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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 4
Question $1 .$
In the figure, if $\mathrm{BD} \perp \mathrm{AC}$ and $\mathrm{CE} \angle \mathrm{AB}$, prove that
(i) $\triangle \mathbf{A E C} \sim \triangle \mathbf{A D B}$
(ii) $\frac{\mathbf{C A}}{\mathbf{A B}}=\frac{\mathbf{C E}}{\mathrm{DB}}$

Solution:
In the figure's $\triangle \mathrm{AEC}$ and $\triangle \mathrm{ADB}$.
We have $\angle \mathrm{AEC}=\angle \mathrm{ADB}=90(\because \mathrm{CE} \angle \mathrm{AB}$ and $\mathrm{BD} \angle \mathrm{AC})$
and $\angle \mathrm{EAC}=\angle \mathrm{DAB}$
[Each equal to $\angle \mathrm{A}$ ]
Therefore by AA-criterion of similarity, we have $\triangle \mathrm{AEC} \sim \triangle \mathrm{ADB}$
(ii) We have
$\triangle \mathrm{AEC} \sim \triangle \mathrm{ADB}[$ As proved above]
$\Rightarrow \frac{C A}{B A}=\frac{E C}{D B} \Rightarrow \frac{C A}{A B}=\frac{C E}{D B}$
Hence proved.

Question $2 .$
In the given figure $\mathrm{AB}\|\mathrm{CD}\| \mathrm{EF}$. If $\mathrm{AB}=6 \mathrm{~cm}, \mathrm{CD}=\mathrm{xcm}, \mathrm{EF}=4 \mathrm{~cm}, \mathrm{BD}=5 \mathrm{~cm}$ and $\mathrm{DE}=\mathrm{y}$ $\mathrm{cm}$. Find $x$ and $y$.

Solution:
In the given figure, $\triangle \mathrm{AEF}$, and $\triangle \mathrm{ACD}$ are similar $\Delta^{\mathrm{s}}$.
$\angle \mathrm{AEF}=\angle \mathrm{ACD}=90^{\circ}$
$\angle \mathrm{A}=\angle \mathrm{A}$ (common)
$\therefore \triangle \mathrm{AEF} \sim \triangle \mathrm{ACD}$ (By AA criterion of similarity)
$$
\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{EF}}{\mathrm{CD}}=\frac{4}{x} \Rightarrow \frac{\mathrm{AC}=\frac{\mathrm{AE} \times \mathrm{CD}}{x \times \mathrm{AE}}}{\mathrm{AC}}=\frac{\mathrm{EF}}{4}
$$
In $\triangle \mathrm{EAB}$ and $\triangle \mathrm{ECD}$,
we have $\angle \mathrm{ECD}=\angle \mathrm{EAB}=90^{\circ}$.
$\angle \mathrm{E}=\angle \mathrm{E}$ (common)
$\therefore \quad \triangle \mathrm{ECD} \sim \triangle \mathrm{EAB}$
$\Rightarrow \quad \frac{\mathrm{CE}}{\mathrm{EA}}=\frac{\mathrm{CD}}{\mathrm{BA}}=\frac{x}{6}$
$\frac{\mathrm{CE}}{\mathrm{EA}}=\frac{x}{6}$
$\mathrm{CE}=\frac{x \times \mathrm{EA}}{6}$

By BPT
$\begin{aligned}
\frac{\mathrm{CE}}{\mathrm{EA}} &=\frac{y}{y+5} \\
\frac{x}{6} &=\frac{y}{y+5}
\end{aligned}$
From (1) and (2), we have
$\begin{aligned}
\mathrm{AC}+\mathrm{CE} &=\frac{x \times \mathrm{AE}}{4}+\frac{x \times \mathrm{AE}}{6} \\
A E &=A E \times x\left[\frac{1}{4}+\frac{1}{6}\right] \\
1 &=x\left(\frac{6+4}{24}\right)=\frac{10 x}{24} \\
\therefore \quad x &=\frac{24}{10}=2.4 \mathrm{~cm}=\frac{12}{5}
\end{aligned}$
Subtstituting $x=2.4 \mathrm{~cm}$ in (3)
We get,
$\begin{aligned}
\frac{2 \cdot 4}{6} &=\frac{y}{y+5} \\
6 y &=2 \cdot 4 y+2 \cdot 4 \times 5 \\
6 y &=2 \cdot 4 y+12 \\
6 y-2 \cdot 4 y &=12 \\
3 \cdot 6 y &=12 \\
y &=\frac{12 \times 10}{3.6 \times 10}
\end{aligned}$

$\begin{aligned}
&=\frac{120}{36}=3.3 \mathrm{~cm} \\
x &=2.4 \mathrm{~cm} \\
y &=3.3 \mathrm{~cm}
\end{aligned}$

Question $3 .$
$\mathrm{O}$ is any point inside a triangle $\mathrm{ABC}$. The bisector of $\angle \mathrm{AOB}, \angle \mathrm{BOC}$ and $\angle \mathrm{COA}$ meet the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ in point $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ respectively.
Show that $\mathrm{AD} \times \mathrm{BE} \times \mathrm{CF}=\mathrm{DB} \times \mathrm{EC} \times \mathrm{FA}$
Solution:
In $\triangle \mathrm{AOB}, \mathrm{OD}$ is the bisector of $\angle \mathrm{AOB}$.

