# NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2

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The solutions to the chapter titled Congruence of Triangles are provided in a simple PDF format on the Extramarks website. The Ch 7 Maths Class 10 Ex 7.2 covers the themes of right-angled triangle congruence and the criteria for triangle congruence. It is advised that Class 7 students complete the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 based on Congruence of Triangles in order to solidify the principles and be ready to answer questions that are frequently asked in the examination. When it comes to examination preparation, using the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 is thought to be the best choice for students. There are numerous exercises in this chapter. On the Extramarks website, in PDF format, the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 are offered. Students can study these solutions straight from the website or mobile application, or they can download them as needed.

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## NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2

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### Access NCERT Solution for Class 7 Mathematics Chapter 7- Congruence of Triangles

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The Class 7 Maths Chapter 7 Exercise 7.2 Solution explains that a triangle has three vertices, three sides, and three angles. Triangles are categorised as equilateral, isosceles, and scalene based on similarities in the side measurements. In this instance, a comparison is made between the same triangle’s sides and angles. In order to compare two distinct triangles, a different set of guidelines is used. Congruent figures are two figures that are alike. These figures are exact replicas of one another. When an object’s counterpart is put over another congruent object, the two appear to be the same figure. Congruent triangles are the same as identical triangles in that they have the same side and angle measurements. Students who face difficulty in understanding the concept can refer to the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 provided by the Extramarks website.

### NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

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Q.1

$\begin{array}{l}\mathrm{Which}\mathrm{congruence}\mathrm{criterion}\mathrm{do}\mathrm{you}\mathrm{use}\mathrm{in}\mathrm{the}\mathrm{following}?\\ \left(\mathrm{a}\right)\mathrm{Given}:\mathrm{AC}=\mathrm{DF}\\ \mathrm{AB}=\mathrm{DE}\\ \mathrm{BC}=\mathrm{EF}\\ \end{array}$

S
o,ΔABCΔDEF $\begin{array}{l}\left(\mathrm{b}\right)\mathrm{Given}:\mathrm{ZX}=\mathrm{RP}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{RQ}=\mathrm{ZY}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PRQ}=\mathrm{XZY}\\ \mathrm{So},\mathrm{\Delta PQR}\cong \mathrm{\Delta XYZ}\end{array}$ $\begin{array}{l}\left(\mathrm{c}\right)\mathrm{Given}:\angle \mathrm{MLN}=\angle \mathrm{FGH}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{NML}=\angle \mathrm{GFH}\\ \mathrm{ML}=\mathrm{FG}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\mathrm{\Delta LMN}\cong \mathrm{\Delta GFH}\end{array}$ $\begin{array}{l}\left(\mathrm{d}\right)\mathrm{Given}:\mathrm{EB}=\mathrm{DB}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{AE}=\mathrm{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}=\angle \mathrm{C}=90°\\ \mathrm{So},\mathrm{\Delta ABE}\cong \mathrm{\Delta CDB}\end{array}$ Ans.

$\begin{array}{l}\left(\text{a}\right)\text{}\overline{)\text{SSS}},\text{as the sides of}\mathrm{\Delta }\text{ABC are equal to the sides of}\mathrm{\Delta }\text{DEF}.\\ \left(\text{b}\right)\text{}\overline{)\text{SAS}},\text{as two sides and the angle included between these}\\ \text{sides of}\mathrm{\Delta }\text{PQR are equal to two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{XYZ}.\\ \left(\text{c}\right)\text{}\overline{)\text{ASA}},\text{as two angles and the side included between these}\\ \text{angles of}\mathrm{\Delta }\text{LMN are equal to two angles and the side}\\ \text{included between these angles of}\mathrm{\Delta }\text{GFH}.\\ \left(\text{d}\right)\text{}\overline{)\text{RHS}},\text{as in the given two right-angled triangles},\text{one side}\\ \text{and the hypotenuse are respectively equal.}\end{array}$

