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

Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions and Answers.

Gujarat Board Textbook Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions

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

Think, Discuss and Write (Page 137)

Question 1.
When two triangles, say ABC and PQR are given, there are, in all, six possible matchings or correspondences. Two of them are:
(i) ABC ā†” PQR
(ii) ABC ā†” QRP
Find the other four correspondences by using two cutouts of triangles. Will all these correspondences lead to congruence? Think about it.
Solution:
The other 4 correspondences are:
(i) ABC ā†” PRQ
(ii) ABC ā†” QPR
(iii) ABC ā†” RPQ
(iv) ABC ā†” RQP
Yes, all these correspondences may lead to congruence.

Try These (Page 140)

Question 1.
In the following figures, 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:
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 1
Solution:
(i) We have āˆ†ABC and āˆ†PQR, such that
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 2
Because the three sides of āˆ†ABC are equal to the three sides of āˆ†PQR, therefore the two triangles are congruent. [By SSS congruency rule] Now from the above equality relations, we have A ā†” P, B ā†” Q, C ā†”
āˆ“ āˆ†ABC ā‰… āˆ†PQR

(ii) We have āˆ†DEF and āˆ†LMN, such that
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 3
i.e. the three sides of ADEF are equal to three sides of ALMN. So, by SSS congruency rule, the two triangles are congruent. From the above equality relations, we have
D ā†” N, E ā†” M and F ā†” L
āˆ“ āˆ†DEF ā‰… āˆ†NML

(iii) We have āˆ†ABC and āˆ†PQR, such that
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 4
Since AB ā‰  QR, therefore āˆ†ABC and āˆ†PQR are not congruent.

(iv) We have āˆ†ABD and āˆ†ADC, such that AB = 3.5cm
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 5
Since, three sides of āˆ†ABD are equal to the three sides of āˆ†ACD, therefore the two triangles are congruent.
From the above equality relations, we have
A ā†” A, B ā†” C and D ā†” D
āˆ“ āˆ†ABD ā‰… AACD

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

Question 2.
In the adjoining figure, AB = AC and D is the mid-point of \(\overline{B C}\)
(i) State the three pairs of equal parts in āˆ†ADB and āˆ†ADC.
(ii) Is āˆ†ADB ā‰… āˆ†ADC? Give reasons.
(iii) Is āˆ B = āˆ C? Why?
Solution:
Here, AB = AC and D is the mid-point \(\overline{B C}\).
i.e. BD = DC
(i) We have; āˆ†ADB and āˆ†ADC, such that
AB = AC (Given)
AD = AD (Common)
BD = DC
(āˆµ D is the mid-point of BC)

(ii) āˆµ The three sides of āˆ†ADB are equal to the three sides of āˆ†ADC. Therefore, by SSS rule of congruency, the two triangles are congruent.
From the above equality relations, we have
A ā†” A, D ā†” D, B ā†” C
āˆ“ āˆ†ADB ā‰… āˆ†ADC

(iii) āˆµ āˆ†ADB ā‰… āˆ†ADC
āˆ“ Their corresponding parts are equal, i.e.
B ā†” C
or āˆ B = āˆ C

Question 3.
In the adjoining figure, AC = BD and AD = BC. Which of the following statements is meaningfully written?
(i) āˆ†ABC ā‰… āˆ†ABD
(ii) āˆ†ABC ā‰… ABAD
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 6
Solution:
Here, AC BD and AD = BC.
In the given triangles, we have
AB = AB [Common]
BD = AC [Given]
AD = BC [Given]
i.e. the three sides of āˆ†ABD are equal to the three sides of āˆ†ABC. So, the two traingles are congruent. From the above equality relations, we have A B, B A, D C
āˆ“ (i) āˆ†ABC ā‰… āˆ†ABD is not true or meaningless.
(ii) āˆ†ABC ā‰… āˆ†BAD is true of meaningful.