In $\triangle B O C, O E$ is the bisector of $\angle B O C$
$\therefore \frac{\mathrm{OB}}{\mathrm{OC}}=\frac{\mathrm{BE}}{\mathrm{EC}}$
In $\triangle \mathrm{COA}, \mathrm{OF}$ is the bisector of $\angle \mathrm{COA}$.
$\therefore \frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{CF}}{\mathrm{FA}}$
Multiplying the corresponding sides of (1), (2) and (3), we get
$1=\frac{\mathrm{AD}}{\mathrm{DB}} \times \frac{\mathrm{BE}}{\mathrm{EC}} \times \frac{\mathrm{CF}}{\mathrm{FA}}$
$\Rightarrow \mathrm{DB} \times \mathrm{EC} \times \mathrm{FA}=\mathrm{AD} \times \mathrm{BE} \times \mathrm{CF}$
Hence proved.

Question $4 .$
In the figure, $\mathrm{ABC}$ is a triangle in which $\mathrm{AB}=\mathrm{AC}$. Points $\mathrm{D}$ and $\mathrm{E}$ are points on the side $\mathrm{AB}$ and $\mathrm{AC}$ respectively such that $\mathrm{AD}=\mathrm{AE}$. Show that the points $\mathrm{B}, \mathrm{C}, \mathrm{E}$ and $\mathrm{D}$ lie on a same circle.

Solution:
In order to prove that the points $\mathrm{B}, \mathrm{C}, \mathrm{E}$ and $\mathrm{D}$ are concyclic, it is sufficient to show that $\angle \mathrm{ABC}+$ $\angle \mathrm{CED}=180^{\circ}$ and $\angle \mathrm{ACB}+\angle \mathrm{BDE}=180^{\circ}$

In $\triangle \mathrm{ABC}$, we have $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{AD}=\mathrm{AE}$.
$\begin{aligned}
&\Rightarrow \mathrm{AB}-\mathrm{AD}=\mathrm{AC}-\mathrm{AE} \\
&\Rightarrow \mathrm{DB}=\mathrm{EC}
\end{aligned}$
Thus we have $\mathrm{AD}=\mathrm{AE}$ and $\mathrm{DB}=\mathrm{EC}$. (By the converse of Thale's theorem)
$\Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \Rightarrow \mathrm{DE} \| \mathrm{BC}$
$\angle \mathrm{ABC}=\angle \mathrm{ADE}$ (corresponding angles)
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BDE}=\angle \mathrm{ADE}+\angle \mathrm{BDE}$ (Adding $\angle \mathrm{BDE}$ on both sides)
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BDE}=180^{\circ}$
$\Rightarrow \angle \mathrm{ACB}+\angle \mathrm{BDE}=180^{\circ}(\because \mathrm{AB}=\mathrm{AC} \therefore \angle \mathrm{ABC}=\angle \mathrm{ACB})$
Again $\mathrm{DE} \| \mathrm{BC}$
$\Rightarrow \angle \mathrm{ACB}=\angle \mathrm{AED}$
$\Rightarrow \angle \mathrm{ACB}+\angle \mathrm{CED}=\angle \mathrm{AED}+\angle \mathrm{CED}$ (Adding $\angle \mathrm{CED}$ on both sides).
$\Rightarrow \angle \mathrm{ACB}+\angle \mathrm{CED}=180^{\circ}$ and
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{CED}=180^{\circ}(\because \angle \mathrm{ABC}=\angle \mathrm{ACB})$
Thus BDEC is a quadrilateral such that
$\Rightarrow \angle \mathrm{ACB}+\angle \mathrm{BDE}=180^{\circ}$ and
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{CED}=180^{\circ}$
$\therefore \mathrm{BDEC}$ is a cyclic quadrilateral. Hence $\mathrm{B}, \mathrm{C}, \mathrm{E}$, and $\mathrm{D}$ are concyclic points.

