Exercise 4.1 - Chapter 4 Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 4.1
Question 1.
Check whether the which triangles are similar and find the value of $x$.

Solution:
(i) $\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AB}}$ (for similar triangle)
But here,
$\therefore$ They are not similar triangles
(ii) In $\triangle \mathrm{ABC}, \triangle \mathrm{PQC}$,
$\begin{aligned}
\angle \mathrm{ABC} &=\angle \mathrm{PQC}=70^{\circ} \\
\angle \mathrm{C} &=\angle \mathrm{C} \text { (common angles) } \\
\therefore \angle \mathrm{A} &=\angle \mathrm{QPC}(\therefore \text { AAA criterion) }
\end{aligned}$
$\therefore \triangle \mathrm{ABC}$ and $\triangle \mathrm{PQC}$ are similar triangles
$\begin{aligned}
\frac{\mathrm{AB}}{\mathrm{PQ}} &=\frac{\mathrm{BC}}{\mathrm{QC}} \\
\frac{5}{x} &=\frac{6}{3} \\
6 x &=15
\end{aligned}$

Question $2 .$
A girl looks the reflection of the top of the lamp post on the mirror which is $6.6 \mathrm{~m}$ away from the foot of the lamppost. The girl whose height is $1.25 \mathrm{~m}$ is standing $2.5 \mathrm{~m}$ away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post?
Solution:
In the picture $\triangle \mathrm{MLN}, \triangle \mathrm{MGRare} \mathrm{similar} \mathrm{triangles.} \mathrm{}$

$\begin{aligned}
\therefore \frac{\mathrm{GR}}{\mathrm{LN}} &=\frac{\mathrm{MG}}{\mathrm{ML}} \\
\frac{1.25}{h} &=\frac{2.5}{6.6} \\
1.25 \times 6.6 &=2.5 \times h \\
h &=\frac{1.25 \times 6.6}{2.5} \\
h &=\frac{125^{3}}{100} \times \frac{66^{3}}{10} \times \frac{10}{25_{5}} \\
&=\frac{33}{10}=3.3 \mathrm{~m}
\end{aligned}$
$\therefore$ Height of the lamp post is $3.3 \mathrm{~m}$.

Question $3 .$
A vertical stick of length $6 \mathrm{~m}$ casts a shadow $400 \mathrm{~cm}$ long on the ground and at the same time a tower casts a shadow $28 \mathrm{~m}$ long. Using similarity, find the height of the tower.
Solution:
In the picture $\triangle \mathrm{ABC}, \triangle \mathrm{DEC}$ are similar triangles.

$\begin{aligned} \therefore \frac{\mathrm{AB}}{\mathrm{DE}} &=\frac{\mathrm{BC}}{\mathrm{EC}} \quad \mathrm{b} \\ \frac{h}{6} &=\frac{28}{4} \end{aligned}$
$A h=28^{7} \times 6$
$h=42 \mathrm{~m}$
Height of a tower $=42 \mathrm{~m}$

Question $4 .$
Two triangles $Q P R$ and $Q S R$, right angled at $P$ and $S$ respectively are drawn on the same base $Q R$ and on the same side of $Q R$. If $P R$ and $S Q$ intersect at $T$, prove that $P T \times T R=S T \times T Q$.
Solution:
In $\triangle \mathrm{RPQ}$,

$\mathrm{RP}^{2}+\mathrm{PQ}^{2}=\mathrm{QR}^{2}$ $\therefore \mathrm{PQ}^{2}=\mathrm{QR}^{2}-\mathrm{RP}^{2}$
In $\triangle \mathrm{TPQ}$,
$\mathrm{TP}^{2}+\mathrm{PQ}^{2}=\mathrm{QT}^{2}$
$\therefore \mathrm{PQ}^{2}=\mathrm{QT}^{2}-\mathrm{TP}^{2} \ldots \ldots \ldots \ldots(2)$
Equating (1) and (2) we get,
$\mathrm{QR}^{2}-\mathrm{RP}^{2}=\mathrm{QT}^{2}-\mathrm{TP}^{2}$
$\mathrm{RP}=\mathrm{RT}+\mathrm{TP}$
$\therefore \mathrm{QR}^{2}-(\mathrm{RT}+\mathrm{TP})^{2}=\mathrm{QT}^{2}-\mathrm{TP}^{2}$
$\therefore \mathrm{QR}^{2}-\mathrm{RT}^{2}-\mathrm{TP}^{2}-2 \mathrm{RT} \cdot \mathrm{TP}=\mathrm{QT}^{2}-\mathrm{TP}^{2}$
$\mathrm{QR}^{2}=\mathrm{QT}^{2}+\mathrm{RT}^{2}+2 \mathrm{RT} \cdot \mathrm{TP} \ldots \ldots \ldots \mathrm{F} \cdot(5)$

