Exercise 5.6 - Chapter 5 Geometry Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.6$
Miscellaneous Practice Problems
Question $1 .$
Find the value of $x$ if $\angle A O B$ is a right angle.

Solution:
Given that $\angle \mathrm{AOB}=90^{\circ}$
$3 x+2 x=90^{\circ}$
$\begin{aligned}
5 x &=90^{\circ} \\
x &=\frac{90^{\circ}}{5} \\
&=18^{\circ} \\
x &=18^{\circ}
\end{aligned}$

Question $2 .$
In the given figure, find the value of $x$.
Solution:
Since $\angle B O C$ and $\angle A O C$ are linear pair, their sum $=180^{\circ}$

$2 x+23+3 x-48=180^{\circ}$
$5 x-25=180^{\circ}$
$5 x-25+25=180^{\circ}+25$
$x=\frac{205^{\circ}}{5}$
$x=41^{\circ}$

Question $3 .$
Find the value of $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$.

Solution:
$\angle \mathrm{DOB}$ and $\angle \mathrm{BOC}$ are linear pair
$\begin{aligned}
&\therefore \angle \mathrm{DOB}+\angle \mathrm{BOC}=180^{\circ} \\
&\mathrm{x}+3 \mathrm{x}+40=180^{\circ} \\
&4 \mathrm{x}+40=180^{\circ} \\
&4 \mathrm{x}+40-40=180^{\circ}-40^{\circ} \\
&4 \mathrm{x}=140^{\circ} \\
&x=\frac{140^{\circ}}{4} \\
&x=35^{\circ}
\end{aligned}$
Also $\angle B O D$ and $\angle A O C$ are vertically opposite angles.
$\begin{aligned}
&\therefore \angle \mathrm{BOD}=\angle \mathrm{AOC} \\
&\mathrm{x}=\mathrm{z}+10 \\
&35^{\circ}=\mathrm{z}+10 \\
&\mathrm{z}+10-10=35-10 \\
&\mathrm{z}=25^{\circ}
\end{aligned}$
Again $\angle \mathrm{AOD}$ and $\angle \mathrm{AOC}$ are linear pair.
$\therefore \angle \mathrm{AOD}+\angle \mathrm{AOC}=180^{\circ}$

$\begin{aligned}
&y+30+z+10=180^{\circ} \\
&y+30+25+10=180^{\circ} \\
&y+65=180^{\circ} \\
&y+65-65=180^{\circ}-65 \\
&y=115^{\circ} \\
&\therefore x=35^{\circ}, \\
&y=115^{\circ}, \\
&z=25^{\circ}
\end{aligned}$

Question $4 .$
Two angles are in the ratio $11: 25$. If they are linear pair, find the angles. Solution:
Given two angles are in the ratio $11: 25$.
Let the angles be $11 \mathrm{x}$ and $25 \mathrm{x}$.
They are also linear pair
$\begin{aligned}
\therefore 11 \mathrm{x} &+25 \mathrm{x}=180^{\circ} . \\
36 x &=180^{\circ} \\
x &=\frac{180^{\circ}}{36} \\
x &=5^{\circ}
\end{aligned}$
$\therefore$ The angles $11 \mathrm{x}=11 \times 5^{\circ}=55^{\circ}$ and $25 \mathrm{x}=25 \times 5=125^{\circ} .$
$\therefore$ The angles are $55^{\circ}$ and $125^{\circ}$.

Question $5 .$
Using the figure, answer the following questions and justify your answer.
(i) Is $\angle 1$ adjacent to $\angle 2$ ?
(ii) Is $\angle \mathrm{AOB}$ adjacent to $\angle \mathrm{BOE}$ ?
(iii) Does $\angle \mathrm{BOC}$ and $\angle \mathrm{BOD}$ form a linear pair?
(iv) Are the angles $\angle \mathrm{COD}$ and $\angle \mathrm{BOD}$ supplementary.
(v) Is $\angle 3$ vertically opposite to $\angle 1$ ?