Question $5 .$
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of $20 \mathrm{~km} / \mathrm{hr}$ and the second train travels at 30 $\mathrm{km} / \mathrm{hr}$. After 2 hours, what is the distance between them?
Solution:
After 2 hours, let us assume that the first train is at A and the second is at B.
$\begin{aligned}
\mathrm{OA} &=\text { speed } \times \text { time } \\
&=20 \times 2=40 \mathrm{~km} \\
\mathrm{OB} &=\text { speed } \times \text { time } \\
&=30 \times 2=60 \mathrm{~km}
\end{aligned}$
Distance between the trains after 2 hours,
$\begin{aligned}
\mathrm{AB} &=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \text { (pythogoras theorem) } \\
&=\sqrt{40^{2}+60^{2}} \\
&=\sqrt{1600+3600}=\sqrt{5200}=\sqrt{400 \times 13} \\
&=20 \sqrt{13} \\
\mathrm{AB} &=72.11 \mathrm{~km} \text { or } \mathrm{AB}=20 \sqrt{13} \mathrm{~km} .
\end{aligned}$

Question 6.
$\mathrm{D}$ is the mid point of side $\mathrm{BC}$ and $\mathrm{AE} \perp \mathrm{BC}$. If $\mathrm{i} \mathrm{} \mathrm{BC} \mathrm{a}, \mathrm{AC}=\mathrm{b}, \mathrm{AB}=\mathrm{c}, \mathrm{ED}=\mathrm{x}, \mathrm{AD}=\mathrm{p}$ and $\mathrm{AE}=\mathrm{h}$ , prove that
(i) $b^{2}=p^{2}+a x+\frac{a^{2}}{4}$
(ii) $\mathrm{c}^{2}=\mathrm{p}^{2}-\mathrm{ax}+\frac{a^{2}}{4}$
(iii) $\mathrm{b}^{2}+\mathrm{c}^{2}=2 \mathrm{p}^{2}+\frac{a^{2}}{2}$
Solution:
From the figure, $D$ is the mid point of $B C$.

We have $\angle \mathrm{AED}=90^{\circ}$
$\therefore \angle \mathrm{ADE}<90^{\circ}$ and $\angle \mathrm{ADC}>90^{\circ}$
i.e. $\angle \mathrm{ADE}$ is acute and $\angle \mathrm{ADC}$ is obtuse,
(i) In $\triangle \mathrm{ADC}, \angle \mathrm{ADC}$ is obtuse angle.
$\begin{aligned}
&\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}{ }^{2}+2 \mathrm{DC} \times \mathrm{DE} \\
&\Rightarrow \mathrm{AC}^{2}=\mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}{ }^{2}+2 \cdot \frac{1}{2} \mathrm{BC} \cdot \mathrm{DE} \\
&\Rightarrow \mathrm{AC}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{BC} \mathrm{C}^{2}+\mathrm{BC} \cdot \mathrm{DE} \\
&\Rightarrow \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} \mathrm{BC}^{2} \\
&\Rightarrow \mathrm{b}^{2}=\mathrm{p}^{2}+\mathrm{ax}+\frac{1}{4} \mathrm{a}^{2}
\end{aligned}$
Hence proved.
(ii) In $\triangle \mathrm{ABD}, \angle \mathrm{ADE}$ is an acute angle.
$\begin{aligned}
&\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}-2 \mathrm{BD} \cdot \mathrm{DE} \\
&\Rightarrow \mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}-2 \times \frac{1}{2} \mathrm{BC} \cdot \mathrm{DE} \\
&\Rightarrow \mathrm{AB}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{BC}^{2}-\mathrm{BC} \cdot \mathrm{DE} \\
&\Rightarrow \mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} \mathrm{BC}^{2} \\
&\Rightarrow \mathrm{c}^{2}=\mathrm{p}^{2}-\mathrm{ax}+\frac{1}{4} \mathrm{a}^{2}
\end{aligned}$
Hence proved.
(iii) From (i) and (ii) we get.
$\mathrm{AB}^{2}+\mathrm{AC}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$
i.e. $\mathrm{c}^{2}+\mathrm{b}^{2}=2 \mathrm{p}^{2}+\frac{a^{2}}{2}$
Hence it is proved.

Question $7 .$
A man whose eye-level is $2 \mathrm{~m}$ above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground $20 \mathrm{~m}$ from the tree and finds that if he stands at a point $\mathrm{C}$ which is $4 \mathrm{~m}$ from the mirror $\mathrm{B}$, he can see the reflection of the top of the tree. How height is the tree?
Solution:
From the figure; $\triangle \mathrm{DAC}, \triangle \mathrm{FBC}$ are similar triangles and $\triangle \mathrm{ACE} \& \triangle \mathrm{ABF}$ are similar triangles.

