Exercise 4.4 - Chapter 4 Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.4$
Question $1 .$
The length of the tangent to a circle from a point $\mathrm{P}$, which is $25 \mathrm{~cm}$ away from the centre is $24 \mathrm{~cm}$.
What is the radius of the circle?
Solution:
$\begin{aligned}
&24^{2}+r^{2}=25^{2} \\
&576+r^{2}=625 \\
&r^{2}=625-576 \\
&=49
\end{aligned}$

Question $2 .$
$\triangle \mathrm{LMN}$ is a right angled triangle with $\angle \mathrm{L}=90^{\circ}$. A circle is inscribed in it. The lengths of the sides containing the right angle are $6 \mathrm{~cm}$ and $8 \mathrm{~cm}$. Find the radius of the circle.
Solution:
$\Delta \mathrm{LMN}$,
By Pythagoras theorem,

$\begin{aligned}
\mathrm{MN}^{2} &=\mathrm{LN}^{2}+\mathrm{LM}^{2} \\
&=8^{2}+6^{2}=100 \\
\mathrm{MN} &=10
\end{aligned}$
Now, Area of $\Delta \mathrm{LMN}=$ Area of $\Delta \mathrm{OLM}+$ Area of $\Delta \mathrm{OMN}+$ Area of $\Delta \mathrm{ONL}$.
$\begin{aligned}
\Rightarrow \frac{1}{2} \times \mathrm{LM} \times \mathrm{LN}=& \frac{1}{2} \mathrm{LM} \times r+\frac{1}{2} \times \mathrm{MN} \times r \\
&+\frac{1}{2} \times \mathrm{NL} \times r \\
\Rightarrow \quad \frac{1}{2} \times 6 \times 8=& \frac{1}{2} \times 6 r+\frac{1}{2} \times 10 r+\frac{1}{2} \times 8 r \\
24 &=3 r+5 r+4 r \\
12 r &=24 \Rightarrow r=2 \mathrm{~cm}
\end{aligned}$

Question $3 .$
A circle is inscribed in $\triangle \mathrm{ABC}$ having sides $8 \mathrm{~cm}, 10 \mathrm{~cm}$ and $12 \mathrm{~cm}$ as shown in figure, Find $\mathrm{AD}$, $B E$ and CF.

(a) a-iii b-ii c-iv $\mathrm{d}-i$
(b) $\mathrm{a}-i v \mathrm{~b}-i i i \mathrm{c}-i \quad \mathrm{~d}-i i$
(c) a - iii b-iv c - $i \quad \mathrm{~d}-i i$
(d) $\mathrm{a}-i i \quad \mathrm{~b}-i i i \mathrm{c}-i \quad \mathrm{~d}-i v$
Solution:
We know that the tangents drawn from are external point to a circle are equal.
Therefore $\mathrm{AD}=\mathrm{AF}=\mathrm{x}$ say.
$\mathrm{BD}=\mathrm{BE}=\mathrm{y}$ say and
$\mathrm{CE}=\mathrm{CF}=\mathrm{z}$ say
Now, $\mathrm{AB}=12 \mathrm{~cm}, \mathrm{BC}=8 \mathrm{~cm}$, and $\mathrm{CA}=10 \mathrm{~cm}$.
$x+y=12, y+z=8$ and $z+x=10$
$(x+y)+(y+z)+(z+x)=12+8+10$
$2(x+y+z)=30$
$x+y+z=15$
Now, $x+y=12$ and $x+y+z=15$
$12+z=15 \Rightarrow z=3$
$y+z=8$ and $x+y+z=15$
$x+8=15 \Rightarrow x=7$
and $z+x=10$ and $x+y+z=15$
$10+y=15 \Rightarrow y=5$
Hence, $\mathrm{AD}=\mathrm{x}=7 \mathrm{~cm}, \mathrm{BE}=\mathrm{y}=5 \mathrm{~cm}$
and $C F=z=3 \mathrm{~cm}$.
 

