Exercise 4.3 - Chapter 4 Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Question 1.
A man goes $18 \mathrm{~m}$ due east and then $24 \mathrm{~m}$ due north. Find the distance of his current position from the starting point?
Solution:
Using Pythagoras theorem

$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$=(18)^{2}+(24)^{2}$
$=324+576$
$=900$
$\mathrm{AC}=\sqrt{900}=30 \mathrm{~m}$
$\therefore$ The distance from the starting point is $30 \mathrm{~m}$.
 

Question $2 .$
There are two paths that one can choose to go from Sarah's house to James house. One way is to take $\mathrm{C}$ street, and the other way requires to take A street and then B street. How much shorter is the direct path along $\mathrm{C}$ street? (Using figure).

Solution:
By using Pythagoras theorem
$A C^{2}=A B^{2}+B C^{2}$ $=2^{2}+(1.5)^{2}$ $=4+2.25$ $=6.25$ $\mathrm{AC}=2.5$ miles.
If one chooses $\mathrm{C}$ street the distance from James house to Sarah's house is $2.5$ miles If one chooses $\mathrm{A}$ street and $\mathrm{B}$ street he has to go $2+1.5=3.5$ miles.
$2.5<3.5,3.5-2.5=1$ Through $\mathrm{C}$ street is shorter by $1.0$ miles.
$\therefore$ The direct path along $\mathrm{C}$ street is shorter by 1 mile.
 

Question $3 .$
To get from point A to point B you must avoid walking through a pond. You must walk $34 \mathrm{~m}$ south and $41 \mathrm{~m}$ east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
By using Pythagoras

$\begin{aligned}
&\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \\
&=34^{2}+41^{2} \\
&=1156+1681 \\
&=2837 \\
&\mathrm{AC}=53.26 \mathrm{~m}
\end{aligned}$
Through B one must walk $34+41=75 \mathrm{~m}$ walking through a pond one must comes only $53.2 \mathrm{~m}$
$\therefore$ The difference is $(75-53.26) \mathrm{m}=21.74 \mathrm{~m}$
$\therefore$ To the nearest, one can save $21.74 \mathrm{~m}$.
 

Question $4 .$
In the rectangle $W X Y Z, X Y+Y Z=17 \mathrm{~cm}$, and $X Z+Y W=26 \mathrm{~cm}$. Calculate the length and breadth of the rectangle?

$\begin{aligned}
&\text { Solution: } \\
&\mathrm{XY}+\mathrm{YZ}=17 \mathrm{~cm} \\
&\mathrm{XZ}+\mathrm{YW}=26 \mathrm{~cm} \ldots \ldots \ldots(\mathrm{l}) \\
&(2) \Rightarrow \mathrm{XZ}=13, \mathrm{YW}=13 \\
&(\because \text { In rectangle diagonals are equal). } \\
&(1) \Rightarrow \mathrm{XY}=5, \mathrm{YZ}=12 \mathrm{XY}+\mathrm{YZ}=17 \\
&\Rightarrow \mathrm{U} \text { ing } \mathrm{Pythagoras} \text { theorem } \\
&5^{2}+12^{2}=25+144=169=13^{2} \\
&\therefore \text { In } \Delta \mathrm{XYZ}=13^{2}=5^{2}+12^{2} \text { it is verified } \\
&\therefore \text { The length is } 12 \mathrm{~cm} \text { and the breadth is } 5 \mathrm{~cm} \text {. } \\
&\text { Question } 5 \text {. } \\
&\text { The hypotenuse of a right triangle is } 6 \mathrm{~m} \text { more than twice of the shortest side. If the third side is } 2 \\
&\mathrm{~m} \text { less than the hypotenuse, find the sides of the triangle? } \\
&\text { Solution: } \\
&\text { Let a is the shortest side. } \\
&\mathrm{c} \text { is the hypotenuse } \\
&\mathrm{b} \text { is the third side. }
\end{aligned}$

$\begin{aligned}
c &=2 a+6 \\
b &=c-2 \\
&=2 a+6-2 \\
&=2 a+4 \\
c^{2} &=a^{2}+b^{2} \\
&(\text { Using pythagoras theorem }) \\
&=a^{2}+(2 a+4)^{2} \\
(2 a+6)^{2} &=a^{2}+(2 a)^{2}+2(2 a) 4+4^{2} \\
24+2(2 a)(6)+& 6^{2}=a^{2}+(2 a)^{2}+16 a+16 \\
a^{2}+16 a-24 a+16 &-36=0 \\
a^{2}-8 a-20 &=0 \\
(a-10)(a+2) &=0 \\
a &=10,-2 . \\
b &=2 a+4=2(10)+4=24 \mathrm{~m} \\
c &=2 a+6=2(10)+6=26 \mathrm{~m}
\end{aligned}$
$\therefore$ The sides of the triangle are $10 \mathrm{~m}, 24 \mathrm{~m}, 26 \mathrm{~m}$.
Verification $26^{2}=10^{2}+24^{2}$
$676=100+576=676$

Question $6 .$
$5 \mathrm{~m}$ long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point $4 \mathrm{~m}$ high. If the foot of the ladder is moved $1.6 \mathrm{~m}$ towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
Let the distance by which top of the slide moves upwards be assumed as ' $x$ '.

