Exercise 5.3 - Chapter 5 Geometry 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 5.3$
Question $1 .$
In the figure, given that $\angle 1=\angle 2$ and $\angle 3 \equiv \angle 4$. Prove that $\Delta \mathrm{MUG} \equiv \triangle T U B$.

Answer:

Question $2 .$
From the figure, prove that $\triangle \mathrm{SUN} \sim \Delta \mathrm{RAY}$.

Answer:
Proof: from the $\triangle \mathrm{SUN}$ and $\triangle \mathrm{RAY}$
$\begin{aligned}
&\mathrm{SU}=10 \\
&\mathrm{UN}=12 \\
&\mathrm{SN}=14 \\
&\mathrm{RA}=5 \\
&\mathrm{AY}=6 \\
&\mathrm{RY}=7
\end{aligned}$
We have $\frac{\mathrm{SU}}{\mathrm{RA}}=\frac{10}{5}=\frac{2}{1}$
$\begin{aligned}
&\frac{\mathrm{UN}}{\mathrm{AY}}=\frac{12}{6}=\frac{2}{1} \\
&\frac{\mathrm{SN}}{\mathrm{RY}}=\frac{14}{7}=\frac{2}{1}
\end{aligned}$
From (1), (2) and (3) we have
$\frac{\mathrm{SU}}{\mathrm{RA}}=\frac{\mathrm{UN}}{\mathrm{AY}}=\frac{\mathrm{SN}}{\mathrm{RY}}=\frac{2}{1}$
The sides are proportional $\therefore \triangle \mathrm{SUN} \sim \triangle \mathrm{RAY}$
 

Question $3 .$
The height of a tower is measured by a mirror on the ground at $R$ by which the top of the tower's reflection is seen. Find the height of the tower. If $\triangle \mathrm{PQR} \sim \triangle \mathrm{STR}$

Answer:
The image and its reflection make similar shapes
$\therefore \triangle \mathrm{PQR} \sim \triangle \mathrm{STR}$
$\frac{\mathrm{PQ}}{\mathrm{ST}}=\frac{\mathrm{QR}}{\mathrm{TR}}=\frac{\mathrm{PR}}{\mathrm{SR}}$
$\frac{\mathrm{PQ}}{\mathrm{ST}}=\frac{\mathrm{QR}}{\mathrm{TR}}$
$\begin{aligned}
&\Rightarrow \frac{h}{8}=\frac{60}{10} \\
&h=\frac{60}{10} \times 8 \\
&=48 \text { fect }
\end{aligned}$
$\therefore$ Height of the tower $=48$ feet.
 

Question $4 .$
Find the length of the support cable required to support the tower with the floor.

From the figure, by Pythagoras theorem,
$\begin{aligned}
&x^{2}=20^{2}+15^{2} \\
&=400+225=625 \\
&x^{2}=25^{2} \Rightarrow x=25 \mathrm{ft} .
\end{aligned}$
$\therefore$ The length of the support cable required to support the tower with the floor is $25 \mathrm{ft}$.
 

Question $5 .$
Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason.
Answer:

Take the sides of a right angled triangle $\triangle \mathrm{ABC}$ as
$a=7$ inches
$\mathrm{b}=25$ inches
$\mathrm{c}=\text { ? }$
By Pythagoras theorem,
$\begin{aligned}
&\mathrm{b}^{2}=\mathrm{a}^{2}+\mathrm{c}^{2} \\
&25^{2}=7^{2}+\mathrm{c}^{2} \\
&\Rightarrow \mathrm{c}^{2}=25^{2}-7^{2}=625-49=576 \\
&\therefore \mathrm{c}^{2}=24^{2} \\
&\Rightarrow \mathrm{c}=24 \text { inches }
\end{aligned}$
$\therefore$ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie. 24 inches.
$\therefore$ The TV will not fit into the cabinet.
 

Challenging Problems
Question $6 .$
In the figure, $\angle \mathrm{TMA} \equiv \angle \mathrm{IAM}$ and $\angle \mathrm{TAM} \equiv \angle \mathrm{IMA}$. P is the midpoint of MI and $\mathrm{N}$ is the midpoint of Al. Prove that $\triangle$ PIN $\sim \triangle$ ATM.

Answer:

Question $7 .$
In the figure, if $\angle \mathrm{FEG} \equiv \angle 1$ then, prove that $\mathrm{DG}^{2}=\mathrm{DE}$.DF.

