Exercise 4.2 - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 4.2$
Question 1.
The angles of a quadrilateral are in the ratio $2: 4: 5: 7$. Find all the angles.
Solution:
In a quadrilateral the angles add upto $360^{\circ}$.
Let's call the angles $2 x, 4 x, 5 x, 7 x$
$\begin{aligned}
&2 x+4 x+5 x+7 x=360^{\circ} \\
&18 x=360^{\circ} \\
&x=\frac{360^{\circ}}{18}=20^{\circ} \\
&A=2 x=2 \times 20^{\circ}=40^{\circ} \\
&B=4 x=4 \times 20^{\circ}=80^{\circ} \\
&C=5 x=5 \times 20^{\circ}=100^{\circ} \\
&D=7 x=7 \times 20^{\circ}=140^{\circ}
\end{aligned}$
 

Question $2 .$
In a quadrilateral $\mathrm{ABCD}, \angle \mathrm{A}=12^{\circ}$ and $\angle \mathrm{C}$ is the supplementary of $\angle \mathrm{A}$. The other two angles are $2 x-10$ and $x+4$. Find the value of $x$ and the measure of all the angles.
Solution:
$\angle \mathrm{A}=72^{\circ}$
$\angle \mathrm{C}=180^{\circ}-72^{\circ}(\because$ Supplementary at $\angle \mathrm{A})=108^{\circ}$
The other two angles are $2 x-10$ and $x+4$.
$2 \mathrm{x}-10+\mathrm{x}+4+108^{\circ}+12^{\circ}=360^{\circ}$

$\begin{aligned}
&3 \mathrm{x}+174^{\circ}=360^{\circ} \\
&3 \mathrm{x}=360^{\circ}-174^{\circ} \\
&x=\frac{186}{3}=62^{\circ} \\
&\therefore \angle \mathrm{A}=72^{\circ} \\
&\angle \mathrm{B}=2 \mathrm{x}-10=2(62)-10=124-10=114^{\circ} \\
&\angle \mathrm{C}=108^{\circ} \\
&\angle \mathrm{D}=\mathrm{x}+4=62+4=66^{\circ}
\end{aligned}$

Question $3 .$
$\mathrm{ABCD}$ is a rectangle whose diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at $\mathrm{O}$. If $\angle \mathrm{OAB}=46^{\circ}$, find $\angle \mathrm{OBC}$.
Solution:
$\angle \mathrm{ABC}=90^{\circ}$

$\begin{aligned}
&\angle \mathrm{OAB}+\angle \mathrm{OBC}=90^{\circ} \\
&46^{\circ}+\angle \mathrm{OAB}=90^{\circ} \\
&\angle \mathrm{ORC}=90^{\circ}-46^{\circ}=44^{\circ}
\end{aligned}$

Question $4 .$
The lengths of the diagonals of a Rhombus are $12 \mathrm{~cm}$ and $16 \mathrm{~cm}$. Find the side of
the rhombus.
Solution:
Let $\mathrm{ABCD}$ be a rhombus with $\mathrm{AC}$ and $\mathrm{BD}$ as its diagonals. We know that the diagonals of a rhombus bisect each other at right angles. Let $\mathrm{O}$ be the intersecting point of both the diagonals
Let $\mathrm{AC}=16 \mathrm{~cm}$ and $\mathrm{BD}=12 \mathrm{~cm}$

$\begin{aligned}
&\mathrm{OA}=\frac{\mathrm{AC}}{2}=\frac{16}{2}=8 \mathrm{~cm} \\
&\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{12}{2}=6 \mathrm{~cm}
\end{aligned}$
use Pythagoras theorem, we have
$\begin{aligned}
&\mathrm{AB}^{2}=\mathrm{OA}^{2}+\mathrm{OB}^{2} \\
&\mathrm{AB}^{2}=100 \\
&\therefore \mathrm{AB}=10 \mathrm{~cm}
\end{aligned}$

Question $5 .$
Show that the bisectors of angles of a parallelogram form a rectangle.
Solution:
Given ABCD is a parallelogram. Draw the angular bisectors AP, BP, CR and DR of the angles $\angle \mathrm{A}, \angle \mathrm{B}, \angle \mathrm{C}$ and $\angle \mathrm{D}$ respectively.
Now to prove: $\mathrm{PQRS}$ is a rectangle.
Proof: A rectangle is a parallelogram with one angle $90^{\circ}$.
First we will prove PQRS is a parallelogram.
Now $A B \| C D$ and $A D$ is transversal. $[\therefore$ Interior angles on the same side of transversal are supplementary]
[Opposite sides of a parallelogram are parallel]

