NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions

Class 7 Maths NCERT Solutions Chapter 7 Congruence of Triangles InText Questions

Try These (Page 140 -141)

Question 1.
In the figure, lengths of the sides of the triangles are indicated. By applying the SSS congruence rule, a state which pairs of triangles are congruent. In the case of congruent triangles, write the result in symbolic form:
Solution:
(i)

In ΔABC and ΔPQR, we have
AB = PQ = 1.5 cm
BC = QR = 2.5 cm
CA = RP = 2.2 cm
By SSS criterion of congruence,
ΔABC ≅ ΔPQR

(ii)

In Δ DEF and Δ LMN, we have
DE = MN = 3.2 cm
DF = LN = 3.5 cm
EF = LM = 3 cm
So, by SSS criterion of congruence,
Δ DEF ≅ Δ NML

(iii)

In ΔABC and ΔPQR, we have
AC = PR = 5 cm
BC = PQ = 4 cm But AB ≠ QR
So, the SSS congruence rule is not applicable.
∴ Δ ABC and Δ PQR are not congruent.

(iv)

In Δ ADB and Δ ADC, we have
AD = AD (Common)
DB = DC = 2.5 cm and, BA = CA = 3.5 cm
∴ By SSS criterion of congruence,
Δ ADB ≅ Δ ADC

Question 2.
In the figure, AB = AC, and D is the mid-point of BC.

(i) State the three pairs of equal parts in ΔADB and ΔADC.
Solution:
Three pairs of equal parts in ΔADB and ΔADC are
AD = AD [Common]
AB = AC [Given]
DB = DC [∵ D is mid-point of BC]

(ii) Is ΔADB ≅ ΔADC ? Give reasons.
Solution:
By SSS criterion of congruence, .
ΔADB ≅ ΔADC

(iii) Is ∠B = ∠C? Why?
Solution:
Yes, ∠B = ∠C [∵ Corresponding parts of congruent triangles are equal]

Question 3.
In the figure, AC = BD and AD = BC. Which of the following statements is meaningfully written?

(i)ΔABC ≅ ΔABD
Solution:
Statement ΔABC ≅ ΔABD is not meaningfully written as
AB = AB    [Correct]
BC = BD    [Incorrect]
CA = DA     [Incorrect]

(ii)ΔABC ≅ ΔBAD
Statement ΔABC ≅ ΔBAD is written meaningfully as
AB = AB    [Common]
BC = AD    [Given]
CA = DB      [Given]

Try These (Page 143 -144)

Question 1.
Which angle is included between the sides. , and   Δ DEF?
Solution:
The angle included between the sides. , and   Δ DEF is ∠DEF.

Question 2.
By applying the 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?
Solution:
Here, we want to establish that ΔPQR ≅ ΔFED [By SAS congruence rule]
Given that, PQ = FE and RP = DF
So, the additional information needed to establish the congruence is ∠P = ∠F.

Question 3.
In the figure, measures of some parts of the triangles are indicated. By applying the SAS congruence rule, state the pairs of congruent triangles, if any, in each case. In the case of congruent triangles, write them in symbolic form.

(i)

In Δ ABC and Δ DEF, we have
AB =DE = 2.5 cm,
AC = DF = 2.8 cm
But ∠A ≠ ∠D
So, the SAS congruence rule is not applicable.
Therefore, Δ ABC is not congruent to Δ DEF.

(ii)

In Δ ACB and Δ RPQ, we have
AC = RP = 2.5 cm
∠C = ∠P = 35°
BC = PQ = 3 cm
∴ By SAS criterion of congruence, ΔACB ≅ ΔRPQ

(iii)

In Δ DEF and ΔPQR, we get
DF = PQ = 3.5 cm
EF = QR = 3 cm
∠F = ∠Q = 40°
∴ By SAS criterion on congruence, Δ DEF ≅ Δ PQR

(iv)

In ΔRSP and ΔPQR, we have
RS = PQC = 3.5 cm)
∠PRS = ∠RPQ = 30°
RP = PR    [Common]
∴ By SAS criterion of congruence, ΔRSP ≅ ΔPQR

Question 4.
In the figure, and,  bisect each other at O.

(i) State the three pairs of equal parts in two triangles AOC and BOD.
Solution:
Three pairs of equal parts in two triangles AOC and BOD
AO = OB
CO = OD
∠AOC = ∠BOD

(ii)Which of the following statements are true?
ΔAOC ≅ ΔDOB
ΔAOC ≅ ΔBOD
Solution:
In ΔAOC and ΔBOD, we have
OA = OB
OC = OD
∠AOC ≅ ∠BOD [vertically opposite angle]
So, by SAS congruence rule, two triangles are congruent.
The correspondence is A ↔ B, O ↔ O, and C ↔ D.
In symbolic form, Δ AOC ≅ ΔBOD
Hence, statement (b) i.e., ΔAOC ≅ ΔBOD is true.

Try These (Page 145 -146)

Question 1.
What is the side included between the angles M and N of ΔMNP?
Solution:
The side MN is included between the angles M and N of ΔMNP.

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)
Solution:
To establish ΔDEF ≅ ΔMNP, by using the ASA congruence rule the additional information needed is about the equality of sides DF and MP.

Question 3.
In the figure, measures of some parts are indicated. By applying the ASA congruence rule, a state which pairs of triangles are congruent. In the case of congruence, write the result in symbolic form.

(i)

Solution:
In Δ ABC and Δ EFD, we have
∠A = ∠F = 40°
AB = EF = 3.5 cm
∠B = ∠E = 60°
∴ By ASA criterion of congruence, ΔABC ≅ ΔEFD

(ii)

Solution:
Given triangles are not congruent.