Question $4 .$
$P Q$ is a tangent drawn from a point $P$ to a circle with centre $O$ and $Q O R$ is a diameter of the circle such that $\angle \mathrm{POR}=120^{\circ}$. Find $\angle \mathrm{OPQ}$.
Solution:
$\angle \mathrm{POR}+\angle \mathrm{POQ}=180^{\circ}$ (straight angle $\left.=180^{\circ}\right)$
$\therefore 120+\angle \mathrm{POQ}=180^{\circ}$
$\angle \mathrm{POQ}=60^{\circ}$
$\angle \mathrm{OQP}=90^{\circ}\left(\because\right.$ radius is $\perp^{\mathrm{r}}$ to the tangent at the point of contact)

$\begin{aligned}
&\therefore \angle \mathrm{POQ}+\angle \mathrm{PQO}+\angle \mathrm{OPQ}=180^{\circ}\left(\because \text { sum of the } 3 \text { angles of a triangle is } 180^{\circ}\right) \\
&\therefore 60+90+\angle \mathrm{OPQ}=80^{\circ} \\
&\angle \mathrm{OPQ}=180^{\circ}-150^{\circ}=30^{\circ}
\end{aligned}$

Question $5 .$
A tangent $\mathrm{ST}$ to a circle touches it at $\mathrm{B} . \mathrm{AB}$ is a chord such that $\angle \mathrm{ABT}=65^{\circ}$. Find $\angle \mathrm{AOB}$, where "O" is the centre of the circle.
Solution:
In the figure,
$\begin{aligned}
&\angle \mathrm{OBT}=90^{\circ}(\because \mathrm{OB} \text {-radius, } \mathrm{BT}-\text { Tangent }) \\
&=115^{\circ} \\
&\therefore \angle \mathrm{OBA}=90^{\circ}-65^{\circ}
\end{aligned}$
$\begin{aligned}
&\angle \mathrm{OAB}=25^{\circ}(\mathrm{OA}=\mathrm{OB}) \\
&\therefore \angle \mathrm{AOB}=180^{\circ}-50^{\circ} \\
&=130^{\circ}
\end{aligned}$
Brain Capacity                            Evolutionery Human
i) $900 \mathrm{cc}$                  (A) Homo sapienS
ii) $650-800 \mathrm{cc}$          $(B)$ Homo erectus
iii) $350-450 \mathrm{cc}$          (C) Homo habilis
iv) $1300-1600 \mathrm{cc}$      (D) Australopithecus
(a) $a-i v \quad b-i \quad c-i i \quad d-i i i$
(b) $a-i i \quad b-i v \quad c-i i i \quad d-i$
(c) $a-i i \quad b-i i i \quad c-i v \quad d-i$
(d) $a-i i i \cdot b-i \quad c-i i \quad d-i v$
 

Question 6.
In figure, $\mathrm{O}$ is the centre of the circle with radius $5 \mathrm{~cm}$. $\mathrm{T}$ is a point such that $\mathrm{OT}=13 \mathrm{~cm}$ and $\mathrm{OT}$ intersects the circle $\mathrm{E}$, if $\mathrm{AB}$ is the tangent to the circle at $\mathrm{E}$, find the length of $\mathrm{AB}$.

Solution:
$\begin{aligned}
&\therefore \mathrm{ET}^{2}=\mathrm{AT}^{2}-\mathrm{AE}^{2}=(\mathrm{AT}+\mathrm{AE})(\mathrm{AT}-\mathrm{AE}) \\
&\therefore \mathrm{ET}^{2}=(\mathrm{AT}+\mathrm{AP})(\mathrm{AT}-\mathrm{AE})(\because \mathrm{AE}=\mathrm{AP}) \\
&\therefore 8 \times 8=12 \times(\mathrm{AT}-\mathrm{AE})
\end{aligned}$
$\begin{aligned}
\therefore \quad(\mathrm{AT}-\mathrm{AE}) &=\frac{64}{12}=\frac{16}{3} \\
\mathrm{AT}+\mathrm{AE} &=\mathrm{AT}+\mathrm{AP}=\mathrm{PT}=12 \ldots
\end{aligned}$
Adding (3) and (4),
$2 \mathrm{AT}=\frac{16}{3}+12$

$\begin{aligned}
&\text { AT }=\frac{8}{3}+\frac{18}{3}=\frac{26}{3} \\
&\qquad \mathrm{AE}=\mathrm{AT}-\mathrm{AE} \\
&\qquad \frac{26}{3}-\frac{16}{3}=\frac{10}{3} \\
&\qquad \mathrm{~EB}=\frac{10}{3}
\end{aligned}$