In $\triangle \mathrm{QSR}$,
$\mathrm{QS}^{2}+\mathrm{SR}^{2}=\mathrm{QR}^{2}$
$\mathrm{SR}^{2}=\mathrm{QR}^{2}-\mathrm{SR}^{2}$
In $\triangle \mathrm{TSR}$,
$\mathrm{ST}^{2}+\mathrm{SR}^{2}=\mathrm{TR}^{2}$

$\therefore \mathrm{SR}^{2}=\mathrm{TR}^{2}-\mathrm{TS}^{2}$
Equating (3) and (4) we get
$\begin{aligned}
&\mathrm{QR}^{2}-\mathrm{SQ}^{2}=\mathrm{TR}^{2}-\mathrm{TS}^{2} \\
&\mathrm{SQ}=\mathrm{QT}+\mathrm{TS} \\
&\therefore \mathrm{QR}^{2}-(2 \mathrm{~T}+\mathrm{TS})^{2}=\mathrm{TR}^{2}-\mathrm{TS}^{2} \\
&\mathrm{QR}^{2}-2 \mathrm{~T}^{2}-\mathrm{TS}^{2}-2 \mathrm{QT} \cdot \mathrm{TS}=\mathrm{TR}^{2}-\mathrm{TS}^{2} \\
&\therefore 2 \mathrm{R}^{2}=\mathrm{TR}^{2}+\mathrm{QT}^{2}+2 \mathrm{QT} \cdot \mathrm{TS} \ldots \ldots \ldots \ldots
\end{aligned}$
Now equating $(5)-(6)$, we get
$\begin{aligned}
&\mathrm{QT}^{2}+\mathrm{RT}^{2}+2 \mathrm{RT} . \mathrm{TP}=\mathrm{QT}^{2}+\mathrm{RT}^{2}+2 \mathrm{QT} . \mathrm{TS} \\
&\therefore \mathrm{PT} . \mathrm{TR}=\mathrm{ST} . \mathrm{TQ}
\end{aligned}$
Hence proved.

Question $5 .$
In the adjacent figure, $\triangle \mathrm{ABC}$ is right angled at $\mathrm{C}$ and $\mathrm{DE} \perp \mathrm{AB}$. Prove that $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$ and hence find the lengths of $\mathrm{AE}$ and $\mathrm{DE}$ ?
Solution:
In $\triangle \mathrm{ABC} \& \triangle \mathrm{ADE}$
$\angle \mathrm{A}$ is common \& $\angle \mathrm{C}=\angle \mathrm{E}=90^{\circ}$
$\therefore$ by similarity
$\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$
$\begin{aligned}
\therefore \frac{\mathrm{AB}}{\mathrm{AD}} &=\frac{\mathrm{AC}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{AE}} \\
\frac{13}{3} &=\frac{5}{\mathrm{DE}}=\frac{12}{\mathrm{AE}}
\end{aligned}$

$\begin{aligned}
13 \mathrm{DE} &=3 \times 5 \\
\mathrm{DE} &=\frac{15}{13}
\end{aligned}$
Since $\triangle \mathrm{ABC}$ is a right angled triangle.
$\begin{aligned}
\mathrm{AB}^{2} &=\mathrm{BC}^{2}+\mathrm{AC}^{2} \\
&=12^{2}+5^{2} \\
&=144+25 \\
&=169 \\
\mathrm{AB} &=13 \\
\frac{5}{5} &=\frac{12}{\mathrm{AE}} \\
\frac{13}{5 \mathrm{AE}} &=\frac{15}{13} \times 12 \\
\mathrm{AE} &=\frac{15}{13} \times \frac{12}{5} \\
\mathrm{AE} &=\frac{36}{13}=2.7 \\
\mathrm{DE} &=\frac{15}{13}=1.1
\end{aligned}$
Substituting the values of $\mathrm{DE}$ and $\mathrm{AE}$ in (1) we can prove that
$\begin{aligned}
\frac{\mathrm{AB}}{\mathrm{AD}} &=\frac{\mathrm{AC}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{AE}} \\
\frac{13}{3} &=\frac{5}{1 \cdot 1}=\frac{12}{2 \cdot 7}=4 \cdot 3
\end{aligned}$

It is proved that $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$.
 

