Exercise 5.2 - Chapter 5 Geometry 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 5.2$
Question $1 .$
Fill in the blanks:
(i) If in a $\triangle \mathrm{PQR}, \mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{QR}^{2}$, then the right angle of $\triangle \mathrm{PQR}$ is at the vertex
Answer:
Q

(ii) If ' $l$ ' and ' $m$ ' are the legs and ' $n$ ' is the hypotenuse of a right angled triangle then, $1^{2}=$
Answer:
$n^{2}-m^{2}$

(iii) If the sides of a triangle are in the ratio $5: 12: 13$ then, it is
Answer:
a right angled triangle
$\begin{aligned}
&\text { Hint: } \\
&13^{2}=169 \\
&5^{2}=25 \\
&12^{2}=144 \\
&169=25+144 \\
&\therefore 13^{2}=5^{2}+12^{2}
\end{aligned}$
(iv) The medians of a triangle cross cach other at
Answer:
Centroid
(v) The centroid of a triangle divides each medians in the ratio
Answer:
$2: 1$

Question $2 .$
Say True or False.
(i) $8,15,17$ is a Pythagorean triplet.
Answer:
True
Hint:
$17^{2}=289$
$15^{2}=225$

$\begin{aligned}
&8^{2}=64 \\
&64+225=289 \Rightarrow 17^{2}=15^{2}+8^{2}
\end{aligned}$
(ii) In a right angled triangle, the hypotenuse is the greatest side.
Answer:
True

(iii) In any triangle the centroid and the incentre are located inside the triangle.
Answer:
True
(iv) The centroid, orthocentre, and incentre of a triangle are collinear.
Answer:
True
(v) The incentre is equidistant from all the vertices of a triangle.
Answer:
False
 

Question $3 .$
Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(i) $8,15,17$
Answer:
Take $\mathrm{a}=8 \mathrm{~b}=15$ and $\mathrm{c}=17$
Now $\mathrm{a}^{2}+\mathrm{b}^{2}=8^{2}+15^{2}=64+225=289$
$172=289=c^{2}$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}$
By the converse of Pythagoras theorem, the triangle with given measures is a right
angled triangle.
$\therefore$ Ans: yes

(ii) $12,13,15$
Answer:
(ii) $12,13.15$
Take $\mathrm{a}=12, \mathrm{~b}=13$ and $\mathrm{c}=15$
Now $\mathrm{a}^{2}+\mathrm{b}^{2}=12^{2}+13^{2}=144+169=313$
$15^{2}=225 \neq 313$
By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
$\therefore$ Ans: No.
(iii) $30,40,50$
Answer:
Take $\mathrm{a}=30, \mathrm{~b}=40$ and $\mathrm{c}=50$
Now $\mathrm{a}^{2}+\mathrm{b}^{2}=30^{2}+40^{2}=900+1600=2500$
$\mathrm{C}^{2}=50^{2}=2500$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}$
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
$\therefore$ Ans: yes
(iv) $9,40,41$
Answer:
Take $\mathrm{a}=9, \mathrm{~b}=40$ and $\mathrm{c}=41$
Now $\mathrm{a}^{2}+\mathrm{b}^{2}=9^{2}+40^{2}=81+1600=1681$
$\mathrm{c}^{2}=41^{2}=1681$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}$
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
$\therefore$ Ans: yes

(v) $24,45,51$
Answer:
Take $a=24, b=45$ and $c=51$
Now $\mathrm{a}^{2}+\mathrm{b}^{2}=24^{2}+45^{2}=576+2025=2601$
$\begin{aligned}
&\mathrm{c}^{2}=5 \mathrm{l}^{2}=2601 \\
&\therefore \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}
\end{aligned}$
By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
$\therefore$ Ans: yes

Question $4 .$
Find the unknown side in the following triangles.
(i)

Answer:
From $\triangle \mathrm{ABC}$, by Pythagoras theorem
$\mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{AC} \mathrm{C}^{2}$
Take $\mathrm{AB}^{2}+\mathrm{AC}^{2}=9^{2}+40^{2}=81+1600=1681$
$\begin{aligned}
&\mathrm{BC}^{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}=1681=41^{2} \\
&\mathrm{BC}^{2}=41^{2} \Rightarrow \mathrm{BC}=41 \\
&\therefore \mathrm{x}=41
\end{aligned}$

