These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions
NCERT In-text Question Page No. 140 and 141
Question 1.
Lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:
Answer:
(i) InΔABC and ΔPQR such that
AB = 1.5 cm; PQ =1.5 cm
∴ AB = PQ
BC = 2.5 cm; QR = 2.5 cm
∴ BC = QR
AC = 2.2 cm and PR = 2.2 cm
∴ AC = PR
∴ The three sides of ΔABC are equal to the three sides of ΔPQR.
∴ ΔABC ≅ ΔPQR
( By SSS congruence criterion)
(ii) In ΔDEF and ΔLMN
DE = 3.2 cm, MN = 3.2
∴ DE = NM
EF = 3 cm; LM = 3 cm
∴ EF = LM
DF = 3.5 cm, LN = 3.5 cm
∴ DF = LN
∴ The three sides of ΔDEF are equal to three sides of ALMN.
ΔDEF ≅ ΔNML
(By SSS congruence criterion)
(iii) In ΔABC and ΔPQR,
∴ AC = 5 cm, PR = 5 cm
∴ AC = PR
BC = 4 cm, PQ = 4 cm
∴ BC = PQ
AB = 2 cm, QR = 2.5 cm
AB ≠ QR
AB ≠ QR
ΔABC and ΔPQR are not congruent,
(iv) We have ΔABD and ΔADC., such that
AB = 3.5 cm, AC = 3.5 cm
∴ AB = AC
BD = 2.5 cm, CD = 2.5 cm
AD is common ( AD = AD)
since three sides of ΔABD are equal to the three sides of ΔACD
∴ The two triangles are congruent.
ΔABD ≅ ΔACD
(SSS Congruence criterion)
Question 2.
In the given figure, AB = AC and D is the mid- point of \(\overline{\mathrm{BC}}\)
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? Give reasons.
(iii) Is ∠B = ∠C ? Why?
Answer:
(i) Here, AB = AC and D is the midpoint of \(\overline{\mathrm{BC}}\)
∴ BD = DC
In ΔADB and ΔADC,
AB = AC, BD = DC
AD = AD (common)
(ii) The three sides of ΔADB are equal to the three sides of ΔADC
By SSS rule of congruence criterion
ΔADB ≅ ΔADC
(iii) Since ΔADB ≅ ΔADC,
∠B = ∠C (CPCT)
Question 3.
In the given figure, AC = BD and AD = BC. Which of the following statement is meaningfully written?
(i) ΔABC ≅ ΔABD
(ii) ΔABC ≅ ΔBAD
Answer:
Have AC = BD and AD = BC
In the given triangles ABC and ABD
AB = AB (common)
BD = AC (Given)
AD = BC (Given)
Three sides of ΔABD are equal to the three sides of ΔABC .So the two triangles are congruent,
∴ A ↔ B; B ↔ A; D ↔ C
(i) ΔABC ≅ ΔABD is not true or meaningless.
(ii) ΔABC ≅ ΔBAD is true or meaningful.
NCERT In-text Question Page No. 143 & 144
Question 1.
Which angle is included between the sides \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\) is ΔDEF.
Answer:
In ADEF, the angle between \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{EF}}\), is ∠DEF.
Question 2.
By applying SAS congruence rule, you want to establish that ΔPQR ≅ ΔFED. It is given that PQ = FE and RP = DF. What additional information is needed to establish the congruence?
Answer:
We have ΔPQR = ΔFED
[Using SAS congruence rule]
∴ P ↔ F, Q ↔ E, R ↔ D
and PQ = FE and RP = DF.
Question 3.
The given measures of some parts of the triangles are indicated. By applying SAS congruence rule, state the pairs of congruence triangles, if any in each case. In case of congruent triangles, write them in symbolic form.
Answer:
(i) ΔABC and ΔDEF are congruent
AB = 2.5 cm; DE = 2.5 cm
∴ AB = DE
AC = 2.8 cm, DF = 2.8 cm
∴ AC = DF
∠A = 80°; ∠D= 70°
∴ ∠A ≠∠D
∴ ΔABC and ΔDEF are not congruent.
