Exercise 6.3 (Revised) - Chapter 6 - Triangles - Ncert Solutions class 10 - Maths

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Chapter 6 Triangles NCERT Solutions Class 10 Maths: Simplify Concepts & Ace Exams

Ex 6.3 Question 1.

State which pairs of triangles in the given figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Answer.

(i) In $\Delta \mathrm{s} A B C$ and $\mathrm{PQR}$, we observe that,
$
\angle \mathrm{A}=\angle \mathrm{P}=60^{\circ}, \angle \mathrm{B}=\angle \mathrm{Q}=80^{\circ} \text { and } \angle \mathrm{C}=\angle \mathrm{R}=40^{\circ}
$
$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$
(ii) In $\triangle \mathrm{s} A B C$ and $\mathrm{PQR}$, we observe that,
$
\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{1}{2}
$
$\therefore$ By SSS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$
(iii) In $\Delta s$ LMP and DEF, we observe that the ratio of the sides of these triangles is not equal.

Therefore, these two triangles are not similar.
(iv) In $\triangle \mathrm{s}$ MNL and QPR, we observe that, $\angle \mathrm{M}=\angle \mathrm{Q}=70^{\circ}$

But, $\frac{M N}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}$
$\therefore$ These two triangles are not similar as they do not satisfy SAS criterion of similarity.
(v) In $\triangle \mathrm{s} A B C$ and FDE, we have, $\angle \mathrm{A}=\angle \mathrm{F}=80^{\circ}$

But, $\frac{\mathrm{AB}}{\mathrm{DE}} \neq \frac{\mathrm{AC}}{\mathrm{DF}}[\because \mathrm{AC}$ is not given $]$
$\therefore$ These two triangles are not similar as they do not satisfy SAS criterion of similarity.
(vi) In $\triangle \mathrm{s}$ DEF and $\mathrm{PQR}$, we have, $\angle \mathrm{D}=\angle \mathrm{P}=70^{\circ}$
$\left[\because \angle \mathrm{P}=180^{\circ}-80^{\circ}-30^{\circ}=70^{\circ}\right]$
And $\angle \mathrm{E}=\angle \mathrm{Q}=80^{\circ}$

$\therefore \text { By AAA criterion of similarity, } \triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}$

Ex 6.3 Question 2.

In figure, $\triangle \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$ and $\angle \mathrm{CDO}=70^{\circ}$. Find $\angle \mathrm{DOC}, \angle \mathrm{DCO}$ and $\angle \mathrm{OAB}$.

Answer.

Since $B D$ is a line and $O C$ is a ray on it.
$
\begin{aligned}
& \therefore \angle \mathrm{DOC}+\angle \mathrm{BOC}=180^{\circ} \\
& \Rightarrow \angle \mathrm{DOC}+125^{\circ}=180^{\circ} \\
& \Rightarrow \angle \mathrm{DOC}=55^{\circ}
\end{aligned}
$

In $\Delta \mathrm{CDO}$, we have $\angle \mathrm{CDO}+\angle \mathrm{DOC}+\angle \mathrm{DCO}=180^{\circ}$
$
\begin{aligned}
& \Rightarrow 70^{\circ}+55^{\circ}+\angle \mathrm{DCO}=180^{\circ} \\
& \Rightarrow \angle \mathrm{DCO}=55^{\circ}
\end{aligned}
$

It is given that $\triangle \mathrm{ODC} \sim \triangle \mathrm{OBA}$
$
\begin{aligned}
& \therefore \angle \mathrm{OBA}=\angle \mathrm{ODC}, \angle \mathrm{OAB}=\angle \mathrm{OCD} \\
& \Rightarrow \angle \mathrm{OBA}=70^{\circ}, \angle \mathrm{OAB}=55^{\circ}
\end{aligned}
$

Hence $\angle \mathrm{DOC}=55^{\circ}, \angle \mathrm{DCO}=55^{\circ}$ and $\angle \mathrm{OAB}=55^{\circ}$
Ex 6.3 Question 3.

Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B \|$ | CD intersect each other at the point 0 . Using a similarity criterion for two triangles, show that $\frac{O A}{O C}=\frac{O B}{O D}$.

Answer.

Given: $\mathrm{ABCD}$ is a trapezium in which $\mathrm{AB} \| \mathrm{DC}$.

To Prove: $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$
Proof: In $\triangle \mathrm{s} O A B$ and $O C D$, we have,
$\angle 5=\angle 6$ [Vertically opposite angles]
$\angle 1=\angle 2$ [Alternate angles]
And $\angle 3=\angle 4$ [Alternate angles]
$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{OAB} \sim \triangle \mathrm{ODC}$
Hence, $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$
Ex 6.3 Question 4.