Answer:

Question $8 .$
The diagonals of the rhombus is $12 \mathrm{~cm}$ and $16 \mathrm{~cm}$. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).
Answer:

Here $\mathrm{AO}=\mathrm{CO}=8 \mathrm{~cm}$
$\mathrm{BO}=\mathrm{DO}=6 \mathrm{~cm}$
(athe diagonals of rhombus bisect each other at right angles)
$\begin{aligned}
&\therefore \text { In } \Delta \mathrm{AOB}, \mathrm{AB}^{2}=\mathrm{AO}^{2}+\mathrm{OB}^{2} \\
&=8^{2}+6^{2}=64+36 \\
&=100=10^{2} \\
&\therefore \mathrm{AB}=10
\end{aligned}$
Since it is a rhombus, all the four sides are equal.
$\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$
$\therefore$ Its Perimeter $=10+10+10+10=40 \mathrm{~cm}$
 

Question $9 .$
In the figure, find $A R$.

Answer:
$\triangle \mathrm{AFI}, \Delta \mathrm{FRI}$ are right triangles.
By Pythagoras theorem,
$\begin{aligned}
&\mathrm{AF}^{2}=\mathrm{AI}^{2}-\mathrm{FI}^{2} \\
&=25^{2}-15^{2} \\
&=625-225=400=20^{2} \\
&\therefore \mathrm{AF}=20 \mathrm{ft} \\
&\mathrm{FR}^{2}=\mathrm{RI}^{2}-\mathrm{FI}^{2} \\
&=17^{2}-15^{2}=289-225=64=8^{2} \\
&\mathrm{FR}=8 \mathrm{ft} . \\
&\therefore \mathrm{AR}=\mathrm{AF}+\mathrm{FR} \\
&=20+8=28 \mathrm{ft} .
\end{aligned}$

Question $10 .$
In $\triangle \mathrm{DEF}, \mathrm{DN}, \mathrm{EO}, \mathrm{FM}$ are medians and point $\mathrm{P}$ is the centroid. Find the following.
(i) $\mathrm{IF} \mathrm{DE}=44$, then $\mathrm{DM}=$ ?
(ii) $\mathrm{IFPD}=12$, then $\mathrm{PN}=$ ?
(iii) IfDO =8, then $P D=$ ?
(iv) $\mathrm{IF} 0 \mathrm{E}=36$ then $\mathrm{EP}=$ ?

Answer:
Given DN, EO, FM are medians.
$\begin{aligned}
&\therefore \mathrm{FN}=\mathrm{EN} \\
&\mathrm{DO}=\mathrm{FO} \\
&\mathrm{EM}=\mathrm{DM}
\end{aligned}$
(i) If $\mathrm{DE}=44$,then
$\begin{aligned}
&\mathrm{DM}=\frac{44}{2}=22 \\
&\mathrm{DM}=22 \\
&\text { (ii) If } \mathrm{PD}=12, \mathrm{PN}=? \\
&\frac{P D}{P N}=\frac{2}{1} \\
&\frac{12}{\mathrm{PN}}=\frac{2}{1} \Rightarrow \mathrm{PN}=\frac{12}{2}=6 \\
&\mathrm{PN}=6
\end{aligned}$
$\begin{aligned}
&\text { (iii) If } \mathrm{DO}=8 \text {, then } \\
&\mathrm{FD}=\mathrm{DO}+\mathrm{OF} \\
&=8+8 \\
&\mathrm{FD}=16
\end{aligned}$

(iv) If $\mathrm{OE}=36$
$\begin{aligned}
\text { then } \frac{\mathrm{EP}}{\mathrm{PO}} &=\frac{2}{1} \\
\frac{\mathrm{EP}}{2} &=\mathrm{PO} \\
\mathrm{OE} &=\mathrm{OP}+\mathrm{PE} \\
36 &=\frac{\mathrm{PE}}{2}+\mathrm{PE} \\
36 &=\frac{\mathrm{PE}}{2}+\frac{2 \mathrm{PE}}{2} \\
36 &=\frac{3 \mathrm{PE}}{2} \\
\mathrm{PE} &=\frac{36 \times 2}{3} \\
\mathrm{PE}=24 &=\frac{\mathrm{P}}{2}
\end{aligned}$