Question 6.
In the adjacent figure, $\triangle \mathrm{ACB} \sim \triangle \mathrm{APQ}$. If $\mathrm{BC}=8 \mathrm{~cm}, \mathrm{PQ}=4 \mathrm{~cm}, \mathrm{BA}=6.5 \mathrm{~cm}$ and $\mathrm{AP}=2.8 \mathrm{~cm}$, find $\mathrm{CA}$ and $\mathrm{AQ}$.
Solution:
$\triangle \mathrm{ACB} \sim \triangle \mathrm{APQ}$

$\begin{aligned}
&\frac{\mathrm{AB}}{\mathrm{AQ}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{CA}}{\mathrm{AP}} \\
&\frac{6.5}{\mathrm{AQ}}=\frac{8}{4}=\frac{\mathrm{CA}}{2.8}
\end{aligned}$

From (1)
$\begin{aligned}
\Rightarrow \quad 4 \mathrm{CA} &=8 \times 2.8 \\
\mathrm{CA} &=\frac{22.4}{4}=5.6 \mathrm{~cm}
\end{aligned}$
From (1)
$\begin{aligned}
\Rightarrow \quad 8 \mathrm{AQ} &=6.5 \times 4 \\
\mathrm{AQ} &=\frac{26}{8}=3.25 \mathrm{~cm}
\end{aligned}$

Question 7.
In figure $\mathrm{OPRQ}$ is a square and $\mathrm{MLN}=90^{\circ}$. Prove that
(i) $\triangle \mathrm{LOP} \sim \triangle \mathrm{QMO}$
(ii) $\triangle \mathrm{LOP} \sim \triangle \mathrm{RPN}$
(iii) $\triangle \mathrm{QMO} \sim \triangle \mathrm{RPN}$
(iv) $\mathrm{QR}^{2}=\mathrm{MQ} \times \mathrm{RN}$.

Solution:
(i) In $\triangle \mathrm{LOP} \& \Delta \mathrm{QMO}$, we have
$\angle \mathrm{OLP}=\angle \mathrm{MQO}$ (each equal to $90^{\circ}$ )
and $\angle \mathrm{LOP}=\angle \mathrm{OMQ}$ (corresponding angles)
$\Delta \mathrm{LOP} \sim \triangle \mathrm{QMO}$ (by AA criterion of similarity)
(ii) In $\triangle L O P \& \triangle \mathrm{PRN}$, we have
$\angle \mathrm{PLO}=\angle \mathrm{NRP}$ (each equal to $\left.90^{\circ}\right)$
$\angle \mathrm{LPO}=\angle \mathrm{PNR}$ (corresponding angles)
$\triangle \mathrm{LOP} \sim \triangle \mathrm{RPN}$
(iii) In $\Delta \mathrm{QMO} \& \Delta \mathrm{RPN}$.
Since $\Delta \mathrm{LOP} \sim \Delta \mathrm{QMO}$ and $\Delta \mathrm{LOP} \sim \Delta \mathrm{RPN}$
$\angle \mathrm{QMO} \sim \triangle \mathrm{RPN}$
(iv) We have
$\triangle \mathrm{QMO} \sim \triangle \mathrm{RPN}$ (using (iii))
$\frac{\mathrm{MQ}}{\mathrm{RP}}=\frac{\mathrm{QO}}{\mathrm{RN}}(\because \mathrm{PROQ}$ is a square $)$
$\mathrm{QR}^{2}=\mathrm{MQ} \times \mathrm{RN} .[\mathrm{RP}=\mathrm{QO}, \mathrm{QO}=\mathrm{QR}]$
 

Question $8 .$
If $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$ such that area of $\triangle \mathrm{ABC}$ is $9 \mathrm{~cm}^{2}$ and the area of $\triangle \mathrm{DEF}$ is $16 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.1$
$\mathrm{cm}$. Find the length of EF.
Solution:
Since the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
$\begin{aligned}
\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}} &=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}} \\
\frac{9}{16} &=\frac{(2.1)^{2}}{\mathrm{EF}^{2}} \\
\Rightarrow \quad \frac{3}{4} &=\frac{2.1}{\mathrm{EF}} \\
3 \mathrm{EF} &=8.4 \\
\mathrm{EF} &=\frac{8.4}{3}=2.8 \mathrm{~cm} .
\end{aligned}$
Note: Taking square root on both sides we get
$\mathrm{EF}=2.8 \mathrm{~cm} \text {. }$

Question $9 .$
Two vertical poles of heights $6 \mathrm{~m}$ and $3 \mathrm{~m}$ are erected above a horizontal ground $\mathrm{AC}$. Find the value of $\mathrm{y}$.