In figure, $\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$ and $\angle \mathbf{1}=\angle \mathbf{2}$. Show that $\Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}$.

Answer.

We have, $\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$
$
\Rightarrow \frac{Q T}{Q R}=\frac{P R}{Q S}
$

Also, $\angle 1=\angle 2$ [Given]

$\therefore \mathrm{PR}=\mathrm{PQ}$ $\qquad$ (2) $[\because$ Sides opposite to equal $\angle \mathrm{s}$ are equal $]$

From eq.(1) and (2), we get
$
\frac{\mathrm{QT}}{\mathrm{QR}}=\frac{\mathrm{PR}}{\mathrm{QS}} \Rightarrow \frac{\mathrm{PQ}}{\mathrm{QT}}=\frac{\mathrm{QS}}{\mathrm{QR}}
$

In $\Delta \mathrm{s}$ PQS and TQR, we have,
$
\frac{\mathrm{PQ}}{\mathrm{QT}}=\frac{\mathrm{QS}}{\mathrm{QR}} \text { and } \angle \mathrm{PQS}=\angle \mathrm{TQR}=\angle \mathrm{Q}
$
$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{PQS} \sim \Delta \mathrm{TQR}$
Ex 6.3 Question 5.

$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of a $\triangle \mathrm{PQR}$ such that $\angle \mathbf{P}=\angle$ RTS. Show that $\triangle \mathbf{R P Q} \sim \Delta \mathbf{R T S}$.

Answer.

In $\Delta \mathrm{s}$ RPQ and RTS, we have

$
\begin{aligned}
& \angle \mathrm{RPQ}=\angle \mathrm{RTS} \text { [Given] } \\
& \angle \mathrm{PRQ}=\angle \mathrm{TRS} \text { [Common] } \\
& \therefore \text { By AA-criterion of similarity, } \\
& \triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}
\end{aligned}
$
Ex 6.3 Question 6.

In the given figure, if $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$, show that $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$.

Answer.

It is given that $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$
$
\begin{aligned}
& \therefore A B=A C \text { and } A E=A D \\
& \Rightarrow \frac{A B}{A D}=\frac{A C}{A E} \\
& \Rightarrow \frac{A B}{A C}=\frac{A D}{A E} \ldots \ldots \ldots .(1)
\end{aligned}
$
$\therefore$ In $\triangle \mathrm{s}$ ADE and $\mathrm{ABC}$, we have,
$
\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}[\text { from eq.(1) }]
$

And $\angle \mathrm{BAC}=\angle \mathrm{DAE}[$ Common $]$
Thus, by SAS criterion of similarity, $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$
Ex 6.3 Question 7.

In figure, altitude $A D$ and $C E$ of a $\triangle A B C$ intersect each other at the point $P$. Show that:

$\text { (i) } \triangle \text { AEP } \sim \Delta \text { CDP }$

(ii) $\triangle \mathrm{ABD} \sim \triangle \mathrm{CBE}$
(iii) $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$
(iv) $\triangle$ PDC $\sim \triangle$ BEC

Answer.

(i) In $\Delta \mathrm{s}$ AEP and CDP, we have,
$\angle \mathrm{AEP}=\angle \mathrm{CDP}=90^{\circ}[\because \mathrm{CE} \perp \mathrm{AB}, \mathrm{AD} \perp \mathrm{BC}]$
And $\angle \mathrm{APE}=\angle \mathrm{CPD}[$ Vertically opposite]
$\therefore$ By AA-criterion of similarity, $\triangle$ AEP $\sim \triangle$ CDP
(ii) In $\triangle \mathrm{s} A \mathrm{ABD}$ and $\mathrm{CBE}$, we have,
$\angle \mathrm{ADB}=\angle \mathrm{CEB}=90^{\circ}$
And $\angle \mathrm{ABD}=\angle \mathrm{CBE}[$ Common]
$\therefore$ By AA-criterion of similarity, $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$
(iii) In $\Delta \mathrm{s}$ AEP and ADB, we have,
$\angle \mathrm{AEP}=\angle \mathrm{ADB}=90^{\circ}[\because \mathrm{AD} \perp \mathrm{BC}, \mathrm{CE} \perp \mathrm{AB}]$
And $\angle \mathrm{PAE}=\angle \mathrm{DAB}[$ Common]
$\therefore$ By AA-criterion of similarity, $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$
(iv) In $\Delta \mathrm{s}$ PDC and BEC, we have,
$\angle \mathrm{PDC}=\angle \mathrm{BEC}=90^{\circ}[\because \mathrm{CE} \perp \mathrm{AB}, \mathrm{AD} \perp \mathrm{BC}]$
And $\angle \mathrm{PCD}=\angle \mathrm{BEC}[$ Common]
$\therefore$ By AA-criterion of similarity, $\Delta$ PDC $\sim \Delta$ BEC

Ex 6.3 Question 8.