$\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ}$
Multiplying by $\frac{1}{2}$
$\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{D}=\frac{1}{2} \times 180^{\circ}$
$\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{D}=90^{\circ}$
$\angle \mathrm{DAS}+\angle \mathrm{ADS}=90^{\circ}$
Now from $\triangle \mathrm{ADS}$
$\begin{aligned}
\angle \mathrm{DAS}+\angle \mathrm{ADS}+\angle \mathrm{DSA} &=180^{\circ} \\
90^{\circ}+\angle \mathrm{DSA} &=180^{\circ} \\
\angle \mathrm{DSA} &=180^{\circ}-90^{\circ} \\
\angle \mathrm{DSA} &=90^{\circ}
\end{aligned}$

Also lines AP and DR intersects
So $\angle \mathrm{PSR}=\angle \mathrm{DS} \mathrm{A}$
$\therefore \angle \mathrm{PSR}=90^{\circ}[\because$ Vertically opposite angles $]$
Similarly we can prove that $\angle \mathrm{SPQ}=90^{\circ}, \angle \mathrm{PQR}=90^{\circ}$ and $\angle \mathrm{SRQ}=90^{\circ}$
$\therefore \angle \mathrm{PSR}=\angle \mathrm{PQR}$ and $\angle \mathrm{SPQ}=\angle \mathrm{SRQ}$
$\therefore$ Both pair of opposite angles of $\mathrm{PQRS}$ is a parallelogram.
Also $\angle \mathrm{PSR}=\angle \mathrm{PQR}=\angle \mathrm{SPQ}=\angle \mathrm{SRQ}=90^{\circ}$
$\therefore \mathrm{PQRS}$ is a parallelogram with one angle $90^{\circ}$.
$\therefore$ PQRS is a rectangle. Hence proved.
 

Question $6 .$
If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that the area of the triangle is equal to half of the area of parallelogram.
Solution:
Given: $\triangle \mathrm{ABE}$ and parallelogram $\mathrm{ABCD}$ have the same base and are between the same parallel lines (i.e) $l_{1} \| l_{2}$.

Perpendicular distance between $\mathrm{l}_{1}$ and $\mathrm{l}_{2}=\mathrm{P}$ (say).
Prove that: area of $(\triangle \mathrm{ABE})=\frac{1}{2} \times$ area of the parallelogram $A B C D$
Proof: Area of $\triangle \mathrm{ABE}=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times \mathrm{AB} \times\left(\right.$ Perpendicular distance between $\mathrm{l}_{1}$
$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{P} \ldots . .(1)$
Area of parallelogram $A B C D=$ base $\times$ height.
$\therefore$ Area of parallelogram $\mathrm{ABCD}=\mathrm{AB} \times \mathrm{P} \ldots(2)$
From (1) and (2),
Area of $\triangle \mathrm{ABE}=\frac{1}{2} \times$ Area of parallelogram $\mathrm{ABCD}$.
Hence proved.

Question 7.
Iron rods $a, b, c, d, e$, and $f$ are making a design in a bridge as shown in the figure. If a $\|b, c\| d, e \| f$, find the marked angles between
(i) $b$ and $c$
(ii) $\mathrm{d}$ and $\mathrm{e}$
(iii) $d$ and $f$
(iv) $\mathrm{c}$ and $f$.
Solution:
Since $1, \mathrm{~m}$ are two parallel lines and $\mathrm{PQ}, \mathrm{RS}, \mathrm{TU}, \mathrm{VW}$ are transversal.
Then $\angle 1=\angle Q O R$ [vertically opposite angles]

$\angle 1=30^{\circ}\left[\therefore \angle \mathrm{QOR}=30^{\circ}\right]$
Also, PQ and TU are parallel and $\mathrm{m}$ and 1 are transversal.