Solution:
(i) Yes, $\angle 1$ is adjacent to $\angle 2$.
Because they both have the common vertex ' $\mathrm{O}$ ' and the common arm $\mathrm{OA}$. Also their interiors do not overlap.
(ii) No, $\angle \mathrm{AOB}$ and $\angle \mathrm{BOB}$ are not adjacent angles because they have overlapping interiors.
(iii) No, $\angle \mathrm{BOC}$ and $\angle \mathrm{BOD}$ does not form a linear pair.
Because $\angle B O C$ itself a straight angle, so the sum of $\angle B O C$ and $\angle B O D$ exceed $180^{\circ}$.
(iv) Yes, the angles $\angle \mathrm{COD}$ and $\angle \mathrm{BOD}$ are supplementary $\angle \mathrm{COD}+\angle \mathrm{BOD}=180^{\circ}$, $[\because$ linear pair of angles]
$\therefore \angle \mathrm{COD}$ and $\angle \mathrm{BOD}$ are supplementary.
(v) No. $\angle 3$ and $\angle 1$ are not formed by intersecting lines. So they are not vertically opposite angles.
 

Question 6 .
In the figure $\mathrm{POQ}, \mathrm{ROS}$ and $\mathrm{TOU}$ are straight lines. Find the $\mathrm{x}^{\circ}$.

Solution:
Given TOU is a straight line.
$\therefore$ The sum of all angles formed at a point on a straight line is $180^{\circ}$
$\angle \mathrm{TOR}+\angle \mathrm{ROP}+\angle \mathrm{POV}+\angle \mathrm{VOU}=180^{\circ}$.
$36^{\circ}+47^{\circ}+45^{\circ}+x^{\circ}=180^{\circ}$.
$128^{\circ}+x^{\circ}=180^{\circ}$
$128^{\circ}+x^{\circ}-128^{\circ}=180^{\circ}-128^{\circ}$
$x=52^{\circ}$

Question $7 .$
In the figure $\mathrm{AB}$ is parallel to $\mathrm{DC}$. Find the value of $\angle 1$ and $\angle 2$. Justify your answer.

Solution:
Given AB \| DC
AB and CD are parallel lines Taking CE as transversal we have.
$\angle 1=30^{\circ},[\because$ alternate interior angles $]$
Taking DE as transversal
$\angle 2=80^{\circ}$. [ $\because$ alternate interior angles]
$\angle 1=30^{\circ}$ and $\angle 2=80^{\circ}$
Justification:
CDE is a triangle
$\angle \mathrm{AEC}+\angle \mathrm{CED}+\angle \mathrm{DEB}=\angle \mathrm{DEC}+\angle \mathrm{ECD}+\angle \mathrm{CDE}$
$\begin{array}{r}
\angle \mathrm{AEC}+\angle \mathrm{CED}+\angle \mathrm{DEB}= \\
\angle 1+80^{\circ}=30^{\circ}+\angle 2
\end{array}$
$\therefore 30^{\circ}+80^{\circ}=30^{\circ}+80^{\circ}$

Question $8 .$
In the figure $\mathrm{AB}$ is parallel to $\mathrm{CD}$. Find $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$.

Solution:
Given AB || CD
$\therefore Z=42$ ( $\because$ Alternate interior angles)
Also $\mathrm{y}=42^{\circ}$ [vertically opposite angles]
Also $\mathrm{x}^{\circ}+63^{\circ}+\mathrm{z}^{\circ}=180^{\circ}$
$x^{\circ}+63^{\circ}+42^{\circ}=180^{\circ}$
$x^{\circ}+105^{\circ}=180^{\circ}$
$x^{\circ}+105^{\circ}-105^{\circ}=180^{\circ}-105^{\circ}$
$x^{\circ}=75^{\circ}$
$\therefore \mathrm{x}=75^{\circ}$
$y=42^{\circ}$
$z=42^{\circ}$

Question $9 .$
Draw two parallel lines and a transversal. Mark two alternate interior angles $\mathrm{G}$ and $\mathrm{H}$. If they are supplementary, what is the measure of each angle?
Solution:
$\mathrm{l}$ and $\mathrm{m}$ are parallel lines and $\mathrm{n}$ is the transversal.
$\angle \mathrm{G}$ and $\angle \mathrm{H}$ are alternate interior angles.
$\angle \mathrm{G}=\angle \mathrm{H} \ldots . .$ (1)
Given $\angle \mathrm{G}$ and $\angle \mathrm{G}$ are Suplementary