$\mathrm{E}$ is a point on the side $A D$ produced of a parallelogram $A B C D$ and $B E$ intersects $C D$ at F. Show that $\triangle \mathrm{ABE} \sim \Delta \mathrm{CFB}$.

Answer.

$\text {In} \Delta \mathrm{s} \text { ABE and CFB, we have, }$

$
\begin{aligned}
& \angle \mathrm{AEB}=\angle \mathrm{CBF}[\text { Alt. } \angle \mathrm{s}] \\
& \angle \mathrm{A}=\angle \mathrm{c} \text { [opp. } \angle \mathrm{s} \text { of a } \| \mathrm{gm}]
\end{aligned}
$
$\therefore$ By AA-criterion of similarity, we have
$
\triangle \mathrm{ABE} \sim \Delta \mathrm{CFB}
$
Ex 6.3 Question 9.

In the given figure, $A B C$ and $A M P$ are two right triangles, right angles at $B$ and $M$ respectively. Prove that:

(i) $\triangle \mathrm{ABC} \sim \Delta \mathrm{AMP}$
(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

Answer.

(i) In $\Delta s A B C$ and AMP, we have,
$\angle \mathrm{ABC}=\angle \mathrm{AMP}=90^{\circ}$ [Given]
$\angle \mathrm{BAC}=\angle \mathrm{MAP}$ [Common angles]
$\therefore$ By AA-criterion of similarity, we have
$\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$

(ii) We have $\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$ [As prove above]
$
\Rightarrow \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}
$
Ex 6.3 Question 10.

$\mathrm{CD}$ and $\mathrm{GH}$ are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides $A B$ and $F E$ at $\triangle A B C$ and $\triangle E F G$ respectively. If $\triangle A B C \sim \triangle F E G$, show that:
(i) $\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$
(ii) $\triangle \mathbf{D C B} \sim \triangle \mathrm{HE}$
(iii) $\triangle \mathbf{D C A} \sim \Delta$ HGF

Answer.

We have, $\triangle \mathrm{ABC} \sim \Delta \mathrm{FEG}$

$
\begin{aligned}
& \Rightarrow \angle \mathrm{A}=\angle \mathrm{F} \ldots \ldots \ldots .(1) \\
& \text { And } \angle \mathrm{C}=\angle \mathrm{G} \\
& \Rightarrow \frac{1}{2} \angle \mathrm{C}=\frac{1}{2} \angle \mathrm{G} \\
& \Rightarrow \angle 1=\angle 3 \text { and } \angle 2=\angle 4 \ldots \ldots \ldots .(2) \\
& \text { [ } \because \mathrm{CD} \text { and } \mathrm{GH} \text { are bisectors of } \angle \mathrm{C} \text { and } \angle \mathrm{G} \\
& \text { respectively] } \\
& \therefore \text { In } \Delta \mathrm{s} \text { DCA and HGF, we have } \\
& \angle \mathrm{A}=\angle \mathrm{F}[\text { From eq.(1)] }
\end{aligned}
$
[ $\because \mathrm{CD}$ and $\mathrm{GH}$ are bisectors of $\angle \mathrm{C}$ and $\angle \mathrm{G}$
respectively]
$\therefore$ In $\triangle \mathrm{s}$ DCA and HGF, we have

$\angle 2=\angle 4[$ [From eq.(2)]
$\therefore$ By AA-criterion of similarity, we have
$\triangle \mathrm{DCA} \sim \triangle \mathrm{HGF}$
Which proves the (iii) part
We have, $\triangle \mathrm{DCA} \sim \Delta \mathrm{HGF}$
$
\begin{aligned}
& \Rightarrow \frac{A G}{F G}=\frac{C D}{G H} \\
& \Rightarrow \frac{C D}{G H}=\frac{A C}{F G}
\end{aligned}
$

Which proves the (i) part
In $\Delta s$ DCA and HGF, we have
$
\begin{aligned}
& \angle 1=\angle 3 \text { [From eq.(2)] } \\
& \angle \mathrm{B}=\angle \mathrm{E}[\because \Delta \mathrm{DCB} \sim \Delta \mathrm{HE}]
\end{aligned}
$

Which proves the (ii) part
Ex 6.3 Question 11.

