Exercise 4.4 - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 4.4$
Question $1 .$
Find the value of $x$ in the given figure.

Solution:
In the cyclic quadrilateral $\mathrm{ABCD}$
$\begin{aligned}
&\angle \mathrm{ABC}=180^{\circ}-120^{\circ}=60^{\circ} \\
&\angle \mathrm{BCA}=90^{\circ} \\
&\therefore \mathrm{x}=\angle \mathrm{BAC}=180^{\circ}-\left(90^{\circ}+60^{\circ}\right)=30^{\circ}
\end{aligned}$

Question $2 .$
In the given figure, $\mathrm{AC}$ is the diameter of the circle with centre $\mathrm{O}$. If $\angle \mathrm{ADE}=30^{\circ} ; \angle \mathrm{DAC}=35^{\circ}$ and $\angle \mathrm{CAB}$ $=40^{\circ}$.

Find
(i) $\angle \mathrm{ACD}$
(ii) $\angle \mathrm{ACB}$
(iii) $\angle \mathrm{DAE}$
Solution:
(i) $\angle \mathrm{ACD}=180^{\circ}-\left(90^{\circ}+35^{\circ}\right)=180^{\circ}-125^{\circ}=55^{\circ}$
(ii) $\angle \mathrm{ACB}=180^{\circ}-\left(90^{\circ}+40^{\circ}\right)=180^{\circ}-130^{\circ}=50^{\circ}$
$\begin{aligned}
&\text { (iii) } \angle \mathrm{ADC}=90^{\circ} \\
&\angle \mathrm{CAE}=180^{\circ}-120^{\circ}=60^{\circ} \\
&\therefore \angle \mathrm{DAE}=60^{\circ}-35^{\circ}=25^{\circ}
\end{aligned}$

Question $3 .$
Find all the angles of the given cyclic quadrilateral $\mathrm{ABCD}$ 

Solution:
In the cyclic quadrilateral $\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$
$\begin{aligned}
2 y+A+4 y-4=180^{\circ} \\
6 y &=180^{\circ} \\
y &=\frac{180^{\circ}}{6}=30^{\circ} \\
\angle \mathrm{B}+\angle \mathrm{D} &=6 x-4+7 x+2 \\
13 x-2 &=180^{\circ} \\
13 x &=180+2=182^{\circ} \\
x &=\frac{182}{13}=14^{\circ} \\
\therefore \angle \mathrm{A} &=2(30)+4^{\circ}=64^{\circ} \\
\angle \mathrm{B} &=6(14)-4^{\circ}=84-4^{\circ}=80^{\circ} \\
\angle \mathrm{C} &=4(30)-4=120-4=116^{\circ} \\
\angle \mathrm{D} &=7(14)+2=98+2=100^{\circ}
\end{aligned}$

Question $4 .$
In the given figure, $\mathrm{ABCD}$ is a cyclic quadrilateral where diagonals intersects at $\mathrm{P}$ such that $\angle \mathrm{DBC}=40^{\circ}$ and $\angle B A C=60^{\circ}$ find
(i) $\angle \mathrm{CAD}$
(ii) $\angle \mathrm{BCD}$

Solution:
$\begin{aligned}
&\angle \mathrm{DBC}=40^{\circ} \\
&\angle \mathrm{BAC}=60^{\circ} \\
&\angle \mathrm{CAD}=\angle \mathrm{CBD}-40^{\circ}
\end{aligned}$
( ' angles in the same segment are equal)

(ii) $\quad \angle \mathrm{BCD}+\angle \mathrm{BAD}=180^{\circ}$
( $\because$ In cyclic quadrilateral opposite angles are supplementary)
$\begin{aligned}
\angle \mathrm{BCD}+(\angle \mathrm{BAD}+\angle \mathrm{CAD}) &=180^{\circ} \\
\angle \mathrm{BCD}+\left(60^{\circ}+40^{\circ}\right) &=180^{\circ} \\
\angle \mathrm{BCD} &=180^{\circ}-100^{\circ}=80^{\circ}
\end{aligned}$