(ii) In ΔDEF and ΔPQR
EF = 3 cm; QR = 3 cm
∴ EF = QR
FD = 3.5 cm; PQ = 3.5 cm
∴ FD = PQ
∠F = 40°, ∠Q = 40°
∴ ∠F = ∠Q
∴ Two sides of ΔADE and their included angle are equal to the corresponding sides and their included angle of ΔPQR
∴ The two triangles are congruent.
ΔDEF ≅ ΔPQR
(SAS Congruence rule)
(iii) In ΔABC and ΔPQR.
AC = 2.5 cm, PR = 2.5 cm
∴ AC = PR
BC = 3 cm; PQ = 3 cm
∴ BC = PQ
∠C = 35° = ∠P = 35°
∴ ∠C = ∠P
∴ Two sides of ΔABC and their included angle are equal to the two corresponding sides and their included angle of ΔPQR.
∴ ΔABC ≅ ΔRQP
(iv) In the ΔPRS and ΔQPR
PQ = 3.5 cm, SR = 3.5 cm
∴ PQ = SR
PR = PR (PR is common)
∠QPR = 30°∠PRS = 30°
∴ Two sides of ΔQPR and their included angle are equal to the corresponding sides and their included angle of ΔPRS.
∴ ΔPQR ≅ ΔRPS
(SAS congruence rule)
Question 4.
In the given Figure, \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements are true?
(a) ΔAOC ≅ ΔDOB
(b) ΔAOC ≅ ΔBOD
Answer:
Since \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) bisect each other at O.
AO = BO and CO = DO
∠AOC = ∠BOD
(vertically opposite angles are equal)
(i) In ΔAOC and ΔBOD
we have, AO = BO and CO = DO
∠AOC = ∠BOD
(ii) From the above given relation
ΔAOC ≅ ΔBOD
(By SAS congruence rule)
(a) The statement ΔAOC ≅ ΔDOB is false.
(b) The statement ΔAOC ≅ ΔBOD is true.
NCERT In-text Question Page No. 145 & 146
Question 1.
What is the side included between the angles M and N of ΔMNP?
Answer:
In AMNP, the side included between M and N is \(\overline{\mathrm{MN}}\).
Question 2.
You want to establish ΔDEF ≅ ΔMNP, using the ASA congruence rule. You are given that ∠D = ∠M and ∠F = ∠P. What information is needed to establish the congruence? (Draw a rough figure and then try!)
Answer:
To establish ΔDEF ≅ ΔMNP, using ASA congruence rule, we need the side containing ∠D and ∠F to be equal to the side containing ∠M and ∠P.
i. e. We need \(\overline{\mathrm{DF}}=\overline{\mathrm{MP}}\).
Question 3.
In the given figure, PL⊥OB and PM⊥OA such that PL = PM. Prove that ΔPLO ≅ ΔPMO
Answer:
In ΔPLO and ΔPMO, we have
∠PLO = ∠PMO = 90°
(Given)
OP = OP (Hypotenuse – common side)
PL = PM (Given)
By using RHS Congruency, we get
ΔPLO ≅ ΔPMO
Question 4.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.
ΔDEF
(i) ∠D = 60°, ∠F = 80°, ∠DF = 5 cm
(ii) ∠D = 60°, ∠F = 80°, DF = 6cm
(iii) ∠E = 80°, ∠F = 30°, EF = 5 cm
ΔPQR
(i) ∠Q = 60°, ∠R = 80°, QR = 5cm
(ii) ∠Q = 60°, ∠R = 80°, QP = 6cm
(iii) ∠P = 80°, PQ = 5 cm, ∠R = 30°
Answer:
(i) Since ∠D = ∠Q (each = 60°)
∠F = ∠R (each = 80°)
Included side DF = Included side QR [each = 5 cm]
∴ Using ASA congruence rule, we can say that two triangles are congruent.
Also, D ↔ Q, F ↔ R and E ↔ P
∴ ΔDEF ≅ ΔQPR
(ii) Here, QP is not the included side.
∴ The given triangles are not congruent.
(iii) Here, PQ is not the included side.
∴ The given triangles are not congruent.
