Exercise 4.2 - Chapter 4 Geometry Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 4.2
Question $1 .$
Given that $\triangle \mathrm{ABC}=\Delta \mathrm{DEF}$ (i) List all the corresponding congruent sides
(ii) List all the corresponding congruent angles.
Solution:
Given $\triangle \mathrm{ABC} \cong \mathrm{DEF}$.

(i) Corresponding congruent sides.
$\overline{A B}=\overline{D E} ; \overline{B C}=\overline{E F} ; \overline{A C}=\overline{D F}$
(ii) Corresponding congruent angles.
$\angle \mathrm{ABC}=\angle \mathrm{DEF} ; \angle \mathrm{BCA}=\angle \mathrm{EFD} ; \angle \mathrm{CAB}=\angle \mathrm{FDE}$
 

Question 2 .
If the given two triangles are congruent, then identify all the corresponding sides and also write the congruent angles.

Solution:
Given $\triangle \mathrm{PQR} \cong \Delta \mathrm{LNM}$
(i) (a) Corresponding sides
$\overline{P Q}=\overline{L N} ; \overline{P Q}=\overline{L M} ; \overline{R Q}=\overline{M N}$
(b) Corresponding angles
$\angle \mathrm{RPQ}=\angle \mathrm{NLM} ; \angle \mathrm{PQR}=\angle \mathrm{LNM} ; \angle \mathrm{PRQ}=\angle \mathrm{LMN}$
(ii) Given $\Delta \mathrm{PQR} \cong \Delta \mathrm{NML}$
(a) Corresponding angles
$\overline{Q R}=\overline{L M} ; \overline{R P}=\overline{L N} ; \overline{P Q}=\overline{M N}$
(b) Corresponding angles
$\angle \mathrm{PQP}=\angle \mathrm{NMN} ; \angle \mathrm{QRP}=\angle \mathrm{MLN} ; \angle \mathrm{RPQ}=\angle \mathrm{LNM}$
 

Question $3 .$
Find the unit digit of expanded form.

(i) $\angle \mathrm{A}$ and $\angle \mathrm{G}$
(ii) $\angle \mathrm{B}$ and $\angle \mathrm{E}$
(iii) $\angle \mathrm{B}$ and $\angle \mathrm{G}$
(iv) $\overline{A C}$ and $\overline{G F}$
(v) $\overline{B A}$ and $\overline{F G}$
(vi) $\overline{E F}$ and $\overline{B C}$
Solution:
Given $\triangle \mathrm{ABC} \cong \Delta \mathrm{EFG}$. Also from given triangles.
$\overline{A B}=\overline{F G} \overline{B C}=\overline{G F} \overline{A C}=\overline{E F}$
Also $\angle \mathrm{A}=\angle \mathrm{F} \angle \mathrm{B}=\angle \mathrm{G} \angle \mathrm{C}=\angle \mathrm{E}$
Answer:
(i) $\angle \mathrm{A}$ and $\angle \mathrm{G}$ are not corresponding angles.
(ii) $\angle \mathrm{B}$ and $\angle \mathrm{E}$ are not corresponding angles.
(iii) $\angle \mathrm{B}$ and $\angle \mathrm{G}$ are corresponding angles.
(iv) $\overline{A C}$ and $\overline{G F}$ are not corresponding sides.
(v) $\overline{B A}$ and $\overline{F G}$ are corresponding sides.
(vi) $\overline{E F}$ and $\overline{B C}$ are not corresponding sides.
 

Question $4 .$
State whether the two triangles are congruent or not. Justify your answer.

Solution:
(i) Let the given triangle be $\triangle \mathrm{ABC} . \overline{A D}$ divides $\triangle \mathrm{ABC}$ into two parts giving $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$. In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$

$\begin{aligned}
&\overline{A B}=\overline{A C} \text { (given) } \\
&\overline{B D}=\overline{A D} \text { (common side) } \\
&\angle \mathrm{BAD}=\angle \mathrm{CAD} \text { (included angles) } \\
&\therefore \text { By SAS criterion } \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} .
\end{aligned}$
(ii) Let the given triangles in the figure be $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DCB}$. In both the triangles

$\begin{aligned}
&\overline{B C}=\overline{B C}(\text { Common side }) \\
&\overline{A B}=\overline{D C} \\
&\overline{A C}=\overline{B D} \\
&\therefore \text { By SSS Criterion } \triangle \mathrm{ABC} \cong \Delta \mathrm{DCB}
\end{aligned}$
(iii) Let the given triangles be $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CDE}$.

