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

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:

Solution:

(i) We have āABC and āPQR, such that

Because the three sides of āABC are equal to the three sides of āPQR, 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

ā“ āABC ā
āPQR

(ii) We have āDEF and āLMN, such that

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

ā“ āDEF ā
āNML

(iii) We have āABC and āPQR, such that

Since AB ā QR, therefore āABC and āPQR are not congruent.

(iv) We have āABD and āADC, such that AB = 3.5cm

Since, three sides of āABD are equal to the three sides of āACD, therefore the two triangles are congruent.

From the above equality relations, we have

A ā A, B ā C and D ā D

ā“ āABD ā
AACD

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 āADB and āADC.

(ii) Is āADB ā
āADC? Give reasons.

(iii) Is ā B = ā C? Why?

Solution:

Here, AB = AC and D is the mid-point \(\overline{B C}\).

i.e. BD = DC

(i) We have; āADB and āADC, such that

AB = AC (Given)

AD = AD (Common)

BD = DC

(āµ D is the mid-point of BC)

(ii) āµ The three sides of āADB are equal to the three sides of āADC. 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

ā“ āADB ā
āADC

(iii) āµ āADB ā
āADC

ā“ Their corresponding parts are equal, i.e.

B ā C

or ā B = ā C

Question 3.

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

(i) āABC ā
āABD

(ii) āABC ā
ABAD

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 āABD are equal to the three sides of āABC. So, the two traingles are congruent. From the above equality relations, we have A B, B A, D C

ā“ (i) āABC ā
āABD is not true or meaningless.

(ii) āABC ā
āBAD is true of meaningful.

Think, Discuss and Write (Page 141)

Question 1.

ABC is an isosceles triangle with AB = AC. Take a trace-copy of āABC and also name it as āABC.

(i) State the three pairs of equal parts in āABC and āACB.

(ii) Is āABC ā
āACB? Why or why not?

(iii) Is ā B = ā C Why or why not?

Solution: We have āABC is an isosceles triangle with

AB = AC.

(i) Now, in āABC and āACB,

BC = BC [Common]

AB = AC [Given]

AC = AB [Construction]

(ii) Yes, all the three sides of āABC are equal to the three sides of āACB and A ā A,B ā C and C ā B.

(iii) Yes

āµ B ā C

ā“ ā B = ā C.

Try These (Page 143)

Question 1.

Which angle is included between the sides \(\overline{D E}\) and \(\overline{E F}\) is āDEF.

Solution:

In āDEF, the angle between \(\overline{D E}\) and \(\overline{E F}\) is ā DEF.

Question 2.

By applying SAS congruence rule, you want to establish that āPQR ā
āFED. It is given that PQ = FE and RP = DF. What additional informaHon is needed to establish the congruence?

Solution:

We have āPQR ā
āFED

[Using SAS congruence rule]

ā“ P ā F, Q ā E, R ā D

and PQ = FE and RP = DF.

Since in SAS congruence rule, included angle between PQ and RP is equal to the included angle between EF and DF. i.e. ā P = ā F

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.

Solution:

(i) āABC and āDEF:

ā“ āABC and āDEF are not congruent.

(ii) āABC and āPQR:

āµ Two sides of āABC and their included angle are equal to the two corresponding sides and their included angle of āPQR. Therefore, the two triangles are congruent.

Also, we have C ā P, A ā R and B ā Q

ā“ āABC ā
āRQP

(iii) āDEF and āPQR

āµ Two sides of āDEF and their included angle are equal to two corresponding sides and their included angle of āPQR.

ā“ The two triangles are congruent by SAS congruence rule.

Also, F ā Q, D ā P, and E ā R

ā“ āDEF ā
āPRQ

āPRS ā
āQPR

(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 ā QPR

= ā PRS

i.e. two sides of āPQR and their included angle are equal to two corresponding sides and their included angle of āPRS.

ā“ Using SAS congruence rule, the two triangles are congruent.

Now, P ā R, Q ā S.

ā“ āPQR ā
āRSP

Question 4.

In the adjoining figure, \(\overline{A B}\) and \(\overline{C D}\) 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 A are true?

(a) āAOC ā
āDOB

(b) āAOC ā
āBOD

Solution:

Since \(\overline{A B}\) and \(\overline{C D}\) bisect each other at O.

ā“ AO = BO amd CO = DO

Also, vertically opposite angles ā AOC = ā BOD

(i) āAOC and āBOD:

We have AO = BO and CO = DO

ā AOC = ā BOD

(ii) From above equality relations, we have two sides and their included angle of āAOC are equal to two corresponding sides and their included angle of āBOD.

ā“ By SAS congruence rule, the two triangles are congruent.

Also, O ā O, A ā B, C ā D

ā“ āAOC ā
āBOD

(a) The statement āAOC ā
āDOB is false.

(b) The statement āAOC ā
āBOD is true.

Try These (Page 145)

Question 1.

What is the side included between the angles M and N of āMNP

Solution:

In āMNP, the side included between M and N is \(\overline{M N}\).

Question 2.

You want to establish āDEF ā
āMNP, using the ASA congruence rule. You are given that ā D = ā M and ā F = ā P. What information is needed to establish the congruence? (Draw a rough figure and then try!)

Solution:

To establish āDEF ā
āMNP, using ASA congruence rule, we need the side containing ā D and ā F to be equal to the side containing ā M and ā P.

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.

Solution:

(i) āABC and āDEF:

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}\) = ā F

\(\left.\begin{array}{l}

\angle \mathrm{B}=60^{\circ} \\

\angle \mathrm{E}=60^{\circ}

\end{array}\right\} \Rightarrow \angle \mathrm{B}\) = ā E

ā“ The two triangles are congruent (using the ASA congruence rule).