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

Think, Discuss and Write (Page 141)

Question 1.
ABC is an isosceles triangle with AB = AC. Take a trace-copy of āˆ†ABC and also name it as āˆ†ABC.
(i) State the three pairs of equal parts in āˆ†ABC and āˆ†ACB.
(ii) Is āˆ†ABC ā‰… āˆ†ACB? Why or why not?
(iii) Is āˆ B = āˆ C Why or why not?
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 7
Solution: We have āˆ†ABC is an isosceles triangle with
AB = AC.
(i) Now, in āˆ†ABC and āˆ†ACB,
BC = BC [Common]
AB = AC [Given]
AC = AB [Construction]
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 8

(ii) Yes, all the three sides of āˆ†ABC are equal to the three sides of āˆ†ACB and A ā†” A,B ā†” C and C ā†” B.

(iii) Yes
āˆµ B ā†” C
āˆ“ āˆ B = āˆ C.

Try These (Page 143)

Question 1.
Which angle is included between the sides \(\overline{D E}\) and \(\overline{E F}\) is āˆ†DEF.
Solution:
In āˆ†DEF, the angle between \(\overline{D E}\) and \(\overline{E F}\) is āˆ DEF.

Question 2.
By applying SAS congruence rule, you want to establish that āˆ†PQR ā‰… āˆ†FED. It is given that PQ = FE and RP = DF. What additional informaHon is needed to establish the congruence?
Solution:
We have āˆ†PQR ā‰… āˆ†FED
[Using SAS congruence rule]
āˆ“ P ā†” F, Q ā†” E, R ā†” D
and PQ = FE and RP = DF.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 9
Since in SAS congruence rule, included angle between PQ and RP is equal to the included angle between EF and DF. i.e. āˆ P = āˆ F

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

Question 3.
In the following figures, measures of some parts of the triangles are indicated. By applying SAS congruence rule, state the pairs of congruent triangles, if any, in each case. In case of congruent triangles, write them in symbolic form.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 10
Solution:
(i) āˆ†ABC and āˆ†DEF:
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 14
āˆ“ āˆ†ABC and āˆ†DEF are not congruent.

(ii) āˆ†ABC and āˆ†PQR:
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 15
āˆµ Two sides of āˆ†ABC and their included angle are equal to the two corresponding sides and their included angle of āˆ†PQR. Therefore, the two triangles are congruent.
Also, we have C ā†” P, A ā†” R and B ā†” Q
āˆ“ āˆ†ABC ā‰… āˆ†RQP

(iii) āˆ†DEF and āˆ†PQR
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 16
āˆµ Two sides of āˆ†DEF and their included angle are equal to two corresponding sides and their included angle of āˆ†PQR.
āˆ“ The two triangles are congruent by SAS congruence rule.
Also, F ā†” Q, D ā†” P, and E ā†” R
āˆ“ āˆ†DEF ā‰… āˆ†PRQ
āˆ†PRS ā‰… āˆ†QPR

(iv) \(\left.\begin{array}{l}
P Q=3.5 \mathrm{~cm} \\
S R=3.5 \mathrm{~cm}
\end{array}\right\}\)
PR = PR [common] ā‡’ PQ = SR

\(\left.\begin{array}{l}
\text { Included } \angle \mathrm{QPR}=30^{\circ} \\
\text { Included } \angle \mathrm{PRS}=30^{\circ}
\end{array}\right\}\) ā‡’ Included āˆ QPR
= āˆ PRS
i.e. two sides of āˆ†PQR and their included angle are equal to two corresponding sides and their included angle of āˆ†PRS.
āˆ“ Using SAS congruence rule, the two triangles are congruent.
Now, P ā†” R, Q ā†” S.
āˆ“ āˆ†PQR ā‰… āˆ†RSP