ie $\quad \angle \mathrm{G}+\angle \mathrm{H}=180^{\circ}$
$\angle \mathrm{H}+\angle \mathrm{H}=180^{\circ}($ from $(1))$
$2 \angle \mathrm{H}=180^{\circ}$
ie $\begin{aligned} \angle \mathrm{G}+\angle \mathrm{H} &=180^{\circ} \\ \angle \mathrm{H}+\angle \mathrm{H} &=180^{\circ}(\text { from }(1)) \\ 2 \angle \mathrm{H} &=180^{\circ} \\ \angle \mathrm{H} &=\frac{180^{\circ}}{2} \\ \angle \mathrm{H} &=90^{\circ} \\ \angle \mathrm{G}=\angle \mathrm{H} &=90^{\circ} \end{aligned}$
$\angle \mathrm{H}=\frac{180^{\circ}}{2}$
$\begin{aligned} \angle \mathrm{H} &=90^{\circ} \\=\angle \mathrm{H} &=90^{\circ} \end{aligned}$

Question $10 .$
A plumber must install pipe 2 parallel to pipe 1 . He knows that $\angle 1$ is 53 . What is the measure of $\angle 2 ?$
Solution:
Given $\angle 1=53^{\circ}$

Clearly $\angle 1$ and $\angle 2$ are interior angles on the same side of the transversal and so they are supplementary.

$\begin{aligned}
&\angle 1+\angle 2=180^{\circ} \\
&53^{\circ}+\angle 2=180^{\circ} \\
&53^{\circ}+\angle 2-53^{\circ}=180^{\circ}-53^{\circ} \\
&\angle 2=127^{\circ}
\end{aligned}$

Challenge Problems
Question $11 .$
Find the value of $y$.
Solution:
Cleary POQ is a straight line"
Sum of all angles formed at a point on a straight line is $180^{\circ}$
$\therefore \angle \mathrm{POT}+\angle \mathrm{TOS}+\angle \mathrm{SOR}+\angle \mathrm{ROQ}=180^{\circ}$
$60^{\circ}+\left(3 y-20^{\circ}\right)+y^{\circ}+\left(y+10^{\circ}\right)=180^{\circ}$
$60^{\circ}+3 y-20^{\circ}+y^{\circ}+y^{\circ}+10^{\circ}=180^{\circ}$
$5 y+50^{\circ}=180^{\circ}$
$5 y+50^{\circ}-50^{\circ}=180^{\circ}-50$
$5 \mathrm{y}=130^{\circ}$
$\begin{aligned}
&y=\frac{130^{\circ}}{5} \\
&y=26^{\circ}
\end{aligned}$

Question 12 .
Find the value of $\mathrm{z}$.

Solution:
The sum of angles at a point is $360^{\circ}$.
$\begin{aligned}
&\therefore \angle \mathrm{QOP}+\angle \mathrm{PON}+\angle \mathrm{NOM}+\angle \mathrm{MOQ}=360^{\circ} \\
&3 \mathrm{z}+(2 \mathrm{z}-5)+(\mathrm{z}+10)+(4 \mathrm{z}-25)=360^{\circ} \\
&3 \mathrm{z}+2 \mathrm{z}+\mathrm{z}+4 \mathrm{z}-5+10-25=360^{\circ} \\
&10 \mathrm{z}-20^{\circ}=360^{\circ} \\
&10 \mathrm{z}-20^{\circ}+20=360^{\circ}+2 \\
&10 \mathrm{z}=380^{\circ} \\
&z=\frac{380^{\circ}}{10^{\circ}} \\
&\mathrm{z}=38^{\circ}
\end{aligned}$

Question $13 .$
Find the value of $\mathrm{x}$ and $\mathrm{y}$ if $\mathrm{RS}$ is parallel to $\mathrm{PQ}$.

Solution:
Given RS || PQ
Considering the transversal RU, we have $\mathrm{y}=25^{\circ}$ (corresponding angles)
Considering ST as transversal

Question $14 .$
Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, if one angle exceeds the twice of the other angle by $48^{\circ}$. Find the angles.
Solution:
Let the two parallel lines be $\mathrm{m}$ and $\mathrm{n}$ and 1 be the transversal
Let one of the interior angles on the same side of the transversal be $x^{\circ}$
Then the other will be $2 x+48$.
We know that they are supplementary.

