Exercise 4.3 - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Question 1.
The diameter of the circle is $52 \mathrm{~cm}$ and the length of one of its chord is $20 \mathrm{~cm}$. Find the distance of the chord from the centre.
Solution:
The distance of the chord from the centre $\mathrm{O}$

$\begin{aligned}
\mathrm{OD} &=\sqrt{26^{2}-10^{2}} \\
&=\sqrt{676-100} \\
&=\sqrt{576}=24 \mathrm{~cm}
\end{aligned}$

Question $2 .$
The chord of length $30 \mathrm{~cm}$ is drawn at the distance of $8 \mathrm{~cm}$ from the centre of the circle. Find the radius of the circle.
Solution:
Radius of the circle
$\begin{aligned}
&=\sqrt{8^{2}+15^{2}}=\sqrt{64+225} \\
&=\sqrt{289}=17 \mathrm{~cm}
\end{aligned}$

Question $3 .$
Find the length of the chord $\mathrm{AC}$ where $\mathrm{AB}$ and $\mathrm{CD}$ are the two diameters perpendicular to each other of $\mathrm{a}$ circle with radius $4 \sqrt{2} \mathrm{~cm}$ and also find $\angle O A C$ and $\angle O C A$.

Solution:
$\triangle \mathrm{OAC}$ is an isơeelés triàngle with oné àngle $90^{\circ}$
$\therefore \angle \mathrm{OAC}+\angle \mathrm{OCA}=180^{\circ}-90^{\circ}$

$\begin{aligned}
&2 \angle \mathrm{OAC}=90^{\circ} \\
&\angle \mathrm{OAC}=45^{\circ} \\
&\therefore \angle \mathrm{OCA}=45^{\circ}
\end{aligned}$
$\begin{aligned}
\text { Length of the chord } &=\sqrt{(4 \sqrt{2})^{2}+(4 \sqrt{2})^{2}} \\
&=\sqrt{16 \times 2+16 \times 2} \\
&=\sqrt{64}=8 \mathrm{~cm}
\end{aligned}$

Question $4 .$
A chord is $12 \mathrm{~cm}$ away from the centre of the circle of radius $15 \mathrm{~cm}$. Find the length of the chord.

Solution:

$\begin{aligned}
\mathrm{BD} &=\sqrt{15^{2}-12^{2}} \\
&=\sqrt{225-144} \\
&=\sqrt{81}=9 \mathrm{~cm}
\end{aligned}$
$\therefore$ length of the chord $\mathrm{AB}=9+9=18 \mathrm{~cm}$

Question $5 .$
In a circle, $\mathrm{AB}$ and $\mathrm{CD}$ are two parallel chords with centre $\mathrm{O}$ and radius $10 \mathrm{~cm}$ such that $\mathrm{AB}=16 \mathrm{~cm}$ and $\mathrm{CD}=12 \mathrm{~cm}$ determine the distance between the two chords?
Solution:

The distance between the two chord $\mathrm{FE}=\mathrm{OE}+\mathrm{OF}$
$\begin{aligned} \mathrm{OE} &=\sqrt{10^{2}-6^{2}}=\sqrt{100-36}=\sqrt{64}=8 \mathrm{~cm} \\ \mathrm{OF} &=\sqrt{10^{2}-8^{2}} \\ &=\sqrt{100-64}=\sqrt{36}=6 \mathrm{~cm} \\ \therefore \mathrm{FE} &=8 \mathrm{~cm}+6 \mathrm{~cm}=14 \mathrm{~cm} \end{aligned}$
$\therefore$ Distance between the chords is $14 \mathrm{~cm}$

Question $6 .$
Two circles of radii $5 \mathrm{~cm}$ and $3 \mathrm{~cm}$ intersect at two points and the distance between their centres is $4 \mathrm{~cm}$. Find the length of the common chord.
Solution:

$\begin{aligned} \mathrm{OD} &=\mathrm{DP}=\frac{4_{\mathrm{cm}}}{2}=2 \mathrm{~cm} \\ \mathrm{AD} &=\mathrm{BD}=\sqrt{5^{2}-4^{2}} \\ &=\sqrt{25-16}=\sqrt{9}=3 \mathrm{~cm} \end{aligned}$
The length of the common chord $\mathrm{AB}=\mathrm{AD}+\mathrm{BD}=(3+3) \mathrm{cm}=6 \mathrm{~cm}$
 

Question $7 .$
Find the value of $x^{\circ}$ in the following

Solution

Question $8 .$
In the given figure, $\angle \mathrm{CAB}=25^{\circ}$, find $\angle \mathrm{BDC}, \angle \mathrm{DBA}$ and $\angle \mathrm{COB}$

Solution:

(i) $\angle \mathrm{CAB}=25^{\circ}$
$\therefore \angle \mathrm{BDC}=25^{\circ}$
(ii) $\angle \mathrm{DBA}=\angle \mathrm{DCA}=180-\left(90+25^{\circ}\right)=180^{\circ}-115^{\circ}=65^{\circ}$
(ii) $\angle \mathrm{COB}=2 \angle \mathrm{CAB}=2 \times 25^{\circ}=50^{\circ}$