Here $\overline{A C}=\overline{C E}$ (given)
$\angle B A C=\angle D E C$ (given)
$\angle \mathrm{ACB}=\angle \mathrm{DCE}$ (vertically opposite angles)
Two angles and the included side are equal.
Therefore by ASA criterion $\triangle \mathrm{ABC} \cong \triangle \mathrm{CDE}$.
(iv) Let the two triangles be $\triangle \mathrm{XYZ}$ and $\triangle \mathrm{XYW}$

Here $\angle \mathrm{W}=\angle \mathrm{Z}=90^{\circ}$
$\frac{\overline{X Y}}{X W}=\overline{X Y}$ (Common Hypothenure)
By RHS criterion AX
(v) Let the two triangles be $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$

In both the triangles $\overline{A C}=\overline{A C}$ (common sides)
$\overline{A D}=\overline{B C}$ (given)
$\overline{A B}=\overline{D C}$ (given)
By SSS criterion $\triangle \mathrm{ABC} \cong \triangle \mathrm{ADC}$.

Question $5 .$
To conclude the congruency of triangles, mark the required information in the following figures with reference to the given congruency criterion.

Solution:
(i) In the given triangles one angle is equal and a side is common and so equal.
To satisfy ASA
To satisfy ASA criterion one more angle should be equal such that the common side is the included side of both angles of a triangle.
The figure will be as follows.

(ii) In the two given triangles two sides of one triangle is equal to two sides of the other triangle.

To satisfy SSS criterion the third sides mut be equal.
(iii) The given triangles have one side in common. They are right angled tringles.

(iv) In the given triangles two angles of one triangle is equal to two angles of the other triangles?

(v) In both the triangles one of their sides are equal.

One of their angles are equal or they are vertically opposite angles.
To satisfy SAS criterion, one more side is to be equal such that the angle is the included of the equal sides.

Question $6 .$
For each pair of triangles state the criterion that can be used to determine the congruency?

Solution:
(i) Given two pair of sides are equal and one side is common to both the triangles.
$\therefore$ SSS congruency criterion is used.
(ii) One of the sides and one of the angles are equal.
$\therefore$ One more angle is vertically opposite angle and so it is also equal.
ASA criterion is used.
(iii) From the figure hypotenuse and one side are equal in both the triangles.
RHS congruency criterion is used. ( $\because$ Considering $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BAD}$ )
$\begin{aligned}
&\angle \mathrm{A}=\angle \mathrm{B}=90^{\circ} \\
&\mathrm{AD}=\mathrm{BC} \\
&\mathrm{AB}=\mathrm{AB} \text { (common) } \\
&\therefore \mathrm{AC}=\mathrm{BD} \text { (hypotenuse) }
\end{aligned}$
(iv) By ASA criterion both triangles are congruent.
(v) By ASA criterion both triangles are congruent. Since two angles in one triangle are equal to two corresponding angles of the other triangle. Again one side is common to both triangle and the side is the included side of the angles.
(vi) Two sides are equal. One angle is vertically opposite angles and one equal.
By SAS criterion both triangles are cogruent.

Question $7 .$
I. Construct a triangle $X Y Z$ with the given conditions.
(i) $X Y=6.4 \mathrm{~cm}, Z Y=7.7 \mathrm{~cm}$ and $X Z=5 \mathrm{~cm}$
Solution:

Rough Diagram

Construction:
Step 1: Draw a line. Marked $Y$ and $Z$ on the line such that $Y Z 7.7 \mathrm{~cm}$.
Step 2: With $Y$ as centre drawn an arc of radius $6.4 \mathrm{~cm}$ above the line $Y Z$.
Step 3: With Z as centre, drwan an arc or radius $5 \mathrm{~cm}$ to intersect arc drawn in steps. Marked the point of intersection as $\mathrm{X}$.
Step 3: Joined $Y X$ and $Z X$. Now $X Y Z$ is the required triangle.
(ii) An equilateral triangle of side $7.5 \mathrm{~cm}$
Solution:

Construction:
Step 1: Drawn a line. Marked $\mathrm{X}$ and $\mathrm{Y}$ on the line such that $\mathrm{XY}=7.5 \mathrm{~cm}$.
Step 2: With $\mathrm{X}$ as centre, drawn an arc of radius $7,5 \mathrm{~cm}$ above the line $\mathrm{XY}$.
Step 3: With $Y$ as centre, drawn an arc of radius $7.5 \mathrm{~cm}$ to intersect arc drawn in steps. Marked the point of intersection as $Z$.
Step 4: Joined $X Z$ and $Y Z$. Now $X Y Z$ in the required triangle.
(iii) An isosceles triangle with equal sides $4.6 \mathrm{~cm}$ and third side $6.5 \mathrm{~cm}$
Solution:

Construction: .
Step 1: Drawn a line. Marked $\mathrm{X}$ and $\mathrm{Y}$ on the line such that $\mathrm{XY}=6.5 \mathrm{~cm}$.
Step 2: With $X$ as centre, drawn an arc of radius $4.6 \mathrm{~cm}$ above the line $X Y$
Step 3: with $\mathrm{Y}$ as centre, drawn an arc of radius $4.6 \mathrm{~cm}$ to intersect arc drawn in steps. Marked the point of intersection as $Z$.
Step 4: Joined $X Z$ and $Y Z$. Now $X Y Z$ is the required triangle.
II. Construct a triangle $\mathrm{ABC}$ with given conditions.
(i) $\mathrm{AB}=7 \mathrm{~cm}, \mathrm{AC}=6.5 \mathrm{~cm}$ and $\angle \mathrm{A}=120^{\circ}$.
Solution:

Construction:
Step 1: Drawn a line. Marked $A$ and $B$ on the line such that $\mathrm{AB}=7 \mathrm{~cm}$.
Step 2: At A, drawn a ray AX making an angle of $120^{\circ}$ with AB.
Step 3: With A as centre, drawn an arc of radius $6.5 \mathrm{~cm}$ to cut the ray AX. Marked the point of intersection as $\mathrm{C}$.
Step 4: Joined BC.
$\mathrm{ABC}$ is the required triangle.
(ii) $\mathrm{BC}=8 \mathrm{~cm}, \mathrm{AC}=6 \mathrm{~cm}$ and $\angle \mathrm{C}=40^{\circ}$.
Solution:

Construction:
Step 1: Drawn a line. Marked B and $\mathrm{C}$ on the line such fhat $\mathrm{BC}=8 \mathrm{Cm}$.
Step 2: At C, drawn a ray CY making an angle of $40^{\circ}$ with $B C$.
Rough Diagram
Step 3: With C as centre, drawn an arc of radius $6 \mathrm{~cm}$ to cut the ray CY, marked the point of intersection as $\mathrm{A}$.
Step 4: Joined AB.
$\mathrm{AB}$ is the required triangle.
(iii) An isosceles obtuse triangle with equal sides $5 \mathrm{~cm}$
Solution:

Construction:
Step 1: Drawn a line. Marked $B$ and $C$ on the line such that $B C=5 \mathrm{~cm}$.
Step 2: At B drawn a ray BY making on obtuse angle $110^{\circ}$ with BC.
Step 3: With B as centre, drawn an arc of radius $5 \mathrm{~cm}$ to cut ray BY. Marked the point of intersection as $\mathrm{C}$.
Step 4: Joined BC. $\mathrm{ABC}$ is the required triangle.
III. Construct a triangle PQR with given conditions.
(i) $\angle \mathrm{P}=60^{\circ}, \angle \mathrm{R}=35^{\circ}$ and $\mathrm{PR}=7.8 \mathrm{~cm}$
Solution:

Construction:
Step 1: Drawn a line. Marked $P$ and $R$ on the line such that $P R=7.8 \mathrm{~cm}$.
Step 2: At P, drawn a ray PX making an angle of $60^{\circ}$ with PR.
Step 3: At R, drawn another ray RY making an angle of $35^{\circ}$ with PR. Mark the point of intersection of the rays $P X$ and $R Y$ as $Q$.
$\mathrm{PQR}$ is the required triangle.
(ii) $\angle \mathrm{P}=115^{\circ}, \angle \mathrm{Q}=40^{\circ}$ and $\mathrm{PQ}=6 \mathrm{~cm}$
Solution:

Construction:
Step 1: Drawn a line. Marked $P$ and $Q$ on the line such that $P Q=6 \mathrm{~cm}$.
Step 2: At P, drawn O ray PX making an angle of $115^{\circ}$ with $\mathrm{PQ}$.
Step 3: At Q, drawn another ray QY making an angle of $40^{\circ}$ with PQ. Marked the point of intersection of the rays $\mathrm{PX}$ and $\mathrm{Q} Y$ as $R$.
$\mathrm{PQR}$ is the required triangle.
(iii) $\angle \mathrm{Q}=90^{\circ}, \angle \mathrm{R}=42^{\circ}$ and $\mathrm{QR}=5.5 \mathrm{~cm}$
Solution:

Construction:
Step 1: Drawn a line. Marked $Q$ and $R$ on the line such that $Q R=5.5 \mathrm{~cm}$.
Step 2: At Q, drawn a ray $Q X$ making an angle of $90^{\circ}$ with $Q R$.
Step 3: At R, drawn another ray RY making an angle of $42^{\circ} \mathrm{QR}$. Marked the point of intersection of the rays $Q X$ and RY as $P$.
$\mathrm{PQR}$ is the required triangle.
 

Objective Type Questions
Question $8 .$
If two plans figures are congruent then they have
(i) same size
(ii) same shape
(iii) same angle
(iv) same shape and same size
Answer:
(iv) same shape and same size
 

Question $9 .$
Which of the following methods are used to check the congruence of plane figures?
(i) translation method
(ii) superposition method
(iii) substitution method
(iv) transposition method
Answer:
(ii) superposition method

Question $10 .$
Which of the following rule is not sufficient to verify the congruency of two triangles.
(i) $\mathrm{SSS}$ rule
(ii) SAS rule
(iii) SSA rule
(iv) ASA rule
Answer:
(iii) SSA rule
 

Question 11.
Two students drew a line segment each. What is the condition for them to be congruent?
(i) They should be drawn with a scale.
(ii) They should be drawn on the same sheet of paper.
(iii) They should have different lengths.
(iv) They should have the same length.
Answer:
(iv) They should have the same length.
 

Question $12 .$
In the given figure, $\mathrm{AD}=\mathrm{CD}$ and $\mathrm{AB}=\mathrm{CB}$. Identify the other three pairs that are equal.

(i) $\angle \mathrm{ADB}=\angle \mathrm{CDB}, \angle \mathrm{ABD}=\angle \mathrm{CBD}, \mathrm{BD}=\mathrm{BD}$
(ii) $\mathrm{AD}=\mathrm{AB}, \mathrm{DC}=\mathrm{CB}, \mathrm{BD}=\mathrm{BD}$
(iii) $\mathrm{AB}=\mathrm{CD}, \mathrm{AD}=\mathrm{BC}, \mathrm{BD}=\mathrm{BD}$
(iv) $\angle \mathrm{ADB}=\angle \mathrm{CDB}, \angle \mathrm{ABD}=\angle \mathrm{CBD}, \angle \mathrm{DAB}=\angle \mathrm{DBC}$
Answer:
(i) $\angle \mathrm{ADB}=\angle \mathrm{CDB}, \angle \mathrm{ABD}=\angle \mathrm{CBD}, \mathrm{BD}=\mathrm{BD}$

Question $13 .$
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}, \angle \mathrm{A}=50^{\circ}=\angle \mathrm{P}, \mathrm{PQ}=\mathrm{AB}$, and $\mathrm{PR}=\mathrm{AC}$. By which property $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$ are congruent?
(i) SSS property
(ii) SAS property
(iii) ASA property
(iv) RHS property
Answer:
(ii) SAS property