$\begin{aligned}
\text { Therefore, } \angle 2+\angle \mathrm{QPT} &=180^{\circ} \\
\angle 2 &=180-75^{\circ} \\
\angle 2 &=105^{\circ} \\
\text { Also, } \quad \angle 2+\angle 3 &=180^{\circ} \\
\Rightarrow & \\
105^{\circ}+\angle 3 &=180^{\circ} \\
\angle 3 &=180^{\circ}-105^{\circ} \\
&=75^{\circ}
\end{aligned}$
Also $\angle 3+\angle 4=180^{\circ}$
$\begin{aligned}
&\Rightarrow 75^{\circ}+\angle 4=180^{\circ} \\
&\angle 4=180^{\circ}-75^{\circ}=105^{\circ}
\end{aligned}$

Hence,
(i) $30^{\circ}$
(ii) $105^{\circ}$
(iii) $75^{\circ}$
(iv) $105^{\circ}$

Question 8 .
In the given figure $\angle \mathrm{A}=64^{\circ}, \angle \mathrm{ABC}=58^{\circ}$. If $\mathrm{BO}$ and $\mathrm{CO}$ are the bisectors of $\angle \mathrm{ABC}$ and $\angle \mathrm{ACB}$ respectively of $\triangle \mathrm{ABC}$, find $\mathrm{x}^{\circ}$ and $\mathrm{y}^{\circ}$.
Solution:
$\begin{aligned}
&\text { In } \triangle \mathrm{ABC}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
&\Rightarrow 64^{\circ}+58^{\circ}+\angle \mathrm{C}=180^{\circ} \\
&\Rightarrow 122^{\circ}+\angle \mathrm{C}=180^{\circ} \\
&\Rightarrow \angle \mathrm{C}=180^{\circ}-122^{\circ}=58^{\circ}
\end{aligned}$

Also $\mathrm{BO}$ and $\mathrm{CO}$ are the bisectors of $\angle \mathrm{ABC}$ and $\angle \mathrm{ACB}$ respectively
$\begin{aligned}
\therefore \quad \angle \mathrm{OBC}=\frac{58^{\circ}}{2} &=29^{\circ} \\
\text { and } \angle \mathrm{OCB}=\frac{58^{\circ}}{2} &=29^{\circ} \Rightarrow y=29^{\circ} \\
\text { In } \triangle \mathrm{OBC}=x^{\circ}+y^{\circ}+\mathrm{OBC} &=180^{\circ} \\
x^{\circ}+29^{\circ}+29^{\circ} &=180^{\circ} \\
x^{\circ}+58^{\circ} &=180^{\circ} \\
x^{\circ} &=180^{\circ}-58^{\circ}=122^{\circ} \\
\therefore x &=122^{\circ}, y=29^{\circ}
\end{aligned}$

Question $9 .$
In the given Fig. if $\mathrm{AB}=2, \mathrm{BC}=6, \mathrm{AE}=6, \mathrm{BF}=8, \mathrm{CE}=7$, and $\mathrm{CF}=7$, compute the ratio of the area of quadrilateral ABDE to the area of ACDF.

Given: $\mathrm{AB}=2$,
$\begin{aligned}
&\mathrm{BC}=6 \\
&\mathrm{AE}=6 \\
&\mathrm{BF}=8 \\
&\mathrm{CE}=7 \text { and } \\
&\mathrm{CF}=7
\end{aligned}$
Consider $\triangle \mathrm{AEC}$ and $\triangle \mathrm{BCF}$,
In $\triangle \mathrm{AEC}$,
$\mathrm{AC}=8$,
$\mathrm{AE}=6$,
$\mathrm{CE}=7$
In $\triangle \mathrm{BCF}$,
$\mathrm{BF}=8$,
$\mathrm{BC}=6$,
$\mathrm{CF}=7$
$\therefore \triangle \mathrm{AEC} \cong \triangle \mathrm{BCF}$
$\therefore$ Area of $\triangle \mathrm{AEC}=$ Area of $\triangle \mathrm{BCF}$
Subtract.area of $\triangle B D C$ both sides, we get
Area of $\triangle \mathrm{AEC}$ - Area of $\triangle \mathrm{BDC}=$ Area of $\triangle \mathrm{BCF}$ - Area of $\triangle \mathrm{BDC}$
$\Rightarrow$ Area of quadrilateral $\mathrm{ABDE}=$ Area of $\triangle \mathrm{CDF}$
$\therefore$ The required ratio is $1: 1$
 

Question $10 .$
In the figure, $\mathrm{ABCD}$ is a rectangle and $\mathrm{EFGH}$ is a parallelogram. Using the measurements given in the figure, what is the length $\mathrm{d}$ of the segment that is perpendicular to $\overline{\mathbf{H E}}$ and $\overline{\mathbf{F G}}$ ?