Question 5.
In the given figure ray AZ bisects ∠DAB as well as ∠DCB.
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is ΔBAC ≅ ΔDAC? Give reasons.
(iii) Is AB = AD? justify your answer.
(iv) Is CD = CB? Give reasons.
Answer:
(i) AC is the bisector of ∠DAB as well as of ∠DCB
∴ ∠DAC = ∠BAC and ∠DCA = ∠BCA
In ΔBCA and ΔDAC equal parts are
AC = AC (Common)
∠DAC = ∠BAC (AC is a bisector)
∠DCA = ∠BCA (AC is a bisector)
(ii) From the above relation, the two triangles are congruent (using ASA congruence rule)
A ↔ A C ↔ C and D ↔ B
∴ ΔADC ≅ ΔABC
or
ΔBAC ≅ ΔDAC
(iii) Since ΔBAC ≅ ΔDAC The corresponding parts are equal.
∴ AB = AD
(iv) Since ΔBAC ≅ ΔDAC
∴ CD = CB (Corresponding parts are equal)
NCERT In-text Question Page No. 148
Question 1.
In the given figure measures of some parts of triangles are given. By applying R.H.S congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.
Answer:
(i) In the right ΔPQR and right ΔDEF Hypotenuse PR = Hypotenuse DF (each 6 cm)
PQ = 3 cm, DE = 2.5 cm
PQ ≠ DE
∴ ΔPQR is not congruent to ΔDEF
(ii) Right ΔABC and right ΔABD we have two right triangles ABC and ABD such that
Hypotenuse AB = Hypotenuse BA (common)
AC = 2 cm, BD = 2 cm
∴ AC = BD ∠C = ∠D = 90°
Buy using RHS congruence rule, we can say that the two right triangles are congruent
∴ ΔABC ≅ ΔBAD
(iii) Right ΔABC and Right ΔACD, We have right ΔABC and right ΔACD such that Hypotenuse AC = Hypotenuse AC (common) side \(\overline{\mathrm{AB}}\) = side \(\overline{\mathrm{AD}}\) (each 3.6 cm)
∴ The two right triangles are congruent.
∴ ΔABC ≅ ΔADC
(iv) Right ΔPQS and Right ΔPRS
Hypotenuse PQ = Hypotenuse PR (each 3 cm)
side PS = side PS (common)
∴ using R.H.S congruence rule that two right triangles are congruent.
∴ i.e., ΔPQS ≅ ΔPRS
Question 2.
It is to be established by RHS congruence rule that ΔABC ≅ ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP?
Answer:
We have ΔABC ≅ ΔRPQ
Since, ∠B = 90° ⇒ Side AC is hypotenuse and ∠P = 90° ≅ Side RQ is hypotenuse.
The required information needed is Hypotenuse
AC = Hypotenuse RQ.
Question 3.
In the given figure, BD and CE are altitudes of ΔABC such that BD = CE.
(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE? why or why not?
(iii) Is ∠DCB = ∠EBC? why or why not?
Answer:
(i) Hypotenuse BC = Hypotenuse BC
(common)
side BD = side CE (Given)
∠BEC = ∠BDC= 90°
The above are the three equal parts in ΔCBD and ΔBCE
(ii) ∠D = ∠E and ∠B = ∠C
∴ ΔCBD ≅ ΔBCE (By RHS)
(iii) Since ΔCBD ≅ ΔBCE
∠DCB =∠EBC
( ∵ their corresponding parts are equal)
Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? why or why not?
(iii) Is ∠B = ∠C? why or why not?
(iv) Is BD = CD? why or why not?
Answer:
(i) The three pairs of equal parts in ΔADB and ΔADC are
AD = AD (common)
Hypotenuse AB = Hypotenuse AC
(Given)
∠ADB = ∠ADC (each 90°)
(ii) A ↔ A, B ↔ C, D ↔ D
∴ ΔADB ≅ ΔADC (RHS)
(iii) ΔADB ≅ ΔADC
As, corresponding parts are equal.
∴ ∠B = ∠C
(iv) ΔADB ≅ ΔADC
As corresponding parts are equal.
BD = CD