$\therefore$ From $\Delta \mathrm{ACD}$ and $\Delta \mathrm{BCF}$
$\begin{aligned}
\frac{\mathrm{AD}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BC}} & \Rightarrow \frac{h}{x}=\frac{20+4}{4} \\
\frac{h}{x} &=\frac{24}{4}=6 . \\
h &=6 x .
\end{aligned}$
From $\triangle \mathrm{ACE} \& \Delta \mathrm{ABF}$
$\begin{aligned}
\frac{2}{x}=\frac{20+4}{20} & \Rightarrow \frac{2}{x}=\frac{24^{6}}{20_{5}} \\
6 x &=10
\end{aligned}$
$\therefore$ height of the tree $h=6 x=10 \mathrm{~m}$.
 

Question $8 .$
An emu which is $8 \mathrm{ft}$ tall standing at the foot of a pillar which is $30 \mathrm{ft}$ height. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Solution:
Let $\mathrm{OA}$ (emu shadow) the $\mathrm{x}$ and $\mathrm{AB}=\mathrm{y}$.
$\Rightarrow$ pillar's shadow $=\mathrm{OB}=\mathrm{OA}+\mathrm{AB}$

From basic proportionality theorem,
$\begin{aligned}
\frac{\mathrm{OA}}{\mathrm{OB}} &=\frac{\mathrm{AD}}{\mathrm{BC}} \\
\frac{x}{x+y} &=\frac{8}{30}
\end{aligned}$
Reciprocating on both sides, we get
$\begin{aligned}
&\Rightarrow \frac{x}{x+y}=\frac{30}{8} \Rightarrow 1+\frac{y}{x}=\frac{30}{8} \Rightarrow \frac{y}{x}=\frac{30}{8}-1 \\
&\Rightarrow \frac{y}{x}=\frac{30-8}{8} \Rightarrow \frac{y}{x}=\frac{22}{8} \Rightarrow \frac{y}{x}=\frac{11}{4} \\
&\Rightarrow x=\frac{4}{11} \times y \Rightarrow \text { shadow }=\frac{4}{11} \times \text { distance }
\end{aligned}$

Question $9 .$
Two circles intersect at $\mathrm{A}$ and $\mathrm{B}$. From a point $\mathrm{P}$ on one of the circles lines $\mathrm{PAC}$ and $\mathrm{PBD}$ are drawn intersecting the second circle at $C$ and $D$. Prove that $C D$ is parallel to the tangent at $P$.
Solution:
Let $X Y$ be the tangent at $P$.
TPT: CD is || to XY.
Construction: Join $\mathrm{AB}$.

$\mathrm{ABCD}$ is a cyclic quadilateral.
$\angle \mathrm{BAC}+\angle \mathrm{BDC}=180^{\circ}$
$\angle \mathrm{BDC}=180^{\circ}-\angle \mathrm{BAC}$
Equating (1) and (2)
we get $\angle \mathrm{BDC}=\angle \mathrm{PAB}$
Similarly we get $\angle \mathrm{PBA}=\angle \mathrm{ACD}$
as $X Y$ is tangent to the circle at ' $P$ '
$\angle \mathrm{BPY}=\angle \mathrm{PAB}$ (by alternate segment there)
$\therefore \angle \mathrm{PAB}=\angle \mathrm{PDC}$
$\angle B P Y=\angle P D C$
$X Y$ is parallel of CD.
Hence proved.

Question $10 .$
Let $A B C$ be a triangle and $D, E, F$ are points on the respective sides $A B, B C, A C$ (or their extensions). Let $\mathrm{AD}: \mathrm{DB}=5: 3, \mathrm{BE}: \mathrm{EC}=3: 2$ and $\mathrm{AC}=21$. Find the length of the line segment CF.

Solution

$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5}{3}, \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{3}{2},$
$\mathrm{AC}=21 \Rightarrow \frac{\mathrm{CF}}{\mathrm{FA}}=\frac{\mathrm{CF}}{21-\mathrm{CF}}$
$\therefore$ By Ceva's theorem,
$\frac{\mathrm{BE}}{\mathrm{EC}} \times \frac{\mathrm{CF}}{\mathrm{FA}} \times \frac{\mathrm{AD}}{\mathrm{DB}}=+1$
$\frac{\not \beta}{2} \times \frac{\mathrm{CF}}{21-\mathrm{CF}} \times \frac{5}{\not \underline{p}}=+1$
$\frac{\mathrm{CF}}{21-\mathrm{CF}}=\frac{2}{5}$

$5 \mathrm{CF}=42-2 \mathrm{CF}$
$5 \mathrm{CF}+2 \mathrm{CF}=42$
$7 \mathrm{CF}=42$
$C F=\frac{42}{7}=6$ units