āµ A ā F, B ā E and C ā D

āABC ā
āFED

(ii) āPQR and āDEF:

In āPQR, ā P = ā 180Ā° – (90Ā° + 50Ā°)

= 40Ā°

Also in āDEF, ā F = 180Ā° – 90Ā° – 50Ā°

= 40Ā°

Now in āPQR and āDEF, 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}\)

\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}\)

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

(iii) āPQR and āLMN:

Therefore, using ASA congruence rule, the two triangles are congruent.

Now, R ā L,Q ā N and P ā M

ā“ āPQR ā
āMNL

(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 ā BAD = ā ABC

ā“ Using ASA congruence rule, we say that two triangles are congruent.

Now, A ā B, D ā C.

ā“ āABC ā
āBAD

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.

āDEF

(i) ā D = 60Ā°, ā F = 80Ā°, ā F = 5 cm

(ii) ā D = 60Ā°, ā F = 80Ā°, ā F = 6 cm

(iii) ā F = 80Ā°, ā F = 30Ā°, ā F = 5 cm

āPQR

(i) ā Q = 60Ā°, ā R = 80Ā°, ā R = 5 cm

(ii) ā Q = 60Ā°, ā R = 80Ā°, ā P = 6 cm

(iii) ā P = 80Ā°, ā Q = 5 cm, ā R = 30Ā°

Solution:

(i) āµ ā D = ā Q (each = 60Ā°)

ā F = ā R (each = 80Ā°)

Included side DF = Included side QR [each = 5 cm]

ā“ Using ASA congruence rule, we can say that two triangles are congruent.

Also, D ā Q, F ā R and E ā P

ā“ āDEF ā
āQPR

(ii) Here, QP is not the included side.

ā“ The given triangles are not congruent.

(iii) Here, PQ is not the included side.

ā“ The given triangles are not congruent.

Question 5.

In the adjoining figure, ray AZ bisects ā DAB as well as ā DCB.

(i) State the three pairs of equal parts in triangles BAC and DAC.

(ii) Is āBAC ā
āDAC? Give reasons.

(iii) Is AB = AD? Justify your answer.

(iv) Is CD = CB? Give reasons.

Solution:

(i) āµ AC is the bisector of ā DAB, as well as of ā DCB

ā DAC = ā BAC and ā DCA = ā BCA

Now, in ā BAC and ā DAC, equal parts are:

AC = AC [Common]

ā DAC = ā BAC [AC is a bisector]

ā DCA = ā 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

ā“ āADC ā
āABC

or āBAC ā
āDAC

(iii) āµ āBAC ā
āDAC

ā“ The corresponding parts are equal,

i.e. AB = AD

(iv) āµ C ā C and D ā B

ā“ CD = CB

or we have āBAC ā
āADAC

ā“ The corresponding parts are equal,

i.e. CD = CB.

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.

Solution:

(i) Right āPQR and Right āDEF:

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)

ā“ āPQR is not congruent to āDEF.

(ii) Right āABC and Right āABD:

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

ā“ āABC ā
āBAD

(iii) Right AABC and Right āACD:

We have right āABC and right āACD, such that: Hypotenuse AC = Hypotenuse AC [Common]

Side \(\overline{A B}\) = Side \(\overline{A D}\) [Each = 3.6 cm]

ā“ The two right triangles are congruent.

Now, A ā A, C ā C, B ā D

ā“ āABC ā
āADC

(iv) Right āPQS and Right āPRS:

We have right āPQS and right āPRS, such that: Hypotenuse PQ Hypotenuse PR [each = 3 cm] Side PS = Side PS [Common]

ā“ Using RHS congruence rule, the two right triangles are congruent.

Now, P ā P, S ā S, Q ā R

ā“ āPQS ā
āPRS

Question 2.

It is to be established by RHS congruence rule that āABC = āRPQ. What additional information is needed, if it is given that ā B = ā P = 90Ā° and AB = RP?

Solution:

We have āABC ā
āRPQ

Since, ā B = 90Ā° ā Side AC is hypotenuse

and ā P = 90Ā° ā Side RQ is hypotenuse

āµ AB = RP

ā“ The required information needed is Hypotenuse AC = Hypotenuse RQ.

Question 3.

In the adjoining figure, BD and CE are altitudes of āABC such that BD = CE.

(i) State the three pairs of equal parts in āCBD and āBCE.

(ii) Is āCBD = āBCE? Why or why not?

(iii) Is āDCB = āEBC Why or why not?

Solution:

(i) Hypotenuse BC = Hypotenuse BC

[Common]

Side BD = Side CE [Given]

ā BEC = ā BDC = 90Ā°

The above equality pairs are the required three equal parts in āCBD and āBCE.

(ii) āµ ā D = ā E and ā B ā ā C

ā“ ā CBD ā
āBCE

(iii) āµ āCBD ā
ABCE

ā“ Their corresponding parts are equal.

or ā DCB = ā EBC.

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 āADB and āADC.

(ii) Is āADB = āADC? Why or why not?

(iii) Is ā B = ā C? Why or why not?

(iv) Is BD = CD? Why or why not?

Solution:

(i) The three pairs of equal parts in āADB and āADC are:

AD = AD (Common)

Hypotenuse AB = Hypotenuse AC (Given)

ā ADB = ā ADC (Each = 90Ā°)

(ii) āµ A ā A, B ā C, D ā D

ā“ āADB ā
āADC

(iii) Since, āADB ā
āADC

ā“ Corresponding parts are equal.

ā“ ā B = ā C

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