In the given figure, $E$ is a point on side $C B$ produced of an isosceles triangle $A B C$ with $\mathrm{AB}=\mathrm{AC}$. If $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{EF} \perp \mathrm{AC}$, prove that $\triangle \mathrm{ABD} \sim \triangle \mathrm{ECF}$.

Answer.

Here $\triangle \mathrm{ABC}$ is isosceles with $\mathrm{AB}=\mathrm{AC}$
$
\therefore \angle \mathrm{B}=\angle \mathrm{C}
$

In $\Delta \mathrm{s} A B D$ and ECF, we have
$
\begin{aligned}
& \angle \mathrm{ABD}=\angle \mathrm{ECF}[\because \angle \mathrm{B}=\angle \mathrm{C}] \\
& \angle \mathrm{ABD}=\angle \mathrm{ECF}=90^{\circ}[\because \mathrm{AD} \perp \mathrm{BC} \text { and } \mathrm{EF} \perp \mathrm{AC}] \\
& \therefore \text { By AA-criterion of similarity, we have } \\
& \triangle \mathrm{ABD} \sim \triangle \mathrm{ECF}
\end{aligned}
$
Ex 6.3 Question 12.

Sides $A B$ and $B C$ and median $A D$ of a triangle $A B C$ are respectively proportional to sides $\mathrm{PQ}$ and $\mathrm{QR}$ and median $\mathrm{PM}$ of a $\triangle \mathrm{PQR}$ (see figure). Show that $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.

 

Answer.

Given: $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$ and $\mathrm{PM}$ is the median of $\triangle \mathrm{PQR}$ such that
$
\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}
$

To prove: $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$
Proof: $\mathrm{BD}=\frac{1}{2} \mathrm{BC}$ [Given]
And $Q M=\frac{1}{2} Q R$ [Given]
Also $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$ [Given]
$
\Rightarrow \frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M}
$

$
\begin{aligned}
& \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\
& \therefore \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}[\mathrm{By} \text { SSS-criterion of similarity }] \\
& \Rightarrow \angle \mathrm{B}=\angle \mathrm{Q}[\text { Similar triangles have corresponding angles equal] } \\
& \text { And } \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}} \text { [Given] }
\end{aligned}
$

And $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}$ [Given]
$\therefore$ By SAS-criterion of similarity, we have
$
\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}
$
Ex 6.3 Question 13.

$\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that $\mathrm{CA}^2=\mathbf{C B} . \mathrm{CD}$.

Answer.

In triangles $\mathrm{ABC}$ and $\mathrm{DAC}$,

$
\begin{aligned}
& \angle \mathrm{ADC}=\angle \mathrm{BAC} \text { [Given] } \\
& \text { and } \angle \mathrm{c}=\angle \mathrm{C}[\text { Common] }
\end{aligned}
$
$\therefore$ By AA-similarity criterion,
$
\begin{aligned}
& \triangle \mathrm{ABC} \sim \triangle \mathrm{DAC} \\
& \Rightarrow \frac{\mathrm{AB}}{\mathrm{DA}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{DC}} \\
& \Rightarrow \frac{\mathrm{CB}}{\mathrm{CA}}=\frac{\mathrm{CA}}{\mathrm{CD}}
\end{aligned}
$

$
\Rightarrow C A^2=\mathrm{CB} . \mathrm{CD}
$
Ex 6.3 Question 14.

Sides $A B$ and $A C$ and median $A D$ of a triangle $A B C$ are respectively proportional to sides $P Q$ and $P R$ and median $P M$ of another triangle $P Q R$. Show that $\triangle A B C \sim \triangle P Q R$.

Answer.

Given: $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$ and $\mathrm{PM}$ is the median of $\triangle \mathrm{PQR}$ such that

$
\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M} \quad \ldots \ldots \ldots(1)
$

To prove: $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$
Proof:
Let us extend $\mathrm{AD}$ to point $\mathrm{D}$ such that $\mathrm{AD}=\mathrm{DE}$ and $\mathrm{PM}$ up to point $\mathrm{L}$ such that $\mathrm{PM}=\mathrm{ML}$

Join B to E. C to E,and Q to L, and R to L
We know that medians is the bisector of opposite side
Hence
$
\mathrm{BD}=\mathrm{DC}
$

Also, $\mathrm{AD}=\mathrm{DE}$ (by construction)

Hence in quadrilateral $\mathrm{ABEC}$, diagonals $\mathrm{AE}$ and $\mathrm{BC}$ bisect each other at point $\mathrm{D}$. Therefore, quadrilateral ABEC is a parallelogram.