Answer:
From $\triangle \mathrm{PQR}$, by Pythagoras theorem.
$\begin{aligned}
&\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{QR}^{2} \\
&34^{2}=\mathrm{y}^{2}+30^{2} \\
&\Rightarrow \mathrm{y}^{2}=34^{2}-30^{2} \\
&=1156-900 \\
&=256=16^{2} \\
&\mathrm{y}^{2}=16^{2} \Rightarrow \mathrm{y}=16
\end{aligned}$

Answer:
From $\triangle \mathrm{XYZ}$, by Pythagoras theorem,
$\begin{aligned}
&=\mathrm{YZ}^{2}=\mathrm{XY} \mathrm{Y}^{2}+\mathrm{XZ} \\
&\Rightarrow \mathrm{XY}=\mathrm{YZ}^{2}-\mathrm{XZ} \\
& \\
&\mathrm{Z}^{2}=39^{2}-36^{2} \\
&=1521-1296 \\
&=225=15^{2} \\
&\mathrm{z}^{2}=15^{2} \\
&\Rightarrow \mathrm{z}=15
\end{aligned}$

Question $5 .$
An isosceles triangle has equal sides each $13 \mathrm{~cm}$ and a base $24 \mathrm{~cm}$ in length. Find its height.

Answer:
In an isosceles triangle the altitude dives its base into two equal parts.
Now in the figure, $\triangle \mathrm{ABC}$ is an isosceles triangle with $\mathrm{AD}$ as its height In the figure, $\mathrm{AD}$ is the altitude and $\triangle \mathrm{ABD}$ is a right triangle.
By Pythagoras theorem,
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$
$\Rightarrow \mathrm{AD}^{2}=\mathrm{AB}^{2}-\mathrm{BD}^{2}$
$=13^{2}-12^{2}=169-144=25$
$\mathrm{AD}^{2}=25=5^{2}$
Height: $\mathrm{AD}=5 \mathrm{~cm}$
 

Question $6 .$
Find the distance between the helicopter and the ship.

Answer:
From the figure AS is the distance between the helicopter and the ship.
$\triangle \mathrm{APS}$ is a right angled triangle, by Pythagoras theorem,
$\begin{aligned}
&\mathrm{AS}^{2}=\mathrm{AP}^{2}+\mathrm{PS}^{2} \\
&=80^{2}+150^{2}=6400+22500=28900=170^{2}
\end{aligned}$
$\therefore$ The distance between the helicopter and the ship is $170 \mathrm{~m}$
 

Question $7 .$
In triangle $\mathrm{ABC}$, line $\mathrm{I}$, is a perpendicular bisector of $\mathrm{BC}$. If $\mathrm{BC}=12 \mathrm{~cm}, \mathrm{SM}=8 \mathrm{~cm}$, find $\mathrm{CS}$.

Answer:
Given $\mathrm{l}_{1}$, is the perpendicular bisector of $\mathrm{BC}$.
$\therefore \angle \mathrm{SMC}=90^{\circ} \text { and } \mathrm{BM}=\mathrm{MC}$
$\mathrm{BC}=12 \mathrm{~cm}$
$\Rightarrow \mathrm{BM}+\mathrm{MC}=12 \mathrm{~cm}$
$\mathrm{MC}+\mathrm{MC}=12 \mathrm{~cm}$
$2 \mathrm{MC}=12$
$\mathrm{MC}=\frac{12}{2}$
$\mathrm{MC}=6 \mathrm{~cm}$
Given $\mathrm{SM}=8 \mathrm{~cm}$
By Pythagoras theorem $\mathrm{SC}^{2}=\mathrm{SM}^{2}+\mathrm{MC}^{2}$
$\begin{aligned}
&\mathrm{SC}^{2}=8^{2}+6^{2} \\
&\mathrm{SC}^{2}=64+36 \\
&\mathrm{CS}^{2}=100 \\
&\mathrm{CS}^{2}=10^{2} \\
&\mathrm{CS}=10 \mathrm{~cm}
\end{aligned}$

Question 8.
Identify the centroid of $\triangle \mathrm{PQR}$.

Answer:
In $\triangle \mathrm{PQR}, \mathrm{PT}=\mathrm{TR} \Rightarrow \mathrm{QT}$ is a median from vertex $\mathrm{Q}$.
$\mathrm{QS}=\mathrm{SR} \Rightarrow \mathrm{PS}$ is a median from vertex $\mathrm{P}$.
QT and PS meet at $W$ and therefore $W$ is the centroid of $\triangle P Q R$.
 

