∆AOB≅∆COD [RHS]
AE=BD 
[ CPCT ]
Hence proved .
8. From Fig. 11.24, show that the triangles are congruent and make pairs of equal angles.
Photo
Solution: In ∆ACB 
and ∆QRP 
, We have
AC=QP=8 cm 
AB=QR=10 cm 
BC=PR=5 cm 
∆ACB≅∆QRP 
[SSS]
So, ∠A=∠Q , ∠B=∠R , ∠C=∠P 
[CPCT]
CHECK YOUR PROGRESS 11.2
1. In Fig. 11.31, PQ = PR and SQ = SR. Prove that ∠PQS = ∠PRS.
Photo
Solution: Given, PQ = PR and SQ = SR.
To prove that ∠PQS = ∠PRS.
Construction : We join PS .
Proof: In ∆PQS 
and ∆PRS 
,We have
PQ=PR 
[given]
SQ=SR 
[given]
PS=PS 
[common side]
∆PQS≅∆PRS 
[SSS]
∠PQS = ∠PRS 
[CPCT]
Hence prove .
2. Prove that ΔABC is an isosceles triangle, if the altitude AD bisects the base BC (Fig. 11.32).
Solution: Given, ABC is a triangle such that the altitude AD bisects the base BC .
To prove : ΔABC is an isosceles triangle.
Proof: ∆ABD 
and ∆ACD , we have
BD=CD 
[given]
∠ADB=∠ADC=90°
AD=AD [common side]
∆ABD≅∆ACD 
[RHS]
∠ABD=∠ACD 
[CPCT]
So, ΔABC is an isosceles triangle.
3. If the line l in Fig. 11.33 is parallel to the base BC of the isosceles ΔABC, find the angles of the triangle.
Solution: Since, the line l is parallel to the base BC .
∠C=65° 
[Alternative interior angle]
∠B=∠C=65° 
[ ABC is an isosceles triangle ]
In ΔABC, we have,
∠B+∠C+∠BAC=180° 
⇒65°+65°+∠BAC=180° 
⇒130°+∠BAC=180° 
⇒∠BAC=180°-130° 
⇒∠BAC=50° 
Thus, ∠B=∠C=65° and ∠BAC=50° 
.
4. ΔABC is an isosceles triangle such that AB = AC. Side BA is produced to a point D such that AB = AD. Prove that ∠ΒCD is a right angle.
Solution: Given, ΔABC is an isosceles triangle such that AB = AC. Side BA is produced to a point D such that AB = AD.
To prove: ∠ΒCD is a right angle.
Proof: In ∆ABC
, we have
AB=AC
∠ACB=∠ABC
∠ACB=∠DBC ………(i)
In ∆ACD
, we have
AC=AD 
∠ACD=∠ADC
∠ACD=∠BDC………(ii)
Adding (i) and (ii) , we get
∠ACB+∠ACD=∠DBC+∠BDC 
⇒∠BCD=∠DBC+∠BDC 
⇒∠BCD+∠BCD=∠DBC+∠BDC+∠BCD 
[Adding both side ∠BCD
]
⇒2∠BCD=180°
⇒∠BCD=180°2
⇒∠BCD=90°
Hence proved.
5. In Fig. 11.35. D is the mid point of BC and perpendiculars DF and DE to sides AB and AC respectively are equal in length. Prove that ΔABC is an isosceles triangle.
Photo
Solution: D is the mid point of BC and perpendiculars DF and DE to sides AB and AC respectively are equal in length.
To prove: ABC is an isosceles triangle.
Proof: Since, D is the mid point of BC .
So, BD = CD
In ∆BDF 
and ∆CDE 
, we have
BD=CD 
[given]
∠BFD=∠CED=90° 
[given]
DF=DE 
[given]
∆BDF≅∆CDE 
[SAS]
∠B=∠C
[CPCT]
i.e., AB = AC
So, ABC is an isosceles triangle.
Hence proved.
6. In Fig. 11.36, PQ = PR, QS and RT are the angle bisectors of ∠Q and ∠R respectively. Prove that QS = RT.
Photo fig. 11.36
Solution: Given, PQ = PR, QS and RT are the angle bisectors of ∠Q and ∠R respectively.
To prove: QS = RT.
Proof: Since, QS and RT are the angle bisectors of ∠Q and ∠R respectively.
∠PQS=∠SQR=12∠Q 
And ∠PRT=∠TRQ=12∠R 
In ∆PQS and ∆PRT 
, we have
PQ=PR 
[given]
∠PQS=∠PRT 
[given]
∠P=∠P 
[common angle]
∆PQS≅∆PRT 
[SAS]
QS=RT 
[CPCT]
7. ΔPQR and ΔSQR are isosceles triangles on the same base QR (Fig. 11.37). Prove that ∠PQS = ∠PRS.
Photo
Solution: Given, ΔPQR and ΔSQR are isosceles triangles on the same base QR .
To prove : ∠PQS = ∠PRS.
Proof: ΔPQR is an isosceles triangle.
PQ = PR 
∠PQR=∠PRQ 
………… (i)
ΔSQR is an isosceles triangle .
QS=RS 
∠RQS=∠QRS 
…………….. (ii)
Adding (i) and (ii) , we get
∠PQR+∠RQS=∠PRQ+∠QRS 
⇒∠PQS=∠PRS 
Hence proved.
8. In ΔABC, AB = AC (Fig. 11.38). P is a point in the interior of the triangle such that ∠ΑΒP = ∠ΑCP . Prove that AP bisects ∠BAC.
Photo
Solution: Given, ΔABC and AB = AC . P is a point in the interior of the triangle such that ∠ΑΒP = ∠ΑCP .
To prove : AP bisects ∠BAC.
Proof: In ∆ABP 
and ∆ACP 
, we have
AB=AC [given]
∠ABP=∠ACP 
[given]
AP=AP 
[Common side]
∆ABP≅∆ACP 
[RHS]
∠BAP=∠CAP 
[ CPCT ]
So, AP bisects ∠BAC.
Hence proved .
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