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

Question 4.
In the adjoining figure, \(\overline{A B}\) and \(\overline{C D}\) bisect each other at O.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 11
(i) State the three pairs of equal parts in two triangles AOC and BOD.
(ii) Which of the following statements A are true?
(a) āˆ†AOC ā‰… āˆ†DOB
(b) āˆ†AOC ā‰… āˆ†BOD
Solution:
Since \(\overline{A B}\) and \(\overline{C D}\) bisect each other at O.
āˆ“ AO = BO amd CO = DO
Also, vertically opposite angles āˆ AOC = āˆ BOD

(i) āˆ†AOC and āˆ†BOD:
We have AO = BO and CO = DO
āˆ AOC = āˆ BOD

(ii) From above equality relations, we have two sides and their included angle of āˆ†AOC are equal to two corresponding sides and their included angle of āˆ†BOD.
āˆ“ By SAS congruence rule, the two triangles are congruent.
Also, O ā†” O, A ā†” B, C ā†” D
āˆ“ āˆ†AOC ā‰… āˆ†BOD
(a) The statement āˆ†AOC ā‰… āˆ†DOB is false.
(b) The statement āˆ†AOC ā‰… āˆ†BOD is true.

Try These (Page 145)

Question 1.
What is the side included between the angles M and N of āˆ†MNP
Solution:
In āˆ†MNP, the side included between M and N is \(\overline{M N}\).

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

Question 2.
You want to establish āˆ†DEF ā‰… āˆ†MNP, using the ASA congruence rule. You are given that āˆ D = āˆ M and āˆ F = āˆ P. What information is needed to establish the congruence? (Draw a rough figure and then try!)
Solution:
To establish āˆ†DEF ā‰… āˆ†MNP, using ASA congruence rule, we need the side containing āˆ D and āˆ F to be equal to the side containing āˆ M and āˆ P.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 12
i.e. We need \(\overline{D F}\) = \(\overline{M P}\)

Question 3.
In the following figures, measures of some parts are indicated. By applying ASA congruence rule, state which pairs of triangles are congruent. In case of congruence, write the result in symbolic form.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 13
Solution:
(i) āˆ†ABC and āˆ†DEF:
We have
\(\left.\begin{array}{l}
A B=3.5 \mathrm{~cm} \\
E F=3.5 \mathrm{~cm}
\end{array}\right\}\) ā‡’ AB = FE

\(\left.\begin{array}{l}
\angle \mathrm{A}=40^{\circ} \\
\angle \mathrm{F}=40^{\circ}
\end{array}\right\} \Rightarrow \angle \mathrm{A}\) = āˆ F

\(\left.\begin{array}{l}
\angle \mathrm{B}=60^{\circ} \\
\angle \mathrm{E}=60^{\circ}
\end{array}\right\} \Rightarrow \angle \mathrm{B}\) = āˆ E
āˆ“ The two triangles are congruent (using the ASA congruence rule).
āˆµ A ā†” F, B ā†” E and C ā†” D
āˆ†ABC ā‰… āˆ†FED

(ii) āˆ†PQR and āˆ†DEF:
In āˆ†PQR, āˆ P = āˆ 180Ā° – (90Ā° + 50Ā°)
= 40Ā°
Also in āˆ†DEF, āˆ F = 180Ā° – 90Ā° – 50Ā°
= 40Ā°
Now in āˆ†PQR and āˆ†DEF, we have
\(\left.\begin{array}{l}
P R=3.3 \mathrm{~cm} \\
E F=3.5 \mathrm{~cm}
\end{array}\right\} \Rightarrow \mathrm{PR} \neq \mathrm{EF}\)

\(\left.\begin{array}{l}
\angle \mathrm{R}=50^{\circ} \\
\angle \mathrm{E}=50^{\circ}
\end{array}\right\} \Rightarrow \angle \mathrm{R}=\angle \mathrm{E}\) \(\left.\begin{array}{l}
\angle \mathrm{P}=40^{\circ} \\
\angle \mathrm{F}=40^{\circ}
\end{array}\right\} \Rightarrow \angle \mathrm{P}=\angle \mathrm{F}\)

āˆ“ Using the ASA congruence rule, we can say that the two triangles are not congruent.