Solution:
Area of $A B C D=$ length $\times$ breadth.
$=\overline{\mathrm{DC}} \times \overline{\mathrm{BC}}=10 \times 8=80 .$
Area of $\triangle \mathrm{AEH}=$ Area of $\triangle \mathrm{CGF}$ [since they are congruent by RHS rule]
Similarly, Area of $\triangle \mathrm{BEF}=$ Area of $\triangle \mathrm{DGH}$
$\therefore$ Area of parallelogram $=\mathrm{EFGH}=$ Area of rectangle $\mathrm{ABCD}-2($ area of $\triangle \mathrm{AEH})-$
2(area of $\triangle B E F$ )
$\begin{aligned}
&=80-2\left(\frac{1}{2} \times 4 \times 3\right)-2 \cdot\left(\frac{1}{2} \times 6 \times 5\right) \\
&=80-12-30=80-42=38 \\
\therefore \text { Area of parallelogram } \mathrm{EFGH}=38 \\
\mathrm{GF} \times d &=38 \\
5 \times d &=38\left[\mathrm{In} \Delta \mathrm{GCF}, \mathrm{GF}^{2}=\mathrm{FC}^{2}+\mathrm{GC}^{2}\right.\\
\Rightarrow \quad \quad \mathrm{GF}^{2}=4^{2}+3^{2} &=16+9=25 \Rightarrow \mathrm{GF}=5] \\
\Rightarrow \quad d &=\frac{38}{5}=7.6
\end{aligned}$

Question $11 .$
In parallelogram $\mathrm{ABCD}$ of the accompanying diagram, line $\mathrm{DP}$ is drawn bisecting $\mathrm{BC}$ at $\mathrm{N}$ and meeting $\mathrm{AB}$ (extended) at $\mathrm{P}$. From vertex $\mathrm{C}$, line $\mathrm{CQ}$ is drawn bisecting side $\mathrm{AD}$ at $\mathrm{M}$ and meeting $\mathrm{AB}$ (extended) at $\mathrm{Q}$. Lines $\mathrm{DP}$ and $\mathrm{CQ}$ meet at O. Show that the area of triangle $\mathrm{QPO}$ is $\frac{9}{8}$ of the area of the parallelogram $A B C D$.

Solution:

Draw $\mathrm{OX} \perp \mathrm{PQ}$
$\begin{aligned}
\text { Area of } \triangle \mathrm{DMN} &=\frac{1}{2} \times \text { base } \times \text { height } \\
&=\frac{1}{2} \mathrm{MN} \times \mathrm{OY} \\
\text { Area of } \mathrm{DMNC} &=\text { Base } \times \text { Height } \\
&=\mathrm{MN} \times 2 \mathrm{OY} \\
&=4(\text { Area of } \Delta \mathrm{OMN}) \\
\text { Arẹa of } \mathrm{ABCD} &=2 \text { Area of } \mathrm{DMNC}
\end{aligned}$

$=4($ Area of $\Delta \mathrm{OMN}) \quad[\because$ From (1) $]$
Arća of ABCD $=2$ Aréa of DMNC
$=2(4$ Area of $\Delta \mathrm{OMN})=8$ Area of $\Delta \mathrm{OMN}$
$\frac{1}{8}$ Aea of $\mathrm{ABCD}=$ Area of $\mathrm{OMN}$
We have $\triangle \mathrm{QMA} \cong \triangle \mathrm{DCM}$
$\triangle \mathrm{BNP} \cong \triangle \mathrm{DCM}$

Area of $\Delta \mathrm{QDP}=$ Area of $\Delta \mathrm{QMA}+$ Area of $\Delta \mathrm{MNO}+$ Area of $\mathrm{MNBS}+$ Area of $\triangle \mathrm{MAB}$
$=$ Area of $\triangle \mathrm{DCM}+$ Area of $\triangle \mathrm{MNO}+$ Area of $\mathrm{MNBA}+$ Area of $\triangle \mathrm{NDC}$ $=2 \mathrm{Area}$ of $\Delta \mathrm{OMN}+$ Area of $\Delta \mathrm{MNO}+4$ Area of $\Delta \mathrm{OMN}+2$ Area of $\Delta \mathrm{OMN}$
$\begin{aligned}
&=9 \text { Area of } \Delta \mathrm{OMN}+\frac{1}{9} \text { Area of } \Delta \mathrm{QOP}+\text { Area } \\
\frac{1}{9} \text { Area of } \Delta \mathrm{QOP} &=\frac{1}{8} \text { Area of } \mathrm{ABCD} \\
\therefore \text { Area of } \Delta \mathrm{QOP} &=\text { Area of } \mathrm{ABCD} . \text { Hence proved. }
\end{aligned}$