NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2
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All students need to have a fundamental understanding of Mathematics. In addition to helping them perform well on their examinations, students will gain longterm advantages by understanding some basic mathematical concepts. The difficulty of Mathematics, however, prevents many students from being interested in studying it. Both academic and practical applications can be found for the subject of Mathematics. Given that it takes a lot of mental effort for students to understand, it may be challenging to ignite their interest in this subject. Mathematical concepts are usually difficult for students to understand.
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 rightangled 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.
Extramarks’ internal subject matter experts, carefully and in accordance with all CBSE regulations, solved the problems and questions from the exercise. Any student in Class 7 who has comprehensive knowledge of all the concepts from the Mathematics textbook and is quite knowledgeable with regard to all the exercises provided in it can easily secure the best scores on the final examination. Students may quickly comprehend the types of questions that may be given in the examination from this chapter and learn the chapter weightage in terms of overall mark distribution with the aid of these NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2. Therefore, students can adequately study for the final examination and score higher marks.
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NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2
Students are exposed to a wide range of concepts owing to their mathematical education. Humans first learnt to count items that are visible to the naked eye by the process of counting, which is the basis or beginning of Mathematics. The two primary subdisciplines of Mathematics are Applied Mathematics and Pure Mathematics, which include Numerical Systems, Geometry, Matrices, Algebra, Combinatorics, Topology, and Calculus (Engineering, Chemistry, Physics, numerical analysis, etc). Mathematical formulas are frequently based on different concepts. One can learn them by constantly using their formulas to respond to queries. Some problems can be quickly solved using mathematical methods.
One of the most crucial mathematical themes is the Congruence of Triangles. It states that all three corresponding sides and all three corresponding angles are equal in size; two triangles are said to be congruent. They are in alignment with one another when moved. Students find it difficult to understand all the concepts theoretically. Therefore, the Extramarks website provides live classes by subject matter experts and also provides interactive videos about every topic so that students can understand the concept indepth. Extramarks also adds graphics and animations to its videos so that the studies become interesting for the students and, thereby, they get encouraged to study the subject.
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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
$\begin{array}{l}\left(\mathrm{b}\right)\mathrm{Given}:\mathrm{ZX}=\mathrm{RP}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{RQ}=\mathrm{ZY}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PRQ}=\mathrm{XZY}\\ \mathrm{So},\mathrm{\Delta PQR}\cong \mathrm{\Delta XYZ}\end{array}$
$\begin{array}{l}\left(\mathrm{c}\right)\mathrm{Given}:\angle \mathrm{MLN}=\angle \mathrm{FGH}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{NML}=\angle \mathrm{GFH}\\ \mathrm{ML}=\mathrm{FG}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\mathrm{\Delta LMN}\cong \mathrm{\Delta GFH}\end{array}$
$\begin{array}{l}\left(\mathrm{d}\right)\mathrm{Given}:\mathrm{EB}=\mathrm{DB}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{AE}=\mathrm{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}=\angle \mathrm{C}=90\xb0\\ \mathrm{So},\mathrm{\Delta ABE}\cong \mathrm{\Delta CDB}\end{array}$
Ans.
$\begin{array}{l}\left(\text{a}\right)\text{}\overline{)\text{SSS}},\text{as the sides of}\mathrm{\Delta}\text{ABC are equal to the sides of}\mathrm{\Delta}\text{DEF}.\\ \left(\text{b}\right)\text{}\overline{)\text{SAS}},\text{as two sides and the angle included between these}\\ \text{sides of}\mathrm{\Delta}\text{PQR are equal to two sides and the angle}\\ \text{included between these sides of}\mathrm{\Delta}\text{XYZ}.\\ \left(\text{c}\right)\text{}\overline{)\text{ASA}},\text{as two angles and the side included between these}\\ \text{angles of}\mathrm{\Delta}\text{LMN are equal to two angles and the side}\\ \text{included between these angles of}\mathrm{\Delta}\text{GFH}.\\ \left(\text{d}\right)\text{}\overline{)\text{RHS}},\text{as in the given two rightangled triangles},\text{one side}\\ \text{and the hypotenuse are respectively equal.}\end{array}$
Q.2
$\begin{array}{l}\text{You want to show that \Delta ART@\Delta PEN,}\\ \left(\text{a}\right)\text{If you have to use SSS criterion, then you need to show}\\ \left(\text{i}\right)\text{AR =}\left(\text{ii}\right)\text{RT =}\left(\text{iii}\right)\text{AT =}\\ \left(\text{b}\right)\text{If it is given that \u2220T =}\angle \text{N and you are to use SAS criterion,}\\ \text{you need to have}\\ \left(\text{i}\right)\text{RT = and}\left(\text{ii}\right)\text{PN =}\\ \left(\text{c}\right)\text{If it is given that AT=PN and you are to use ASA criterion,}\\ \text{you need to have}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left(\text{i}\right)\text{?}\left(\text{ii}\right)\text{\hspace{0.33em}?}\end{array}$
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
$\mathrm{You}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}\angle \mathrm{AMP}\cong \angle \mathrm{AMQ}.\text{\hspace{0.17em}}\mathrm{In}\mathrm{the}\mathrm{following}$
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
$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ABC},\text{\hspace{0.17em}}\angle \text{A}=\text{3}0\xb0,\angle \text{B}=\text{4}0\xb0\text{and}\angle \text{C}=\text{11}0\xb0\\ \text{In}\mathrm{\Delta}\text{PQR},\angle \text{P}=\text{3}0\xb0\text{},\angle \text{Q}=\text{4}0\xb0\text{and}\angle \text{R}=\text{11}0\xb0\\ \text{A student says that}\mathrm{\Delta}\text{ABC}\cong \mathrm{\Delta}\text{PQR by AAA congruence}\\ \text{criterion}.\text{Is he justified}?\text{Why or why not?}\end{array}$
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
$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{figure},\mathrm{the}\mathrm{two}\mathrm{triangles}\mathrm{are}\mathrm{congruent}.\\ \mathrm{The}\mathrm{corresponding}\mathrm{parts}\mathrm{are}\mathrm{marked}.\\ \end{array}$
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.
Here
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{}(\text{2})\\ \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}0\xba.\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{}(\text{ASA criterion})\end{array}