(iii) āˆ†PQR and āˆ†LMN:
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 17
Therefore, using ASA congruence rule, the two triangles are congruent.
Now, R ā†” L,Q ā†” N and P ā†” M
āˆ“ āˆ†PQR ā‰… āˆ†MNL

(iv) AABC and AABD:
We have AB = AB [Common]
\(\left.\begin{array}{l}
\angle \mathrm{BAD}=45^{\circ}+30^{\circ}=75^{\circ} \\
\angle \mathrm{ABC}=45^{\circ}+30^{\circ}=75^{\circ}
\end{array}\right\}\)
ā‡’ āˆ BAD = āˆ ABC
āˆ“ Using ASA congruence rule, we say that two triangles are congruent.
Now, A ā†” B, D ā†” C.
āˆ“ āˆ†ABC ā‰… āˆ†BAD

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

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.
āˆ†DEF
(i) āˆ D = 60Ā°, āˆ F = 80Ā°, āˆ F = 5 cm
(ii) āˆ D = 60Ā°, āˆ F = 80Ā°, āˆ F = 6 cm
(iii) āˆ F = 80Ā°, āˆ F = 30Ā°, āˆ F = 5 cm
āˆ†PQR
(i) āˆ Q = 60Ā°, āˆ R = 80Ā°, āˆ R = 5 cm
(ii) āˆ Q = 60Ā°, āˆ R = 80Ā°, āˆ P = 6 cm
(iii) āˆ P = 80Ā°, āˆ Q = 5 cm, āˆ R = 30Ā°
Solution:
(i) āˆµ āˆ D = āˆ Q (each = 60Ā°)
āˆ F = āˆ R (each = 80Ā°)
Included side DF = Included side QR [each = 5 cm]
āˆ“ Using ASA congruence rule, we can say that two triangles are congruent.
Also, D ā†” Q, F ā†” R and E ā†” P
āˆ“ āˆ†DEF ā‰… āˆ†QPR

(ii) Here, QP is not the included side.
āˆ“ The given triangles are not congruent.

(iii) Here, PQ is not the included side.
āˆ“ The given triangles are not congruent.

Question 5.
In the adjoining figure, ray AZ bisects āˆ DAB as well as āˆ DCB.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 18
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is āˆ†BAC ā‰… āˆ†DAC? Give reasons.
(iii) Is AB = AD? Justify your answer.
(iv) Is CD = CB? Give reasons.
Solution:
(i) āˆµ AC is the bisector of āˆ DAB, as well as of āˆ DCB
āˆ DAC = āˆ BAC and āˆ DCA = āˆ BCA
Now, in āˆ BAC and āˆ DAC, equal parts are:
AC = AC [Common]
āˆ DAC = āˆ BAC [AC is a bisector]
āˆ DCA = āˆ BCA [AC is a bisector]

(ii) From the above equality relation, the two triangles are congruent (using ASA congruence rule).
Also, A ā†” A, C ā†” C and D ā†” B
āˆ“ āˆ†ADC ā‰… āˆ†ABC
or āˆ†BAC ā‰… āˆ†DAC

(iii) āˆµ āˆ†BAC ā‰… āˆ†DAC
āˆ“ The corresponding parts are equal,
i.e. AB = AD

(iv) āˆµ C ā†” C and D ā†” B
āˆ“ CD = CB
or we have āˆ†BAC ā‰… āˆ†ADAC
āˆ“ The corresponding parts are equal,
i.e. CD = CB.

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

Try These (Page 148)

Question 1.
In the following figures, measures of some parts of triangles are given. By applying RHS congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 19
Solution:
(i) Right āˆ†PQR and Right āˆ†DEF:
We have two right triangles PQR and DEF, such that:
Hypotenuse PR = Hypotenuse DF [each 6 cm]
But side PQ (3 cm) ā‰  DE (2.5 cm)
āˆ“ āˆ†PQR is not congruent to āˆ†DEF.