$\begin{aligned}
\therefore x+\left(2 x+48^{\circ}\right) &=180^{\circ} \\
x+2 x+48^{\circ} &=180^{\circ} \\
3 x+48^{\circ} &=180^{\circ} \\
3 x+48^{\circ}-48^{\circ} &=180^{\circ}-48^{\circ} \\
3 x &=132^{\circ} \\
x &=\frac{132^{\circ}}{3} \\
x &=44^{\circ}
\end{aligned}$
$\begin{aligned} \therefore x+\left(2 x+48^{\circ}\right) &=180^{\circ} \\ x+2 x+48^{\circ} &=180^{\circ} \\ 3 x+48^{\circ} &=180^{\circ} \\ 3 x+48^{\circ}-48^{\circ} &=180^{\circ}-48^{\circ} \\ 3 x &=132^{\circ} \\ x &=\frac{132^{\circ}}{3} \\ x &=44^{\circ} \\ \therefore \quad \text { One angle is } x &=44^{\circ} \\ \therefore \quad \text { The angles are } 44^{\circ} \text { and } & 136^{\circ} \end{aligned}$

Question 15 .
In the figure, the lines $\mathrm{GH}$ and $\mathrm{IJ}$ are parallel. If $\angle 1=108^{\circ}$ and $\angle 2=123^{\circ}$, find the value of $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$.
Solution:
Given GH || IZ
$\begin{aligned}
&\angle 1=108^{\circ} \\
&\angle 2=123^{\circ} \\
&\angle 1+\angle \mathrm{KGH}=180 \text { [linear pair] } \\
&108^{\circ}+\angle \mathrm{KGH}=180^{\circ}
\end{aligned}$

$\begin{aligned}
&108^{\circ}+\angle \mathrm{KGH}-108^{\circ}=180^{\circ}-108^{\circ} \\
&\angle \mathrm{KGH}=72^{\circ} \\
&\angle \mathrm{KGH}=\mathrm{x}^{\circ} \text { (corresponding angles if } \mathrm{KG} \text { is a transversal) } \\
&\therefore \mathrm{x}^{\circ}=72^{\circ} \\
&\text { Similarly } \\
&\angle 2+\angle \mathrm{GHK}=180^{\circ}(\because \text { linear pair }) \\
&123^{\circ}+\angle \mathrm{GHK}=180^{\circ} \\
&123^{\circ}+\angle \mathrm{GHK}-123^{\circ}=180^{\circ}-123^{\circ} \\
&\angle \mathrm{GHK}=57^{\circ} \\
&\text { Again } \angle \mathrm{GHK}=\mathrm{y}^{\circ} \text { (corresponding angles if } \mathrm{KH} \text { is a transversal) } \\
&\mathrm{y}=57^{\circ} \\
&\mathrm{x}^{\circ}+\mathrm{y}^{\circ}+\mathrm{z}^{\circ}=180^{\circ}\left(\text { sum of three angles of a triangle is } 180^{\circ}\right) \\
&72^{\circ}+57^{\circ}+\mathrm{z}^{\circ}=180^{\circ} \\
&129^{\circ}+\mathrm{z}^{\circ}=180^{\circ} \\
&129^{\circ}+\mathrm{z}^{\circ}-129^{\circ}=180^{\circ}-129^{\circ} \\
&\mathrm{z}=51^{\circ} \\
&\mathrm{x}=72^{\circ}, \\
&\mathrm{y}=57^{\circ}, \\
&\mathrm{z}=51^{\circ}
\end{aligned}$

Question $16 .$
In the parking lot shown, the lines that mark the width of each space are parallel. If $\angle 1=(2 x-3 y)^{\circ} ; \angle 2=(x+39)^{\circ}$ find $x^{\circ}$ and $y^{\circ} .$

Solution:
From the picture
$\angle 2+65^{\circ}=180^{\circ}$ [Sum of interior angles on the same side of a transversal]
$x+39^{\circ}+65^{\circ}=180^{\circ}$
$x+104^{\circ}=180^{\circ}$
$x+104^{\circ}-104^{\circ}=180^{\circ}-104^{\circ}$
$x=76^{\circ}$
Also from the picture
$\angle 1=65^{\circ}$ [alternate exterior angles]
$\begin{aligned}
&2 x-3 y=65^{\circ} \\
&2(76)-3 y=65^{\circ} \\
&152^{\circ}-3 y=65^{\circ} \\
&152^{\circ}-3 y-152^{\circ}=65-152^{\circ} \\
&-3 y=-87 \\
&y=\frac{-87}{-3} \\
&y=29^{\circ}=x=76^{\circ} ; y=29^{\circ}
\end{aligned}$