$\begin{aligned}
&\mathrm{OB}^{2}=\mathrm{OC}^{2}+\mathrm{BC}^{2} \\
&=6^{2}+8^{2}
\end{aligned}$
$=36+64=100$
$\mathrm{OB}=$ Radius of the larger circle $=\sqrt{100}=10 \mathrm{~cm} .$
 

Question 8.
Two circles with centres $\mathrm{O}$ and $\mathrm{O}^{\prime}$ of radii $3 \mathrm{~cm}$ and $4 \mathrm{~cm}$ respectively intersect at two points $\mathrm{P}$ and Q, such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.
Solution:
Given: $O P=O Q=4$
$O^{\prime} P=O^{\prime} Q=3$

$\mathrm{OO}^{\prime}$ is the perpendicular bisector of chord $\mathrm{PQ}$.
Let $R$ be the point of intersection of $P Q$ and $O O^{\prime}$.
Assume $P R=Q R=x$ and $O R=y$
In OPO', $\mathrm{OP}^{2}+\mathrm{O}^{\prime} \mathrm{P}^{2}=\left(\mathrm{OO}^{\prime}\right)^{2} \Rightarrow \mathrm{OO}^{\prime}$
$=\sqrt{4^{2}+3^{2}}=5$
$\mathrm{OR}=\mathrm{y} \Rightarrow \mathrm{OR}=5-\mathrm{y}$
In $\triangle \mathrm{OPR}, \mathrm{PR}^{2}+\mathrm{OR}^{2}=\mathrm{OP}^{2} \Rightarrow \mathrm{x}^{2}+\mathrm{y}=4^{2}$
In $\Delta \mathrm{O}^{\prime} \mathrm{PR}, \mathrm{PR}^{2}+\mathrm{O}^{\prime} \mathrm{R}^{2}=\mathrm{O}^{\prime} \mathrm{P}^{2} \Rightarrow \mathrm{x}^{2}+(5-\mathrm{y})^{2}=9$
$(1)-(2)=y^{2}-\left(25+y^{2}-10 y\right)=16-9$
$\begin{aligned}
&\Rightarrow \mathrm{y}^{2}-25-\mathrm{y}^{2}+10 \mathrm{y}=7 \\
&\Rightarrow 10 \mathrm{y}=25+7 \Rightarrow 10 \mathrm{y}=32 \\
&\Rightarrow \mathrm{y}=3.2
\end{aligned}$
Substituting $y=3.2$ in (1), we get $x=\sqrt{4^{2}-3.2^{2}}$ $x=2.4$
$P Q=2 x \Rightarrow P Q=4.8 \mathrm{~cm}$

Question $9 .$
Show that the angle bisectors of a triangle are concurrent.
Solution:

In $\triangle \mathrm{ABC}$, let $\mathrm{AD}, \mathrm{BE}$ are two angle bisectors.

They meet at the point ' $\mathrm{O}$ '
We have to prove that $=\frac{A C}{C D}=\frac{A O}{O D}$
Construct $\mathrm{CO}$ to meet the interesecting point $\mathrm{O}$ from $\mathrm{C}$.
In $\Delta \mathrm{ABE}, \quad \frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{BO}}{\mathrm{OE}}$ also $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$ (by
$\therefore \frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{DC}}$ In $\Delta \mathrm{ABD}, \frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AO}}{\mathrm{OD}}$ From $(1) \&(2)$ we get $\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{AO}}{\mathrm{OD}}$
Hence proved.
 

Question $10 .$
In $\triangle \mathrm{ABC}$, with $\angle \mathrm{B}=90^{\circ}, \mathrm{BC}=6 \mathrm{~cm}$ and $A B=8 \mathrm{~cm}, \mathrm{D}$ is a point on $\mathrm{AC}$ such that $\mathrm{AD}=2 \mathrm{~cm}$ and $\mathrm{E}$ is the midpoint of $\mathrm{AB}$. Join $\mathrm{D}$ to $\mathrm{E}$ and extend it to meet at $\mathrm{F}$. Find $\mathrm{BF}$. In the figure $\triangle \mathrm{ABC}, \triangle \mathrm{EBF}$ are similar triangles.
Solution:
Consider DABC, Then D, E, F are respective points on the sides $\mathrm{CA}, \mathrm{AB}$ and $\mathrm{BC}$. By constrution