(iii)

Solution:
In ΔPQR and ΔMNL, we have
∠R=∠L = 60°
QR = NL = 6 cm
∠Q = ∠N= 30°
∴ By ASA criterion of congruence,
ΔPQR ≅ ΔMNL

(iv)

Solution:
In ΔABC and ΔBAD, we have
∠CAB = ∠DBA = 30°
AB = BA [Common]
∠ABC = ∠BAD = 30 ° + 45° = 75°
∴ By ASA criterion of congruence, ΔABC ≅ ΔBAD

Question 4.
Given below are Measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruent rule. In the case of congruence, Write it in symbolic form.

(i) ΔDEF = ∠D = 60°, ∠F = 60°, DF = 5 cm
ΔPQR = ∠Q = 60°, ∠R = 80°, QR = 5 cm
Solution:

In ΔDEF and ΔPQR, we have
∠D = ∠Q = 60°
DF = QR = 5 cm
∠F = ∠R = 80°
∴ By ASA criterion of congruence,ΔDEF ≅ ΔQRP

(ii) ΔDEF = ∠D = 60°, ∠F = 60°, DF = 6 cm
ΔPQR = ∠Q = 60°, ∠R = 80°, QR = 6 cm
Solution:
Sience side included between equal angles in the given triangles is not equal.

∴ Given triangles are not congruent.

(iii) ΔDEF = ∠E = 80°, ∠F = 30°, EF = 5 cm
ΔPQR = ∠P = 80°, ∠R = 30°, PQ = 5 cm
Solution:

In ΔDEF and ΔQPR, we have
∠E = ∠P = 80°
EF = PR = 5 cm
∠F = ∠R = 30°
∴ By ASA criterion of congruence,
ΔDEF ≅ ΔQPR

Question 5.
In the figure, ray AZ bisects ∠DAB as well as ∠DCB.

(i) State the three pairs of equal parts in triangles BAC and DAC.
Solution:
The three pairs of equal parts in triangles BAC and DAC are
∠BAC = ∠DAC,
AC = AC      [Common]

(ii) Is ΔBAC ≅ ΔDAC? Give reasons.
Solution:
Yes, In ΔBAC and ΔDAC, we have,
∠DAC =∠BAC   [Given]
AC = AC            [Common]
∠DCA – ∠BCA   [Given]
∴ By ASA criterion of congruence,
ΔBAC ≅ ΔDAC

(iii) Is AB = AD ? Justify your answer.
Solution:
Yes, AB = AD [∵ Corresponding parts of congruent triangles are equal.]

(iv) Is CD = CB ? Give reasons.
Solution:
Yes, CD = CB [∵ Corresponding parts of congruent triangles are equal.]

Try These (Page 148)

Question 1.
In the figure, measures of some parts of triangles are given. By applying the RHS congruence rule, a state which pairs of triangles are congruent. In the case of congruent triangles, write the result in symbolic form.

(i)

In ΔPQR and ΔDEF, we have
∠Q = ∠E = 90°
hypotenuse PR = hypotenuse DF = 6 cm
PQ ≠ DE
Therefore, RHS congruence rule is not satisfies.
Hence, ΔPQR and ΔDEF are not congruent.

(ii)

In ΔCAB and ΔDBA, we have
∠C = ∠D = 90°
AB = BA = 3.5 cm
CA = DB = 2 cm
∴ By RHS criterion of congruence,
ΔCAB ≅ ΔDBA

(iii)

In ΔABC and ΔADC, we have
∠B = ∠D = 90°
hypotenuse AC = hypotenuse AC
AB = AD = 3.6 cm
∴ By RHS criterion of congruence,
ΔABC ≅ ΔADC

(iv)

In ΔPSQ and ΔPSR, we have
∠PSQ = ∠PSR = 90°
hypotenuse PQ = hypotenuse PR = 3 cm
PS = PS [Common]
∴ By RHS criterion of congruence,
ΔPSQ ≅ ΔPSR

Question 2.
It is to be established by the RHS congruence rule that ΔABC ≅ ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP?
Solution:
To establish by using RHS congruence rule that ΔABC ≅ ΔRPQ. The additional information needed is about equality of hypotenuses, i.e., AC = RQ.

Question 3.
In the figure, BD and CE are altitudes of ΔABC such that BD = CE.

(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
Solution:
The three pairs of equal parts in ΔCBD and ΔBCE
CB = BC        [Common]
∠CDB = ∠BEC and BD = CE

(ii) Is ΔCBD = ΔBCE? Why or why not?
Solution:
Yes, In ΔCBD and ΔBCE, we have
∠BDC = ∠BEC = 90°
CB = BC [Common]
BO = CE
∴ By RHS criterion of congruence,
ΔCBD ≅ ΔBCE

(iii) Is ∠DCB = ∠EBC? Why or why not?
Solution:
Yes, ∠DCB = ∠EBC
∵ Corresponding parts of congruent triangles are equal.

Question 4.
ABC is an isosceles triangle with AB – AC, and AD is one of its altitudes (see adjoining figure).

(i) State the three pairs of equal parts in ΔADB and ΔADC
Solution:
The three pairs of equal parts in ΔADB and ΔADC are
AD = AD [Common]
ZADB = ZADC
AB = AC

(ii) Is ΔADB = ΔADC ? Why or why not?
Solution:
Yes, In ΔADB and ΔADC, we have
∠ADB = ∠ADC [Each 90°]
AB = AC
AD = AD [Common]
∴ By the RHS criterion of congruence,
ΔADB = ΔADC

(iii) Is ∠B = ∠C ? Why or why not?
Solution:
Yes, ∠B = ∠C
(∵ Corresponding parts of congruent triangles are equal.)

(iv) Is BD = CD ? Why or why not?
Solution:
(iv) Yes, BD = CD
(∵ Corresponding parts of congruent triangles are equal.)