Question $17 .$
Draw two parallel lines and a transversal. Mark two corresponding angles $\mathrm{A}$ and $\mathrm{B}$. If $\angle \mathrm{A}=4 \mathrm{x}$, and $\angle B=3 x+7$, find the value of $x$. Explain.
Solution:
Let $\mathrm{m}$ and $\mathrm{n}$ are two parallel lines and $\mathrm{l}$ is the transversal.
$\mathrm{A}$ and $\mathrm{B}$ are corresponding angles.
We know that corresponding angles are equals,
$\begin{aligned}
\therefore 4 x &=3 x+7 \\
4 x-3 x &=3x+7-3x \\
x &=7^{\circ} \\
\therefore \angle \mathrm{A} &=4 \times 7=28^{\circ} \text { and } \angle \mathrm{B}=3(7)+7 \\
&=21+7=28
\end{aligned}$

Question $18 .$
In the figure $\mathrm{AB}$ in parallel to $\mathrm{CD}$. Find $\mathrm{x}^{\circ}, \mathrm{y}^{\circ}$ and $\mathrm{z}^{\circ}$.
Solution:

Given $\mathrm{AB} \| \mathrm{CD}$
Then $\mathrm{AD}$ and $\mathrm{BC}$ are transversals.
$\mathrm{x}=48^{\circ}$, alternate interior angles; $\Lambda \mathrm{D}$ is transversal $\mathrm{y}=60^{\circ}$, alternate interior angles; $\mathrm{BC}$ is transversal
$\angle \mathrm{AEB}+48^{\circ}+\mathrm{y}^{\circ}=180^{\circ},\left(\right.$ sum of angles of a triangle is $\left.180^{\circ}\right)$
$\begin{aligned}
\angle \mathrm{AEB}+48^{\circ}+60^{\circ} &=180^{\circ} \\
\angle \mathrm{AEB}+108^{\circ} &=180^{\circ} \\
\angle \mathrm{AEB}+108^{\circ}-108^{\circ} &=180^{\circ}-108^{\circ} \\
\angle \mathrm{AEB} &=72^{\circ} \\
\therefore \angle \mathrm{AEB}+z^{\circ} &=180^{\circ} \text { [linear pair] } \\
72^{\circ}+z^{\circ} &=180^{\circ} \\
72^{\circ}+z^{\circ}-72^{\circ} &=180^{\circ}-72^{\circ} \\
z^{\circ} &=108^{\circ} \\
x=48^{\circ} ; y=60^{\circ} ; \mathrm{z} &=108^{\circ}
\end{aligned}$

Question $19 .$
Two parallel lines are cut by transversal. If one angle of a pair of corresponding angles can be represented by $42^{\circ}$ less than three times the other. Find the corresponding angles.
Solution:
We know that the corresponding angles are equal.
Let one of the corresponding angles be $x$.
Then the other will be $3 \mathrm{x}-42^{\circ}$.

$\begin{aligned}
\therefore 3 x-42^{\circ} &=x^{\circ} \\
3 x-42^{\circ}+42^{\circ} &=x^{\circ}+42^{\circ} \\
3 x^{\circ} &=x+42^{\circ} \\
3 x^{\circ}-x^{\circ} &=x+42-x \\
2 x &=42^{\circ} \\
x &=\frac{42}{2} \\
x &=21^{\circ}
\end{aligned}$
$\therefore$ The corresponding angles are $21^{\circ}$ each

Question $20 .$
In the given figure, $\angle 8=107^{\circ}$, what is these sum of the angles $\angle 2$ and $\angle 4$.
Solution:
Given $\angle 8=107^{\circ}$
$\angle 2=107^{\circ}$
$[\because \angle 8$ and $\angle 2$ are alternate exterior angles, $\because \angle 8=\angle 2]$

$\lfloor 4=\lfloor 2$ (vertically opposite angles)
$\begin{aligned}
\therefore \quad \angle 4 &=107^{\circ} \\
\therefore \quad \angle 2+\angle 4 &=107^{\circ}+107^{\circ} \\
&=214^{\circ} \\
\quad \angle 2+\angle 4 &=214^{\circ}
\end{aligned}$