Question $9 .$
Name the orthocentre of $\triangle \mathrm{PQR}$.

Answer:
This is a right triangle
$\therefore$ orthocentre $=\mathrm{P}\left[\because\right.$ In right triangle orthocentre is the vertex containing $\left.90^{\circ}\right]$
 

Question $10 .$
In the given figure, $A$ is the midpoint of $Y Z$ and $G$ is the centroid of the triangle $X Y Z$. If the length of $\mathrm{GA}$ is $3 \mathrm{~cm}$, find $X A$.

Answer:
Given $A$ is the midpoint of $Y Z$.
$\therefore \mathrm{ZA}=\mathrm{AY}$
$\mathrm{G}$ is the centroid of $\mathrm{XYZ}$ centroid divides each median in a ratio $2: 1 \Rightarrow \mathrm{XG}: \mathrm{GA}=$
$\begin{aligned}
&2: 1 \\
&\frac{\mathrm{XG}}{\mathrm{GA}}=\frac{2}{1} \\
&\frac{\mathrm{XG}}{3}=\frac{2}{1} \\
&\mathrm{XG}=2 \times 3 \\
&\mathrm{XG}=6 \mathrm{~cm} \\
&\mathrm{XA}=\mathrm{XG}+\mathrm{GA}=6+3 \Rightarrow \mathrm{XA}=9 \mathrm{~cm}
\end{aligned}$

Question 11.
If $I$ is the incentre of $\triangle \mathrm{XYZ}, \angle \mathrm{IYZ}=30^{\circ}$ and $\angle \mathrm{IZY}=40^{\circ}$, find $\angle \mathrm{YXZ}$.

Answer:
Since I is the incentre of AXYZ
$\begin{aligned}
&\angle \mathrm{IYZ}=30^{\circ} \Rightarrow \angle \mathrm{IYX}=30^{\circ} \\
&\angle \mathrm{IZY}=40^{\circ} \Rightarrow \angle \mathrm{IZX}=40^{\circ} \\
&\therefore \angle \mathrm{XYZ}=\angle \mathrm{XYI}+\angle \mathrm{IYZ}=30^{\circ}+30^{\circ}
\end{aligned}$

Objective Type Questions
Question $12 .$
If $\triangle \mathrm{GUT}$ is isosceles and right angled, then $\angle \mathrm{TUG}$ is

(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $45^{\circ}$
(D) $55^{\circ}$
Answer:
(C) $45^{\circ}$
Hint:
$\angle \mathrm{U} \angle \mathrm{T}=45^{\circ}(\because \mathrm{GUT}$ is an isosceles given $)$
$\therefore \angle \mathrm{TUG}=45^{\circ}$
 

Question $13 .$
The hypotenuse of a right angled triangle of sides $12 \mathrm{~cm}$ and $16 \mathrm{~cm}$ is
(A) $28 \mathrm{~cm}$
(B) $20 \mathrm{~cm}$
(C) $24 \mathrm{~cm}$
(D) $21 \mathrm{~cm}$
Answer:
(B) $20 \mathrm{~cm}$
Hint:
Side take $a=12 \mathrm{~cm}$
$\mathrm{b}=16 \mathrm{~cm}$
The hypotenuse $\mathrm{c}^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}$
$\begin{aligned}
&=12^{2}+16^{2} \\
&=144+256 \\
&\mathrm{c}^{2}=400 \Rightarrow \mathrm{c}=20 \mathrm{~cm}
\end{aligned}$

Question $14 .$
The area of a rectangle of length $21 \mathrm{~cm}$ and diagonal $29 \mathrm{~cm}$ is
(A) $609 \mathrm{~cm}^{2}$
(B) $580 \mathrm{~cm}^{2}$
(C) $420 \mathrm{~cm}^{2}$
(D) $210 \mathrm{~cm}^{2}$
Answer:
(C) $420 \mathrm{~cm}^{2}$

Question $15 .$
The sides of a right angled triangle are in the ratio $5: 12: 13$ and its perimeter is 120 units then, the sides are.
(A) $25,36,59$
(B) $10,24,26$
(C) $36,39,45$
(D) $20,48,52$
Answer:
(D) $20,48,52$