From the diagram, $\mathrm{DB}=\mathrm{AB}-\mathrm{AD}$
$=3-1.6 \Rightarrow \mathrm{DB}=1.4 \mathrm{~m}$
also $\mathrm{BE}=\mathrm{BC}+\mathrm{CE}$
$=4+x$
$\therefore \mathrm{DBE}$ is a right angled triangle
$\begin{aligned}
&\mathrm{DB}^{2}+\mathrm{BE}^{2}=\mathrm{DE}^{2} \Rightarrow(1.4)^{2}+(4+\mathrm{x})^{2}=5^{2} \\
&\Rightarrow(4+\mathrm{x})^{2}=25-1.96 \Rightarrow(4+\mathrm{x})^{2}=23.04 \\
&\Rightarrow 4+\mathrm{x}=\sqrt{23.04}=4.8 \\
&\Rightarrow \mathrm{x}=4.8-4 \Rightarrow \mathrm{x}=0.8 \mathrm{~m}
\end{aligned}$

Question $7 .$
The perpendicular $P S$ on the base $Q R$ of $\triangle P Q R$ intersects $Q R$ at $S$, such that $Q S=3 S R$. Prove that $2 \mathrm{PQ}^{2}=2 \mathrm{PR}^{2}+\mathrm{QR}^{2}$
Solution:

In $\triangle \mathrm{PQS}$,
$\mathrm{PQ}^{2}=\mathrm{PS}^{2}+\mathrm{QS}^{2}$
In $\triangle \mathrm{PSR}$,
$\mathrm{PR}^{2}=\mathrm{PS}^{2}+\mathrm{SR}^{2}$
(1) $-(2) \Rightarrow \mathrm{PQ}^{2}-\mathrm{PR}^{2}=\mathrm{QS}^{2}-\mathrm{SR}^{2}$
$\therefore(3)$
$\begin{aligned}
\Rightarrow \mathrm{PQ}^{2}-\mathrm{PR}^{2} &=\frac{9}{16} \mathrm{QR}^{2}-\frac{\mathrm{QR}^{2}}{16} \\
&=\frac{8 \mathrm{QR}^{2}}{16}=\frac{\mathrm{QR}^{2}}{2} \\
2 \mathrm{PQ}^{2}-2 \mathrm{PR}^{2} &=\mathrm{QR}^{2} \\
2 \mathrm{PQ}^{2} &=\mathrm{QR}^{2}+2 \mathrm{PR}^{2}
\end{aligned}$
Hence it proved.

Question 8.
In the adjacent figure, $\mathrm{ABC}$ is a right-angled triangle with right angle at $\mathrm{B}$ and points $\mathrm{D}, \mathrm{E}$ trisect $\mathrm{BC}$. Prove that $8 \mathrm{AE}^{2}=3 \mathrm{AC}^{2}+5 \mathrm{AD}^{2}$

Solution:
Since $D$ and $E$ are the points of trisection of $B C$,
therefore $\mathrm{BD}=\mathrm{DE}=\mathrm{CE}$
Let $\mathrm{BD}=\mathrm{DE}=\mathrm{CE}=\mathrm{x}$
Then $B E=2 x$ and $B C=3 x$
In right triangles $\mathrm{ABD}, \mathrm{ABE}$ and $\mathrm{ABC}$, (using Pythagoras theorem)
We have $A D^{2}=A B^{2}+B D^{2}$
$\Rightarrow \mathrm{AD}^{2}=\mathrm{AB}^{2}+\mathrm{x}^{2} \ldots \ldots \ldots \ldots(1)$
$\mathrm{AE}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2}$
$\Rightarrow \mathrm{AB}^{2}+(2 \mathrm{x})^{2}$
$\Rightarrow \mathrm{AE}^{2}=\mathrm{AB}^{2}+4 \mathrm{x}^{2} \ldots \ldots \ldots(2)$
and $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AB}^{2}+(3 \mathrm{x})^{2}$
$\mathrm{AC}^{2}=\mathrm{AB}^{2}+9 \mathrm{x}^{2}$
Now $8 \mathrm{AE}^{2}-3 \mathrm{AC}^{2}-5 \mathrm{AD}^{2}=8\left(\mathrm{AB}^{2}+4 \mathrm{x}^{2}\right)-3\left(\mathrm{AB}^{2}+9 \mathrm{x}^{2}\right)-5\left(\mathrm{AB}^{2}+\mathrm{x}^{2}\right)$ $=8 \mathrm{AB}^{2}+32 \mathrm{x}^{2}-3 \mathrm{AB}^{2}-27 \mathrm{x}^{2}-5 \mathrm{AB}^{2}-5 \mathrm{x}^{2}$
$=0$
$\therefore 8 \mathrm{AE}^{2}-3 \mathrm{AC}^{2}-5 \mathrm{AD}^{2}=0$
$8 \mathrm{AE}^{2}=3 \mathrm{AC}^{2}+5 \mathrm{AD}^{2}$
Hence it is proved.