Examples - Chapter 6 Triangles class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 6 - Triangles - Ncert Solutions class 10 - Maths

Example 1 :

If a line intersects sides $A B$ and $A C$ of a $\triangle A B C$ at $D$ and $E$ respectively and is parallel to $B C$, prove that $\frac{A D}{A B}=\frac{A E}{A C}$ (see Fig. 6.13).

Solution :
$\mathrm{DE} \| \mathrm{BC}$
(Given)

So,
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
(Theorem 6.1)

or,
$
\frac{D B}{A D}=\frac{E C}{A E}
$
or,
$
\frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{\mathrm{EC}}{\mathrm{AE}}+1
$
or,
$
\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}
$

So,
$
\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}
$

Example 2 :

$\mathrm{ABCD}$ is a trapezium with $\mathrm{AB} \| \mathrm{DC}$. $\mathrm{E}$ and $F$ are points on non-parallel sides $A D$ and $B C$ respectively such that $E F$ is parallel to $A B$ (see Fig. 6.14). Show that $\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}$.

Solution :

Let us join $\mathrm{AC}$ to intersect $\mathrm{EF}$ at $\mathrm{G}$ (see Fig. 6.15).

$\mathrm{AB} \| \mathrm{DC}$ and $\mathrm{EF} \| \mathrm{AB}$ (Given)

So, EF \|DC (Lines parallel to the same line are parallel to each other)

Now, in $\triangle \mathrm{ADC}$,
$
\text { EG } \| \text { DC (As EF } \| \text { DC) }
$

So, $\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AG}}{\mathrm{GC}} \quad$ (Theorem 6.1)
Similarly, from $\triangle C A B$,
$
\frac{C G}{A G}=\frac{C F}{B F}
$

i.e.,
$
\frac{A G}{G C}=\frac{B F}{F C}
$

Therefore, from (1) and (2),
$
\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}
$

Example 3 :

In Fig. 6.16, $\frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$ and $\angle \mathrm{PST}=$ $\angle \mathrm{PRQ}$. Prove that $\mathrm{PQR}$ is an isosceles triangle.

Solution : It is given that $\frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$.

So, $\quad \mathrm{ST} \| \mathrm{QR} \quad$ (Theorem 6.2)
Therefore,
$\angle \mathrm{PST}=\angle \mathrm{PQR}$ (Corresponding angles)

Also, it is given that

So,
$
\begin{aligned}
& \angle \mathrm{PST}=\angle \mathrm{PRQ} \\
& \angle \mathrm{PRQ}=\angle \mathrm{PQR}[\text { From (1) and (2)] }
\end{aligned}
$

Therefore, $\quad \mathrm{PQ}=\mathrm{PR} \quad$ (Sides opposite the equal angles) i.e., $\mathrm{PQR}$ is an isosceles triangle.

$\text { Example } 4 \text { : In Fig. 6.29, if } \mathrm{PQ} \| \mathrm{RS} \text {, prove that } \Delta \mathrm{POQ} \sim \Delta \mathrm{SOR} \text {. }$

Solution : 

$\text { Example } 5 \text { : Observe Fig. } 6.30 \text { and then find } \angle \mathrm{P} \text {. }$

$\text { Solution : In } \triangle \mathrm{ABC} \text { and } \triangle \mathrm{PQR} \text {, }$

$
\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} \text { and } \frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}
$

That is, $\quad \frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,
$\Delta \mathrm{ABC} \sim \Delta \mathrm{RQP}$
(SSS similarity)

$\text { Therefore, } \quad \angle \mathrm{C}=\angle \mathrm{P} \quad \text { (Corresponding angles of similar triangles) }$

$\text { But } \quad \angle \mathrm{C}=180^{\circ}-\angle \mathrm{A}-\angle \mathrm{B} \quad \text { (Angle sum property) }$

$
=180^{\circ}-80^{\circ}-60^{\circ}=40^{\circ}
$

So,
$
\angle \mathrm{P}=40^{\circ}
$

Example 6 : In Fig. 6.31,
$
\mathrm{OA} \cdot \mathrm{OB}=\mathrm{OC} \cdot \mathrm{OD} \text {. }
$

Show that $\angle \mathrm{A}=\angle \mathrm{C}$ and $\angle \mathrm{B}=\angle \mathrm{D}$.

Solution : $\quad \mathrm{OA} \cdot \mathrm{OB}=\mathrm{OC} \cdot \mathrm{OD} \quad$ (Given)

So,
$
\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}
$

Also, we have $\quad \angle \mathrm{AOD}=\angle \mathrm{COB} \quad$ (Vertically opposite angles) (2)
Therefore, from (1) and (2), $\triangle \mathrm{AOD} \sim \triangle \mathrm{COB} \quad$ (SAS similarity criterion)
So,
$
\angle \mathrm{A}=\angle \mathrm{C} \text { and } \angle \mathrm{D}=\angle \mathrm{B}
$
(Corresponding angles of similar triangles)

Example 7 :

A girl of height $90 \mathrm{~cm}$ is walking away from the base of a lamp-post at a speed of $1.2 \mathrm{~m} / \mathrm{s}$. If the lamp is $3.6 \mathrm{~m}$ above the ground, find the length of her shadow after 4 seconds.3

Solution :

Let AB denote the lamp-post and $C D$ the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32). From the figure, you can see that $\mathrm{DE}$ is the shadow of the girl. Let DE be $x$ metres.

Now, $B D=1.2 \mathrm{~m} \times 4=4.8 \mathrm{~m}$.
Note that in $\triangle \mathrm{ABE}$ and $\triangle \mathrm{CDE}$,
$
\begin{array}{ll}
\angle \mathrm{B}=\angle \mathrm{D} & \begin{array}{l}
\text { (Each is of } 90^{\circ} \text { because lamp-post } \\
\text { as well as the girl are standing } \\
\text { vertical to the ground) }
\end{array}
\end{array}
$

and
$\angle \mathrm{E}=\angle \mathrm{E}$
(Same angle)

So,
$\triangle \mathrm{ABE} \sim \triangle \mathrm{CDE}$
(AA similarity criterion)

Therefore, $\quad \frac{B E}{D E}=\frac{A B}{C D}$
i.e., $\quad \frac{4.8+x}{x}=\frac{3.6}{0.9} \quad\left(90 \mathrm{~cm}=\frac{90}{100} \mathrm{~m}=0.9 \mathrm{~m}\right)$

i.e.,
$
4.8+x=4 x
$
i.e.,
$
3 x=4.8
$
i.e.,
$
x=1.6
$

So, the shadow of the girl after walking for 4 seconds is $1.6 \mathrm{~m}$ long.

Example 8 :

In Fig. 6.33, $\mathrm{CM}$ and $\mathrm{RN}$ are respectively the medians of $\triangle A B C$ and $\triangle \mathrm{PQR}$. If $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$, prove that :
(i) $\triangle$ AMC $\sim \Delta$ PNR

(ii) $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{AB}}{\mathrm{PQ}}$
(iii) $\Delta \mathrm{CMB} \sim \Delta \mathrm{RNQ}$

Solution : (i)
$
\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}
$
(Given)

So,
$
\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}
$

and
$
\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q} \text { and } \angle \mathrm{C}=\angle \mathrm{R}
$

But
$
\mathrm{AB}=2 \mathrm{AM} \text { and } \mathrm{PQ}=2 \mathrm{PN}
$
(As $\mathrm{CM}$ and $\mathrm{RN}$ are medians)

$\text { So, from (1), } \quad \frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}$

[Note : You can also prove part (iii) by following the same method as used for proving part (i).]