Solution:
$\triangle \mathrm{PAC}, \triangle \mathrm{QBC}$ are similar 6 triangles
$\begin{aligned}
\therefore \quad \frac{\mathrm{PA}}{\mathrm{QB}} &=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{PC}}{\mathrm{QC}} \\
\Rightarrow \quad(\mathrm{AC}) y &=6 \mathrm{BC}
\end{aligned}$
$\triangle \mathrm{ACR} \& \triangle \mathrm{ABQ}$ are similar triangles.
$\begin{aligned}
\frac{\mathrm{CR}}{\mathrm{QB}} &=\frac{\mathrm{AC}}{\mathrm{AB}} \\
\frac{3}{y} &=\frac{\mathrm{AC}}{\mathrm{AB}} \\
\Rightarrow \quad(\mathrm{AC}) y &=3 \mathrm{AB}
\end{aligned}$

$\begin{aligned}
&(1)=(2) \Rightarrow 6 B C=3 A B \\
&2 B C=A B \\
&\Rightarrow A C=A B+B C \\
&=2 B C+B C \\
&A C=3 B C \\
&\text { Substituting } A C=3 B C \text { in }(1) \text {, we get } \\
&(A C) y=6 B C \\
&3(B C) y=6(B C) \\
&y=\frac{6}{3}=2 m
\end{aligned}$

Question $10 .$
Construct a triangle similar to a given triangle $\mathrm{PQR}$ with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{2}{3}$ ).
Solution:
Given a triangle $\mathrm{PQR}$, we are required to construct another triangle whose sides are $\frac{3}{5}$ of the corresponding sides of the triangle $\mathrm{PQR}$.

Steps of construction:
(1) Draw any ray $Q X$ making an acute angle with $Q R$ on the side opposite to the vertex $P$.
(2) Locate 3 (the greater of 2 and 3 in $\frac{2}{3}$ ) points. $\mathrm{Q}_{1} \mathrm{Q}_{2}, \mathrm{Q}_{2}$ on $\mathrm{QX}$ so that $\mathrm{QQ}_{1}=\mathrm{Q}_{1} \mathrm{Q}_{2}=\mathrm{Q}_{2} \mathrm{Q}_{3}$
(3) Join $Q_{3} R$ and draw a line through $Q_{2}$ (the second point, 2 being smaller of 2 and 3 in $\frac{2}{3}$ ) parallel to $Q_{3} R$ to intersect $Q R$ at $R^{\prime}$.
(4) Draw line through $R^{\prime}$ parallel to the line $R P$ to intersect $Q P$ at $P^{\prime}$.
The $\triangle \mathrm{P}^{\prime} \mathrm{QR}$ ' is the required triangle each of the whose sides is $\frac{2}{3}$ of the corresponding sides of 3 $\triangle \mathrm{PQR}$.

Question $11 .$
Construct a triangle similar to a given triangle $\mathrm{LMN}$ with its sides equal to $\frac{4}{5}$ of the corresponding sides of the triangle LMN
(scale factor $\frac{4}{5}$ ).
Solution:
Given a triangle LMN, we are required to construct another triangle whose sides are $\frac{4}{5}$ of the corresponding sides of the $\triangle \mathrm{LMN}$.