Question $5 .$
In the given figure, $\mathrm{AB}$ and $\mathrm{CD}$ are the parallel chords of a circle with centre $\mathrm{O}$. Such that $\mathrm{AB}=8 \mathrm{~cm}$ and $\mathrm{CD}=6 \mathrm{~cm}$. If $\mathrm{OM} \perp \mathrm{AB}$ and $\mathrm{OL} \perp \mathrm{CD}$ distance between $\mathrm{LM}$ is $7 \mathrm{~cm}$. Find the radius of the circle?

Solution:

In the figure $\mathrm{LM}=7 \mathrm{~cm}$
Let $\mathrm{OM}=(7-x) \mathrm{cm}$
$\begin{aligned}
\mathrm{MB} &=\frac{8}{2}=4 \mathrm{~cm} \\
\mathrm{OB} &=\sqrt{4^{2}+(7-x)^{2}} \\
\mathrm{OD} &=\sqrt{3^{2}+x^{2}} \\
\mathrm{OB} &=\mathrm{OD}(\because \text { radius }) \\
\sqrt{16+(7-x)^{2}} &=\sqrt{3^{2}+x^{2}}
\end{aligned}$
Squaring both sides
$\begin{aligned}
16+(7-x)^{2} &=9+x^{2} \\
16+49-14 x+x^{2} &=9+x^{2} \\
14 x &=65-9 \\
14 x &=56 \\
x &=\frac{56}{14}=4 \\
\therefore \quad \therefore \text { Radius OD } &=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~cm}
\end{aligned}$

Question $6 .$
The arch of a bridge has dimensions as shown, where the arch measure $2 \mathrm{~m}$ at its highest point and its width is $6 \mathrm{~m}$. What is the radius of the circle that contains the arch?

Solution:
$\begin{aligned}
\text { If } C D &=2 \mathrm{~cm} \text { and } \mathrm{R} \text { is the radius, i.e. } \mathrm{OC}=\mathrm{OA}=\mathrm{OB}=\mathrm{R} \\
\therefore \mathrm{OD} &=\mathrm{OC}-\mathrm{DC}=\mathrm{R}-2 \mathrm{~cm} \\
\text { since } \mathrm{AB} &=6 \mathrm{~cm} \quad \Rightarrow \mathrm{AD}=\mathrm{DB}=\frac{6}{2}=3 \mathrm{~cm}
\end{aligned}$

In $\triangle \mathrm{OBD}$,
$\begin{aligned}
\Rightarrow \quad \mathrm{OB}^{2} &=\mathrm{OD}^{2}+\mathrm{BD}^{2} \\
\Rightarrow \quad \mathrm{R}^{2} &=(\mathrm{R}-2)^{2}+3^{2} \\
\Rightarrow \quad \mathrm{R}^{2} &=\mathrm{R}^{2}-4 \mathrm{R}+4+9 \\
&=\mathrm{R}^{2}-4 \mathrm{R}+13 \\
\Rightarrow \quad 4 \mathrm{R} &=13 \\
\Rightarrow \quad \mathrm{R} &=\frac{13}{4}=3.25 \mathrm{~cm}
\end{aligned}$

Question $7 .$
In figure $\angle \mathrm{ABC}=120^{\circ}$, where $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are points on the circle with centre $\mathrm{O}$. Find $\angle \mathrm{OAC}$ ?
Solution:
$\begin{aligned}
\text { reflex } \angle \mathrm{AOC} &=2 \times \angle \mathrm{ABC}=2 \times 120^{\circ}=240^{\circ} \\
\therefore \angle \mathrm{AOC} &=360^{\circ}-240^{\circ}=120^{\circ} \\
\Rightarrow \quad \text { Hence } \angle \mathrm{OAC}+\angle \mathrm{OCA} &=180^{\circ}-120^{\circ}=60^{\circ} \\
2 \angle \mathrm{AOC} &=60^{\circ} \\
\Rightarrow \quad \angle \mathrm{OAC} &=\frac{60^{\circ}}{2}=30^{\circ} \\
\quad[\because \angle \mathrm{OAC}=\angle \mathrm{OCA}]
\end{aligned}$