Q.2

$\begin{array}{l}\text{You want to show that ΔART@ΔPEN,}\\ \left(\text{a}\right)\text{If you have to use SSS criterion, then you need to show}\\ \left(\text{i}\right)\text{AR =}\left(\text{ii}\right)\text{RT =}\left(\text{iii}\right)\text{AT =}\\ \left(\text{b}\right)\text{If it is given that ∠T =}\angle \text{N and you are to use SAS criterion,}\\ \text{you need to have}\\ \left(\text{i}\right)\text{RT = and}\left(\text{ii}\right)\text{PN =}\\ \left(\text{c}\right)\text{If it is given that AT=PN and you are to use ASA criterion,}\\ \text{you need to have}\\ \text{ }\left(\text{i}\right)\text{?}\left(\text{ii}\right)\text{ ?}\end{array}$ Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\text{AR}=\text{PE}\left(\text{ii}\right)\text{RT}=\text{EN}\left(\text{iii}\right)\text{AT}=\text{PN}\\ \left(\text{b}\right)\\ \left(\text{i}\right)\text{RT}=\text{EN}\left(\text{ii}\right)\text{PN}=\text{AT}\\ \left(\text{c}\right)\\ \left(\text{i}\right)\text{}\angle \text{ATR}=\angle \text{PNE}\left(\text{ii}\right)\text{}\angle \text{RAT}=\angle \text{EPN}\end{array}$

Q.3

$\mathrm{You}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}\angle \mathrm{AMP}\cong \angle \mathrm{AMQ}.\text{\hspace{0.17em}}\mathrm{In}\mathrm{the}\mathrm{following}$

proof, supply the missing reasons.

 Steps Reasons (i) PM = QM (i)… $\left(ii\right)\angle PMA=\angle QMA$ (ii)… (iii) AM = AM (iii)… $\left(iv\right)\Delta AMP\cong \Delta AMQ$ (iv)… Ans.

 Steps Reasons (i) PM = QM (i)…Given $\left(ii\right)\angle PMA=\angle QMA$ (ii)…Given (iii) AM = AM (iii)…Common $\left(\mathrm{iv}\right)\mathrm{\Delta AMP}\cong \mathrm{\Delta AMQ}$ $\begin{array}{l}\left(\text{iv}\right)\text{SAS},\text{as the two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{AMP}\\ \text{are equal to two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta }\text{AMQ}\end{array}$

Q.4

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC},\text{\hspace{0.17em}}\angle \text{A}=\text{3}0°,\angle \text{B}=\text{4}0°\text{and}\angle \text{C}=\text{11}0°\\ \text{In}\mathrm{\Delta }\text{PQR},\angle \text{P}=\text{3}0°\text{},\angle \text{Q}=\text{4}0°\text{and}\angle \text{R}=\text{11}0°\\ \text{A student says that}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{PQR by AAA congruence}\\ \text{criterion}.\text{Is he justified}?\text{Why or why not?}\end{array}$

Ans.

$\begin{array}{l}\text{No}.\text{This property represents that these triangles have their}\\ \text{respective angles of equal measure}.\text{However},\text{this gives no}\\ \text{information about their sides}.\text{The sides of these triangles have}\\ \text{a ratio somewhat different than 1}:\text{1}.\\ \text{Therefore},\text{AAA property does not prove the two}\\ \text{triangles are congruent.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{figure},\mathrm{the}\mathrm{two}\mathrm{triangles}\mathrm{are}\mathrm{congruent}.\\ \mathrm{The}\mathrm{corresponding}\mathrm{parts}\mathrm{are}\mathrm{marked}.\\ \end{array}$

We can write ΔRAT? Ans.

$\begin{array}{l}\text{From the figure, we observe that:}\\ \angle \text{RAT}=\text{}\angle \text{WON}\\ \angle \text{ART}=\text{}\angle \text{OWN}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AR}=\text{OW}\\ \text{Therefore},\text{}\mathrm{\Delta }\text{RAT}\cong \mathrm{\Delta }\text{WON},\text{by ASA criterion.}\end{array}$

Q.6

$\mathrm{Complete}\mathrm{the}\mathrm{congruence}\mathrm{statement}:$ Ans.