Steps of construction:
(1) Draw any ray making an acute angle to the vertex $L$.
(2) Locate 5 points (greater of 4 and 5 in $\frac{4}{5}$ ) $\mathrm{M}_{1}, \mathrm{M}_{2}, \mathrm{M}_{3}, \mathrm{M}_{4}$, and $\mathrm{M}_{5}$ and $\mathrm{MX}$ so that $\mathrm{MM}_{1}=$ $\mathrm{M}_{1} \mathrm{M}_{2}=\mathrm{M}_{2} \mathrm{M}_{3}=\mathrm{M}_{3} \mathrm{M}_{4}=\mathrm{M}_{4} \mathrm{M}_{5}$
(3) Join $\mathrm{M}_{5} \mathrm{~N}$ and draw a line parallel to $\mathrm{M}_{5} \mathrm{~N}$ through $\mathrm{M}_{4}$ (the fourth point, 4 being the smaller of 4 and 5 in $\frac{4}{5}$ ) to intersect $\mathrm{MN}$ atN'.
(4) Draw a line through $\mathrm{N}^{1}$ parallel to the line $\mathrm{NL}$ to intersect $\mathrm{ML}$ and $\mathrm{L}$ '. Then $\Delta L$ ' $\mathrm{MN}$ ' is the required triangle each of the whose sides is $\frac{4}{5}$ of the corresponding sides of $\Delta \mathrm{LMN}$.
 

Question $12 .$
Construct a triangle similar to a given triangle $\mathrm{ABC}$ with its sides equal to $\frac{6}{5}$ of the corresponding
sides of the triangle $\mathrm{ABC}$
(scale factor $\frac{6}{4}$ ).
Solution:
$\triangle \mathrm{ABC}$ is the given triangle. We are required to construct another triangle whose sides are $\frac{6}{5}$ of the corresponding sides of the given triangle $\mathrm{ABC}$
Steps of construction:

(1) Draw any ray BX making an acute angle with $B C$ on the opposite side to the vertex $A$.
(2) Locate 6 points (the greater of 6 and 5 in $\frac{6}{5}$ ) $\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}, \mathrm{~B}_{4}, \mathrm{~B}_{5}, \mathrm{~B}_{6}$ so that $\mathrm{BB}_{1}=\mathrm{B}_{1} \mathrm{~B}_{2}=$ $\mathrm{B}_{2} \mathrm{~B}_{3}=\mathrm{B}_{4} \mathrm{~B}_{5}=\mathrm{B}_{5} \mathrm{~B}_{6}$
(3) Join $B_{5}$ (the fifth point, 5 being smaller of 5 and 6 in $\frac{6}{5}$ ) to $C$ and draw a live through $B_{6}$ parallel to $\mathrm{B}_{5} \mathrm{C}$ intersecting the extended line segment $\mathrm{BC}$ at $\mathrm{C}^{1}$.
(4) Draw a line through $\mathrm{C}^{\prime}$ parallel to $\mathrm{CA}$ intersecting the extended line segment $\mathrm{BA}$ at $\mathrm{A}^{\prime}$.
Then $\triangle \mathrm{A}^{\prime} \mathrm{BC} \mathrm{C}^{\prime}$ is the required triangle each of whose sides is $\frac{6}{5}$ of the corresponding sides of the given triangle $\mathrm{ABC}$.
 

Question $13 .$
Construct a triangle similar to a given triangle $\mathrm{PQR}$ with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{7}{3}$ ).
Solution:
Given a triangle $\triangle \mathrm{PQR}$. We have to construct another triangle whose sides are $\frac{7}{3}$ of the corresponding sides of the given $\triangle \mathrm{PQR}$.

Steps of construction:
(1) Draw any ray $Q X$ making an acute angle with $Q R$ on the opposite side to the vertex $P$.
(2) Locate 7 points (the greater of 7 and 3 in $\frac{7}{3}$ ) $Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6}$, and $Q_{7}$ so that $Q_{2}=$ $\mathrm{Q}_{1} \mathrm{Q}_{2}=\mathrm{Q}_{2} \mathrm{Q}_{3}=\mathrm{Q}_{3} \mathrm{Q}_{4}=\mathrm{Q}_{4} \mathrm{Q}_{5}=\mathrm{Q}_{5} \mathrm{Q}_{6}$
$=\mathrm{Q}_{6} \mathrm{Q}_{7}$
(3) Join $Q_{3}$ to $R$ and draw a line segment through $Q_{7}$ parallel to $Q_{3} R$ intersecting the extended line segment $Q R$ at $R$ '.
(4) Draw a line segment through $\mathrm{R}^{\prime}$ parallel to $\mathrm{PR}$ intersecting the extended line segment $\mathrm{QP}$ at $\mathrm{P}^{\prime}$. Then $\triangle \mathrm{P}^{\prime} \mathrm{QR}^{\prime}$ ' is the required triangle each of whose sides is $\frac{7}{3}$ of the corresponding sides of the given triangle.