Question $8 .$
A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius $6 \mathrm{~m}$ ground to nineth standard students for planting sapplings. Four students plant trees at the points $A, B, C$ and $D$ as shown in figure. Here $A B=8 m, C D=10 \mathrm{~m}$ and $A B \perp C D$. If another student places a flower pot at the point $\mathrm{P}$, the intersection of $\mathrm{AB}$ and $\mathrm{CD}$, then find the distance from the centre to $\mathrm{P}$.

Solution:
$\text { In the figure } \quad \begin{aligned}
\mathrm{OA} &=\mathrm{OD}=6 \mathrm{~cm} \\
\mathrm{AB} &=8 \mathrm{~cm} \text { (Chord) } \\
\mathrm{CD} &=10 \mathrm{~cm} \text { (Chord) } \\
\mathrm{OM} &=\sqrt{6^{2}-4^{2}}\left(\because \text { OM bisects the chord and } \perp^{r}\right. \text { to the chord) }\\
&=\sqrt{36-16}=\sqrt{20} \mathrm{~cm} \\
\mathrm{ON} &=\sqrt{6^{2}-5^{2}}=\sqrt{36-25}=\sqrt{11} \mathrm{~cm}
\end{aligned}$
ONPM is a rectangle with all the angles $90^{\circ}$ and with length $\sqrt{20} \mathrm{~cm}$, breadth $\sqrt{11} \mathrm{~cm}$.

We need to find OP which is the diagonal of the rectangle ONPM.
$\begin{aligned}
\therefore O P=& \sqrt{\mathrm{ON}^{2}+\mathrm{NP}^{2}}=\sqrt{\sqrt{11^{2}}+\sqrt{20^{2}}} \\
&(\because \mathrm{OM}=\mathrm{NP}, \text { opposite sides of the rectangle }) \\
=& \sqrt{11+20}=\sqrt{31}=5.56 \mathrm{~cm}=5.6 \mathrm{cr} .
\end{aligned}$

Question $9 .$
In the given figure, $\angle \mathrm{POQ}=100^{\circ}$ and $\angle \mathrm{PQR}=30^{\circ}$, then find $\angle \mathrm{RPO}$.

Solution:
$\begin{aligned} \text { In the figure } \angle \mathrm{POQ} &=100^{\circ} \\ \angle \mathrm{PQR} &=30^{\circ} \\ \angle \mathrm{PQR} &=\frac{1}{2} \angle \mathrm{POQ}-\frac{1}{2} \times 100 \\ &=50^{\circ} \\ \text { In } \Delta \mathrm{OPQ}, \quad \quad \angle \mathrm{OPQ} &=\angle \mathrm{OQP}=\angle \mathrm{POQ}=180^{\circ} \\ 2 \angle \mathrm{OPQ}+100 &=180^{\circ} \\ 2 \angle \mathrm{OPQ} &=180^{\circ} \\ 2 \angle \mathrm{OPQ} &=80^{\circ} \\ \therefore \angle \mathrm{OPQ} &=40^{\circ} \\ \text { In } \triangle \mathrm{PRQ}, \quad 50^{\circ}+(40+x)+30^{\circ} &=180^{\circ} \\(40+x)^{\circ} &=180^{\circ}-80=100^{\circ} \\ x^{\circ} &=100^{\circ}-40^{\circ}=60^{\circ} \\ \therefore \angle \mathrm{RPO} &=x=60^{\circ} \end{aligned}$