Exercise 4.1 - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $4.1$ : Chapter 4 - Geometry - 9th Maths Guide Samacheer Kalvi Solutions 
Question 1.

In the figure, $\mathrm{AB}$ is parallel to $\mathrm{CD}$, find $\mathrm{x}$

Solution:
(i) From the figure
$\angle 1=140^{\circ}$ ( $\therefore$ corresponding angles are equal)
$\begin{aligned}
&\angle 2=40^{\circ}\left(\therefore \angle 1+\angle 2=180^{\circ}\right) \\
&\angle 3=30^{\circ}\left(\because \angle 3+150=180^{\circ}\right) \\
&\angle 4=110^{\circ}\left(\because \angle 2+\angle 3+\angle 4=180^{\circ}\right)
\end{aligned}$
$\therefore \angle \mathrm{x}=70^{\circ}\left(\because \angle 4+\angle \mathrm{x}=180^{\circ}\right)$

(ii) From t
$\angle 1=48^{\circ}$
$\angle 3=108^{\circ}\left(\angle 1+24^{\circ}+\angle 3=180^{\circ}\right)$
$\angle 4=108^{\circ}$ (If two lines are intersect, then the vertically the opposite angles are equal)
$\angle 5=72^{\circ}\left(\because \angle 3+\angle 5=180^{\circ}\right)$
$\begin{aligned}
&\therefore \angle 3+\angle 4+\angle 5=108^{\circ}+108^{\circ}+72^{\circ} \\
&x=288^{\circ}
\end{aligned}$

(iii) From the figure
$\angle \mathrm{D}=53^{\circ}(\because \angle \mathrm{B}$ and $\angle \mathrm{D}$ are alternate interior angles)
Sum of the three angles of a triangle is $180^{\circ}$
$\angle \mathrm{x}^{\circ}=180^{\circ}-\left(38^{\circ}+53^{\circ}\right)$
$=180^{\circ}-91^{\circ}=89^{\circ}$

The angles of a triangle are in the ratio $1: 2: 3$, find the measure of each angle of the triangle.
Solution:
Let the angles be $x, 2 x$ and $3 x$ respectively.

Sum of the three angles of a triangle $=180^{\circ}$
$\begin{aligned}
\therefore \quad x+2 x+3 x &=180^{\circ} \\
6 x=180^{\circ} \Rightarrow x &=\frac{180^{\circ}}{6} \\
\therefore \quad x &=30^{\circ} \\
2 x=2 \times 30 &=60^{\circ} \\
3 x=3 \times 30 &=90^{\circ}
\end{aligned}$
The 3 angles of the triangle are $30^{\circ}, 60^{\circ}, 90^{\circ}$.
 

Question $3 .$
Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say 'how'; if they are not congruent say 'why' and also say if a small modification would make them congruent:

Solution:
(i) Consider $\triangle \mathrm{PQR}$ and $\triangle \mathrm{ABC}$
Given, $R Q=B C$
$\mathrm{PQ}=\mathrm{AB}$
$\triangle \mathrm{ABC}$ is not congruent to $\triangle \mathrm{PQR}$
If $\mathrm{PR} \mathrm{AC}$, then $\triangle \mathrm{ABC} \cong \triangle \mathrm{PQR}$

(ii) Consider $\triangle \mathrm{ABD}$ and $\triangle \mathrm{BCD}$ for the triangles to be congruent. Given, $A B=D C$
$\mathrm{AD}=\mathrm{BC}$ and $\mathrm{AB}$ is common side.
$\therefore$ By SSS rule $\triangle \mathrm{ABD} \cong \triangle \mathrm{BCD}$.

(iii) Consider $\triangle \mathrm{PXY}$ and $\triangle \mathrm{PXz}$,
Given, $X Y-X Z$
$\mathrm{PY}=\mathrm{PZ}$ and $\mathrm{PX}$ is common
$\therefore$ By SSS rule $\triangle \mathrm{PXY} \cong \triangle \mathrm{PXZ}$.

(iv) Consider $\triangle \mathrm{OAB}$ and $\triangle \mathrm{ODC}$,
Given, $\mathrm{OA}=\mathrm{OC}$

$\angle \mathrm{ABO}=\angle \mathrm{ODC}$ and $\angle \mathrm{AOB}=\angle \mathrm{DOC}$ (vertically opposite angles)
$\therefore$ By AAS rule, $\mathrm{AOAB}=\mathrm{AODC}$.
(v) Consider $\triangle \mathrm{AOB}$ and $\triangle \mathrm{DOC}$,
Given, $A O=O C$
$\mathrm{OB}=\mathrm{OD}$

and $\angle \mathrm{AOB}=\angle \mathrm{DOC}$ [vertically opposite angles]
$\therefore$ By $\mathrm{SAS}$ rule, $\triangle \mathrm{AOB}=\triangle \mathrm{DOC}$.
(vi) Consider $\triangle \mathrm{AMB}$ and $\triangle \mathrm{AMC}$,
Given, $\mathrm{AB}=\mathrm{AC}$
$\angle \mathrm{AMB}=\angle \mathrm{AMC}=90^{\circ}$
$\therefore \mathrm{AM}$ is common.
$\therefore$ By RHS rule
$\triangle \mathrm{AMB} \cong \triangle \mathrm{AMC}$.

Question $4 .$
$\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles in which $\mathrm{AB}=\mathrm{DF}, \angle \mathrm{ACB}=70^{\circ}, \angle \mathrm{ABC}=60^{\circ} ; \angle \mathrm{DEF}=$ $70^{\circ}$ and $\angle \mathrm{EDF}=60^{\circ}$. Prove that the triangles are congruent.
Solution:

$\begin{aligned}
\text { In } \triangle \mathrm{ABC}, \angle \mathrm{ACB} &=70^{\circ}, \angle \mathrm{ABC}=60^{\circ} \\
\therefore \quad \angle \mathrm{BAC} &=180^{\circ}-\left(70^{\circ}+60^{\circ}\right) \\
&=180^{\circ}-130^{\circ}=50^{\circ} \\
\text { In } \Delta \mathrm{DEF}, \angle \mathrm{DEF} &=70^{\circ}, \angle \mathrm{EDF}=60^{\circ} \\
\therefore \angle \mathrm{DFE} &=180^{\circ}-\left(70^{\circ}+60^{\circ}\right) \\
&=180^{\circ}-130^{\circ}=50^{\circ} \\
&=\angle \mathrm{F} \\
\angle \mathrm{A} &=\mathrm{DF} \\
\mathrm{AB} &=\angle \mathrm{D} \\
\text { and } \angle \mathrm{B} &=\Delta \mathrm{FDE}
\end{aligned}$

Question $5 .$
Find all the three angles of the $\triangle \mathrm{ABC}$

Solution:

$\begin{aligned}
&\text { Exterior angle }=\text { Sum of the two opposite interior angles. } \\
&\begin{aligned}
4 x-15 &=2 x-5+x+35 \\
4 x &=3 x+30+15 \\
4 x-3 x &=45^{\circ} \\
&=45^{\circ} \\
\therefore \quad \angle \mathrm{A} &=x+35 \\
&=45^{\circ}+35^{\circ}=80^{\circ} \\
\angle \mathrm{B} &=2 x-5 \\
&=2\left(45^{\circ}\right)-5=90^{\circ}-5^{\circ}=85^{\circ} \\
\angle \mathrm{C} &=4 x-15=4(45)-15^{\circ} \\
&=180^{\circ}-15^{\circ}=165^{\circ}
\end{aligned}
\end{aligned}$