D, E, F are collinear.
By menelaus' theorem, $\frac{\mathrm{AE}}{\mathrm{EB}} \times \frac{\mathrm{BF}}{\mathrm{FC}} \times \frac{\mathrm{CD}}{\mathrm{DA}}=1 \ldots(1)$
$\mathrm{FC}=\mathrm{FB}+\mathrm{BC}=\mathrm{BF}+6$
By pythagoras therom $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$=64+36=100$
$\therefore \mathrm{AC}=10, \mathrm{CD}=8$
$(1) \Rightarrow \frac{4}{4} \times \frac{\mathrm{BF}}{\mathrm{BF}+6 \mathrm{~A}} \times \frac{8}{2}=1$

$4 \mathrm{BF}=\mathrm{BF}+6 \Rightarrow \mathrm{BF}=2 \mathrm{~cm}$

Question $11 .$
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.

Solution:
In the figure, let $\mathrm{O}$ be the concurrent point of the angle bisectors of the three angles.
$\begin{aligned}
&\frac{\mathrm{BF}}{\mathrm{FA}}=\frac{\mathrm{OB}}{\mathrm{OA}} \\
&\frac{\mathrm{CD}}{\mathrm{DB}}=\frac{\mathrm{OC}}{\mathrm{OB}} \\
&\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{\mathrm{OA}}{\mathrm{OC}}
\end{aligned}$
Multiplying the corresponding sides of (1), (2)
and (3) we get
$\frac{x}{\not \supset} \times \frac{10^{2}}{\not{\beta}} \times \frac{\not \beta}{A_{2}}=1$
$\frac{x}{2}=1$
$x=2 \mathrm{~cm}$
 

Question $12 .$
Draw a tangent at any point $\mathrm{R}$ on the circle of radius $3.4 \mathrm{~cm}$ and centre at $\mathrm{P}$ ?
Solution:
Radius $=3.4 \mathrm{~cm}$
Centre $=\mathrm{P}$
Tangent at any point $R$.

Construction:
Steps:
(1) Draw a circle with centre $\mathrm{P}$ of radius $3.4 \mathrm{~cm}$.
(2) Take a point $R$ on the circle. Join PR.
(3) Draw $\perp^{\mathrm{r}}$ line $\mathrm{TT}^{1}$ to $\mathrm{PR}$. Which passes through $\mathrm{R}$.
(4) TT1 is the required tangent.

Question $13 .$
Draw a circle of radius $4.5 \mathrm{~cm}$. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem.
Solution:
Construction:
Steps:
(1) With $\mathrm{O}$ as the centre, draw a circle of radius $4.5 \mathrm{~cm}$.
(2) Take a point $R$ on the circle. Through $R$ draw any chord $P R$.
(3) Take a point $Q$ distinct from $P$ and $R$ on the circle, so that $P, Q, R$ are in anti-clockwise direction. Join $\mathrm{PQ}$ and $\mathrm{QR}$.
(4) Through $\mathrm{R}$ drawn a tangent $\mathrm{TT}^{1}$ such that $\angle \mathrm{TRP}=\angle \mathrm{PQR}$.
(5) $\mathrm{TT}^{1}$ is the required tangent.

Question $14 .$
Draw the two tangents from a point which is $10 \mathrm{~cm}$ away from the centre of a circle of radius $5 \mathrm{~cm}$. Also, measure the lengths of the tangents.
Solution:
Radius $=5 \mathrm{~cm}$
The distance between the point from the centre is $10 \mathrm{~cm}$.
Construction:

Steps:
(1) With $\mathrm{O}$ as centre, draw a circle of radius $5 \mathrm{~cm}$.
(2) Draw a line $\mathrm{OP}=10 \mathrm{~cm}$.
(3) Draw a perpendicular bisector of $\mathrm{OP}$ which cuts $\mathrm{OP}$ at $\mathrm{M}$.
(4) With M as centre and $M O$ as radius, draw a circle which cuts previous circle at $A$ and $B$.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA and $\mathrm{PB}=8.7 \mathrm{~cm}$.
Verification:
In the right triangle $\angle \mathrm{POA}$;
$\begin{aligned}
\mathrm{PA} &=\sqrt{\mathrm{OP}^{2}-\mathrm{OA}^{2}} \\
\mathrm{PA} &=\sqrt{10^{2}-5^{2}} \\
&=\sqrt{100-25} \\
&=\sqrt{75} \\
& \cong 8.7 \mathrm{~cm} \text { (approximately) }
\end{aligned}$

Question $15 .$
Take a point which is $11 \mathrm{~cm}$ away from the centre of a circle of radius $4 \mathrm{~cm}$ and draw the two tangents to the circle from that point.
Solution:
Radius $=4 \mathrm{~cm}$
The distance of a point from the center $=11 \mathrm{~cm}$.