$\begin{array}{l}\text{Given that},\\ \text{BC}=\text{BT}\\ \text{TA}=\text{CA}\\ \text{BA is common}.\\ \text{Therefore},\text{}\mathrm{\Delta }\text{BCA}\cong \mathrm{\Delta }\text{BTA}\\ \text{Similarly},\\ \text{PQ}=\text{RS}\\ \text{TQ}=\text{QS}\\ \text{PT}=\text{RQ}\\ \text{Therefore},\text{}\mathrm{\Delta }\text{QRS}\cong \mathrm{\Delta }\text{TPQ.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{squared}\mathrm{sheet},\mathrm{draw}\mathrm{two}\mathrm{triangles}\mathrm{of}\mathrm{equal areas}\\ \mathrm{such}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{congruent}.\\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{not}\mathrm{congruent}.\\ \left(\mathrm{iii}\right)\mathrm{What}\mathrm{can}\mathrm{you}\mathrm{say}\mathrm{about}\mathrm{the}\mathrm{ir perimeters}?\end{array}$

Ans.

(i) Here
,ΔABC and ΔPQR have the same area and are congruent to each other also. Also, the perimeter of both the triangles
will be the same
.
(ii) $\begin{array}{l}\text{Here},\text{the two triangles have the same height and base}.\\ \text{Thus},\text{their areas are equal}.\\ \text{However},\text{these triangles are not congruent to each other}.\\ \text{Also},\text{the perimeter of both the triangles will not be the same.}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{rough}\mathrm{sketch}\mathrm{of}\mathrm{two}\mathrm{triangles}\mathrm{such}\mathrm{that}\mathrm{they}\mathrm{have}\\ \mathrm{five}\mathrm{pairs}\mathrm{of}\mathrm{congruent}\mathrm{parts}\mathrm{but}\mathrm{still}\mathrm{the}\mathrm{triangles}\mathrm{are}\mathrm{not}\\ \mathrm{congruent}.\end{array}$

Ans.

$\begin{array}{l}\text{A pair of triangles with threee equal angles, and two equal}\\ \text{sides are non congruent.}\end{array}$

Below is the example: Q.9

$\begin{array}{l}\mathrm{If}\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta PQR}\mathrm{are}\mathrm{to}\mathrm{be}\mathrm{congruent},\mathrm{name}\mathrm{one}\mathrm{additional}\\ \mathrm{pair}\mathrm{of}\mathrm{corresponding}\mathrm{parts}.\mathrm{What}\mathrm{criterion}\mathrm{did}\mathrm{you}\mathrm{use}?\end{array}$ Ans.

$\begin{array}{l}\text{Here,}\\ \text{BC}=\text{QR}\\ \text{So,}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{PQR}\left(\text{ASA criterion}\right)\end{array}$

Q.10

$\begin{array}{l}\mathrm{Explain},\mathrm{why}\\ \mathrm{\Delta ABC}\cong \mathrm{\Delta FED}\end{array}$ Ans.

$\begin{array}{l}\text{Given that},\\ \angle \text{ABC}=\text{}\angle \text{FED}\dots \dots \dots \dots \left(\text{1}\right)\\ \angle \text{BAC}=\text{}\angle \text{EFD}\dots \dots \dots \dots \text{}\left(\text{2}\right)\\ \text{The two angles of}\Delta \text{ABC are equal to the two respective}\\ \text{angles of}\Delta \text{FED}.\text{Also},\text{the sum of all interior angles of a}\\ \text{triangle is 18}0º.\text{Therefore},\text{third angle of both triangles will}\\ \text{also be equal in measure}.\\ \angle \text{BCA}=\text{}\angle \text{EDF}\dots \dots \dots \dots \left(\text{3}\right)\\ \text{Also},\text{given that},\\ \text{BC}=\text{ED}\dots \dots \dots \dots \dots \dots \left(\text{4}\right)\\ \text{By using equation}\left(\text{1}\right),\text{}\left(\text{3}\right),\text{and}\left(\text{4}\right),\text{we obtain}\\ \overline{)\Delta \text{ABC}\cong \text{}\Delta \text{FED}}\text{}\left(\text{ASA criterion}\right)\end{array}$