(ii) Right āˆ†ABC and Right āˆ†ABD:
We have two right triangles ABC and ABD, such that:
Hypotenuse AB = Hypotenuse BA [Common]
Side AC = Side BD [each = 2 cm]
So, using RHS congruence rule, we can say that the two right triangles are congruent.
Now, A ā†” B, B ā†” A, C ā†” D
āˆ“ āˆ†ABC ā‰… āˆ†BAD

(iii) Right AABC and Right āˆ†ACD:
We have right āˆ†ABC and right āˆ†ACD, such that: Hypotenuse AC = Hypotenuse AC [Common]
Side \(\overline{A B}\) = Side \(\overline{A D}\) [Each = 3.6 cm]
āˆ“ The two right triangles are congruent.
Now, A ā†” A, C ā†” C, B ā†” D
āˆ“ āˆ†ABC ā‰… āˆ†ADC

(iv) Right āˆ†PQS and Right āˆ†PRS:
We have right āˆ†PQS and right āˆ†PRS, such that: Hypotenuse PQ Hypotenuse PR [each = 3 cm] Side PS = Side PS [Common]
āˆ“ Using RHS congruence rule, the two right triangles are congruent.
Now, P ā†” P, S ā†” S, Q ā†” R
āˆ“ āˆ†PQS ā‰… āˆ†PRS

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

Question 2.
It is to be established by RHS congruence rule that āˆ†ABC = āˆ†RPQ. What additional information is needed, if it is given that āˆ B = āˆ P = 90Ā° and AB = RP?
Solution:
We have āˆ†ABC ā‰… āˆ†RPQ
Since, āˆ B = 90Ā° ā‡’ Side AC is hypotenuse
and āˆ P = 90Ā° ā‡’ Side RQ is hypotenuse
āˆµ AB = RP
āˆ“ The required information needed is Hypotenuse AC = Hypotenuse RQ.

Question 3.
In the adjoining figure, BD and CE are altitudes of āˆ†ABC such that BD = CE.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 20
(i) State the three pairs of equal parts in āˆ†CBD and āˆ†BCE.
(ii) Is āˆ†CBD = āˆ†BCE? Why or why not?
(iii) Is āˆ†DCB = āˆ†EBC Why or why not?
Solution:
(i) Hypotenuse BC = Hypotenuse BC
[Common]
Side BD = Side CE [Given]
āˆ BEC = āˆ BDC = 90Ā°
The above equality pairs are the required three equal parts in āˆ†CBD and āˆ†BCE.

(ii) āˆµ āˆ D = āˆ E and āˆ B ā†” āˆ C
āˆ“ āˆ† CBD ā‰… āˆ†BCE

(iii) āˆµ āˆ†CBD ā‰… ABCE
āˆ“ Their corresponding parts are equal.
or āˆ DCB = āˆ EBC.

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

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles InText Questions 21
(i) State the three pairs of equal parts in āˆ†ADB and āˆ†ADC.
(ii) Is āˆ†ADB = āˆ†ADC? Why or why not?
(iii) Is āˆ B = āˆ C? Why or why not?
(iv) Is BD = CD? Why or why not?
Solution:
(i) The three pairs of equal parts in āˆ†ADB and āˆ†ADC are:
AD = AD (Common)
Hypotenuse AB = Hypotenuse AC (Given)
āˆ ADB = āˆ ADC (Each = 90Ā°)

(ii) āˆµ A ā†” A, B ā†” C, D ā†” D
āˆ“ āˆ†ADB ā‰… āˆ†ADC

(iii) Since, āˆ†ADB ā‰… āˆ†ADC
āˆ“ Corresponding parts are equal.
āˆ“ āˆ B = āˆ C

(iv) Also, \(\overline{B D}\) = \(\overline{C D}\)

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