Construction:
Steps:
(1) With centre $\mathrm{O}$, draw a circle of radius $4 \mathrm{~cm}$.
(2) Draw a line $\mathrm{OP}=11 \mathrm{~cm}$.
(3) Draw a $\perp^{\mathrm{r}}$ bisector of $\mathrm{OP}$, which cuts atM.
(4) With M as centre and $\mathrm{MO}$ as radius, draw a circle which cuts previous circle at $\mathrm{A}$ and $\mathrm{B}$.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are $\mathrm{PA}=\mathrm{PB}$ $=10.2 \mathrm{~cm} .$
Verification:
In the right triangle
$\begin{aligned}
\angle \mathrm{OPA}, \mathrm{PA} &=\sqrt{\mathrm{OP}^{2}-\mathrm{OA}^{2}} \\
&=\sqrt{11^{2}-4^{2}} \\
&=\sqrt{121-16} \\
&=\sqrt{105} \\
& \cong 10.2 \mathrm{~cm} \text { (approximately) }
\end{aligned}$
 

Question $16 .$
Draw the two tangents from a point which is $5 \mathrm{~cm}$ away from the centre of a circle of diameter 6 $\mathrm{cm}$. Also, measure the lengths of the tangents.
Solution:
Diameter $=6 \mathrm{~cm}$
Radius $=\frac{6}{2}=3 \mathrm{~cm}$.
The distance between the centre and the point is $5 \mathrm{~cm}$.

Construction:
Steps:
(1) With centre $\mathrm{O}$, draw a circle of radius, $3 \mathrm{~cm}$.
(2) Draw a line $\mathrm{OP}=5 \mathrm{~cm}$.
(3) Draw a bisector of $\mathrm{OP}$, which cuts $\mathrm{OP}$ and $\mathrm{M}$.
(4) With $\mathrm{M}$ as centre and $\mathrm{MO}$ as radius draw a circle which cuts previous circle at $\mathrm{A}$ and $\mathrm{B}$.
(5) Join $\mathrm{AP}$ and $\mathrm{BP}$. $\mathrm{AP}$ and $\mathrm{BP}$ are the required tangents. Thus length of the tangents are $\mathrm{PA}=\mathrm{PB}$ $=4 \mathrm{~cm}$
Verification:
In the right triangle $\triangle \mathrm{OPA}$,
$\begin{aligned}
\mathrm{PA} &=\sqrt{\mathrm{OP}^{2}-\mathrm{OA}^{2}} \\
&=\sqrt{5^{2}-3^{2}} \\
&=\sqrt{25-9}=\sqrt{16} \\
&=4 \mathrm{~cm}
\end{aligned}$

Question $17 .$
Draw a tangent to the circle from the point $P$ having radius $3.6 \mathrm{~cm}$, and centre at $\mathrm{O}$. Point $\mathrm{P}$ is at a distance $7.2 \mathrm{~cm}$ from the centre.
Radius $3.6 \mathrm{~cm}$.
Solution:
Distance from the centre to the point is $7.2 \mathrm{~cm}$.

Construction:
Steps:
(1) Draw a circle of radius $3.6 \mathrm{~cm}$ with centre $\mathrm{O}$.
(2) Draw a line $\mathrm{OP}=7.2 \mathrm{~cm}$.
(3) Draw a perpendicular bisector of OP, which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at $A$ and $B$.
(5) Join $\mathrm{AP}$ and $\mathrm{BP} . \mathrm{AP}$ and $\mathrm{BP}$ are the required tangents. Thus length of the tangents are $\mathrm{PA}=\mathrm{PB}$ $=6.2 \mathrm{~cm}$.
Verification:
In the right triangle.
$\begin{aligned}
\text { OPA, PA } &=\sqrt{\mathrm{OP}^{2}-\mathrm{OA}^{2}} \\
=\sqrt{7.2^{2}-3.6^{2}} &=\sqrt{51.84-12.96} \\
&=\sqrt{38.88} \\
& \cong 6.2 \mathrm{~cm} \text { (approximately) }
\end{aligned}$