Additional Questions - Chapter 5 Geometry Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise $5.1$
Question $1 .$
Can two adjacent angles be supplementary?
Solution:
Yes, In the figure

$\angle \mathrm{AOB}$ and $\angle \mathrm{BOC}$ are adjacent angles.
Also $\angle \mathrm{AOB}+\angle \mathrm{BOC}=180^{\circ}$
$\therefore \angle \mathrm{AOB}$ and $\angle \mathrm{BOC}$ are supplementary

Question $2 .$
Can two obtuse angles form a linear pair?
Solution:
No, the sum of the measures of two obtuse angles is more than $180^{\circ}$.

Question $3 .$
Can two right angles form a linear pair?
Solution:
Yes, because the sum of two right angles is $180^{\circ}$ and form a linear pair.

Question $4 .$
Find $x, y$ and $z$ from the figure.

Solution:
$x=55^{\circ}$ vertically opposite angles
$y+55^{\circ}=180^{\circ}$
$\mathrm{y}=180^{\circ}-55^{\circ}$
$y=125^{\circ}$

Execise $5.2$
Question $1 .$
Can two lines intersect in more than one point?
Solution:
No, two lines cannot intersect in more than one point.
 

Question $2 .$
In the figure EF parallel to $\mathrm{GH}$
Solution:
$\angle \mathrm{EAB}=60^{\circ}$ and $\angle \mathrm{ACD}=105^{\circ}$
Determine (i) $\angle \mathrm{CAF}$ and
(ii) $\angle \mathrm{BAC}$

Solution:
(i) Since EF $\| \mathrm{GH}$ and $\mathrm{AC}$ is a transversal $\Rightarrow \angle \mathrm{CAF}+\angle \mathrm{ACH}=180^{\circ}$
$\Rightarrow \angle \mathrm{CAF}+105^{\circ}=180^{\circ}$.
$=75^{\circ}$
(ii) $\therefore \mathrm{EF} \| \mathrm{GH}$ and $\mathrm{AC}$ is transversal.
$\therefore \angle \mathrm{EAC}=\angle \mathrm{ACH}[\because$ Alternate interior angles $]$
$\begin{aligned}
&\therefore \angle \mathrm{EAC}=\angle \mathrm{ACH}[\because \text { Alternat } \\
&\Rightarrow \angle \mathrm{BAC}=105^{\circ} \\
&\Rightarrow \angle \mathrm{BAC}+\angle \mathrm{BAB}=105^{\circ} \\
&\Rightarrow \angle \mathrm{BAC}+60^{\circ}=105^{\circ}
\end{aligned}$
$\begin{aligned}
&\Rightarrow \angle \mathrm{BAC}=105^{\circ}-60^{\circ} \\
&=45^{\circ} \\
&\therefore \angle \mathrm{CAF}=75^{\circ} \text { and } \angle \mathrm{BAC}=45^{\circ}
\end{aligned}$

Question $3 .$
In the given figure, the arms of two angles are parallel. If $\angle \mathrm{ABC}=70^{\circ}$, then find

(i) $\angle \mathrm{DGC}$
(ii) $\angle \mathrm{DEF}$
Solution:
We have $\mathrm{AB} \| \mathrm{ED}$ and $\mathrm{BC} \| \mathrm{EF}$
(i) $\mathrm{BC}$ is transversal
$\angle \mathrm{DGC}=\angle \mathrm{ABC}$ [corresponding angles]
But $\angle \mathrm{ABC}=70^{\circ}$
$\angle \mathrm{DGC}=70^{\circ}$
(ii) $\mathrm{ED}$ is a transversal to $\mathrm{BC} \| \mathrm{EF}$
$\therefore \angle \mathrm{DEF}=\angle \mathrm{DGC}$ [corresponding]
$\angle \mathrm{DGC}=70^{\circ}$
$\angle \mathrm{DEF}=70^{\circ}$

Exercise $5.6$
Question $1 .$
In the following figure, show that $\mathrm{CD} \| \mathrm{EF}$

Solution:

$\angle B A D=\angle C D A$ But they form a pair of alternate angles
$\begin{aligned}
&\angle \mathrm{BAD}=\angle \mathrm{BAE}+\angle \mathrm{EAD} \\
&=40^{\circ}+30^{\circ}=70^{\circ} . \\
&\text { and } \angle \mathrm{CDA}=70^{\circ}
\end{aligned}$
$\begin{aligned}
&\Rightarrow \triangle \mathrm{B} \| \mathrm{CD} \\
&\text { Also } \angle \mathrm{BAE}+\angle \mathrm{AEF}=40^{\circ}+140^{\circ}=180^{\circ}
\end{aligned}$
$\text { Also } \angle B A E+\angle A E F=40^{\circ}+140^{\circ}=180^{\circ}$
But they form a pair of interior opposite angles.
$\Rightarrow \mathrm{AB} \| \mathrm{EF}$
From (1) and (2), we get
$\mathrm{AB}\|\mathrm{CD}\| \mathrm{EF}$
$\Rightarrow \mathrm{CD} \| \mathrm{EF}$
 

Question 2.
In the adjoining figure, the lines $\overleftrightarrow{A B}$ and $\overleftrightarrow{C D}$ intersect at ' $\mathrm{O}$ '. If $\angle \mathrm{COB}=50^{\circ}$, find the measures of the other three angles.
Solution:
$\begin{aligned}
&\angle \mathrm{COB}=50^{\circ} \\
&\angle \mathrm{AOD}=50^{\circ} \text { (vertically opposite angles) }
\end{aligned}$
Now $\angle \mathrm{AOC}$ and $\angle \mathrm{COB}$ form a linear pair,
Thus $\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ}$
$\begin{aligned}
&\Rightarrow \angle \mathrm{AOC}+50^{\circ}=180^{\circ} \\
&\angle \mathrm{AOC}=180^{\circ}-50^{\circ}=130^{\circ}
\end{aligned}$
Also $\angle \mathrm{AOC}$ and $\angle \mathrm{BOD}$ are vertically opposite angles.
$\therefore \angle \mathrm{BOD}=\angle \mathrm{AOC}=130^{\circ}$
Thus the three angles are
$\begin{aligned}
&\angle \mathrm{AOD}=50^{\circ} \\
&\angle \mathrm{AOC}=130^{\circ} \\
&\angle \mathrm{BOD}=130^{\circ}
\end{aligned}$