$
\begin{aligned}
& \mathrm{AC}=\mathrm{BE} \\
& \mathrm{AB}=\mathrm{EC} \text { (opposite sides of \| |gm are equal) }
\end{aligned}
$

Similarly, we can prove that $\mathrm{PQLR}$ is a parallelogram
$
\begin{aligned}
& P R=Q L \\
& P Q=L R \text { opposite sides of } \| \text { gm are equal ) }
\end{aligned}
$

Given that
$
\begin{aligned}
& \frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M} \\
& \frac{A B}{P Q}=\frac{B E}{Q L}=\frac{A D}{P M}[\text { from (2) (3)] } \\
& \frac{A B}{P Q}=\frac{B E}{Q L}=\frac{2 A D}{2 P M} \\
& \frac{A B}{P Q}=\frac{B E}{Q L}=\frac{A E}{P L}[\text { as } A D=D E, A E=A D+D E=2 A D \\
& \qquad P M=M L \cdot P L=P M+M L=2 P M \\
& \triangle A B E \sim \triangle P Q L \quad \text { (By SSS Similiarity Criteria) }
\end{aligned}
$

We know that corresponding angles of similar triangles are equal.
$
\angle B A E=\angle Q P L(4)
$

Similarly, we can prove that $\triangle A E C \sim \triangle P L R$.

We know that corresponding angles of similar triangles are equal.
$
\angle C A E=\angle R P L \text { (5) }
$

Adding (4) and (5),
$
\begin{aligned}
& \angle B A E+\angle C A E=\angle Q P L+\angle R P L \\
& \angle C A B=\angle R P Q \\
& \text { In } \triangle A B C \text { and } \triangle P Q R
\end{aligned}
$

$
\begin{aligned}
& \frac{A B}{P Q}=\frac{A C}{P R} \\
& \angle C A B=\angle R P Q \\
& \triangle A B C \sim \triangle P Q R
\end{aligned}
$

Hence proved
Ex 6.3 Question 15.

A vertical pole of length $6 \mathrm{~m}$ casts a shadow $4 \mathrm{~m}$ long on the ground and at the same time a tower casts a shadow $28 \mathrm{~m}$ long. Find the height of the tower.

Answer.

Let $\mathrm{AB}$ the vertical pole and $\mathrm{AC}$ be its shadow. Also, let $\mathrm{DE}$ be the vertical tower and $\mathrm{DF}$ be its shadow. Joined $\mathrm{BC}$ and $\mathrm{EF}$.

Let $\mathrm{DE}=x$ meters

Here, $\mathrm{AB}=6 \mathrm{~m}, \mathrm{AC}=4 \mathrm{~m}$ and $\mathrm{DF}=28 \mathrm{~m}$

In the triangles $\mathrm{ABC}$ and $\mathrm{DEF}$,
$
\angle \mathrm{A}=\angle \mathrm{D}=90^{\circ}
$

And $\angle \mathrm{C}=\angle \mathrm{F}$ [Each is the angular elevation of the sun]

$\therefore$ By AA-similarity criterion,
$
\begin{aligned}
& \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF} \\
& \Rightarrow \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}} \\
& \Rightarrow \frac{6}{x}=\frac{4}{28} \\
& \Rightarrow \frac{6}{x}=\frac{1}{7} \\
& \Rightarrow x=42 \mathrm{~m}
\end{aligned}
$

Hence, the height of the tower is 42 meters.
Ex 6.3 Question 16.

If $\mathrm{AD}$ and $\mathrm{PM}$ are medians of triangles $\mathrm{ABC}$ and $\mathrm{PQR}$ respectively, where $\triangle \mathrm{ABC} \sim \Delta$ PQR, prove that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$.

Answer.

Given: AD and PM are the medians of triangles
$\mathrm{ABC}$ and $\mathrm{PQR}$ respectively, where

$\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$
To prove: $\frac{A B}{P Q}=\frac{A D}{P M}$
Proof: In triangles ABD and PQM,
$\angle \mathrm{B}=\angle \mathrm{Q}$ [Given]
And $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}[\because \mathrm{AD}$ and $\mathrm{PM}$ are the medians of $\mathrm{BC}$ and $\mathrm{QR}$ respectively $]$
$
\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}
$
$\therefore$ By SAS-criterion of similarity,
$
\triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}
$

$\begin{aligned}
& \Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M} \\
& \Rightarrow \frac{A B}{P Q}=\frac{A D}{P M}
\end{aligned}$