Exercise 6.2 (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 6.2 Question 1.

$\text {In figure (i) and (ii), } \mathrm{DE} \| \mathrm{BC} \text {. Find } \mathrm{EC} \text { in (i) and } \mathrm{AD} \text { in (ii). }$

Answer.

(i) Since $D E \| B C$,
$
\begin{aligned}
& \therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\
& \Rightarrow \frac{1.5}{3}=\frac{1}{\mathrm{EC}} \\
& \Rightarrow \mathrm{EC}=\frac{3}{1.5} \\
& \Rightarrow \mathrm{EC}=2 \mathrm{~cm}
\end{aligned}
$
(ii)Since $D E \| B C$,
$
\begin{aligned}
& \therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\
& \Rightarrow \frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \mathrm{AD}=\frac{1.8 \times 7.2}{5.4} \\
& \Rightarrow \mathrm{EC}=2.4 \mathrm{~cm}
\end{aligned}
$
Ex 6.2 Question 2.

$E$ and $F$ are points on the sides $P Q$ and $P R$ respectively of a $\triangle P Q R$. For each of the following cases, state whether $E F \| Q$ Q:
(i) $\mathrm{PE}=3.9 \mathrm{~cm}, \mathrm{EQ}=4 \mathrm{~cm}, \mathrm{PF}=3.6 \mathrm{~cm}$ and $\mathrm{FR}=2.4 \mathrm{~cm}$
(ii) $\mathrm{PE}=4 \mathrm{~cm}, \mathrm{QE}=4.5 \mathrm{~cm}, \mathrm{PF}=8 \mathrm{~cm}$ and $\mathrm{RF}=9 \mathrm{~cm}$
(iii) $\mathrm{PQ}=1.28 \mathrm{~cm}, \mathrm{PR}=2.56 \mathrm{~cm}, \mathrm{PE}=0.18 \mathrm{~cm}$ and $\mathrm{PF}=0.36 \mathrm{~cm}$

Answer.

(i)Given: $\mathrm{PE}=3.9 \mathrm{~cm}, \mathrm{EQ}=4 \mathrm{~cm}, \mathrm{PF}=3.6 \mathrm{~cm}$ and $\mathrm{FR}=2.4 \mathrm{~cm}$
Now, $\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{4}=0.97 \mathrm{~cm}$

$
\begin{aligned}
& \text { And } \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=1.2 \mathrm{~cm} \\
& \because \frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}
\end{aligned}
$

Therefore, $\mathrm{EF}$ does not divide the sides $\mathrm{PQ}$ and $\mathrm{PR}$ of $\triangle \mathrm{PQR}$ in the same ratio.
$\therefore \mathrm{EF}$ is not parallel to $\mathrm{QR}$.
(ii) Given: $\mathrm{PE}=4 \mathrm{~cm}, \mathrm{QE}=4.5 \mathrm{~cm}, \mathrm{PF}=8 \mathrm{~cm}$ and $\mathrm{RF}=9 \mathrm{~cm}$

Now, $\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4.5}=\frac{8}{9} \mathrm{~cm}$
And $\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9} \mathrm{~cm}$
$\because \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}$
Therefore, EF divides the sides $\mathrm{PQ}$ and $\mathrm{PR}$ of $\triangle \mathrm{PQR}$ in the same ratio.
$\therefore$ EF is parallel to $\mathrm{QR}$.
(iii) Given: $\mathrm{PQ}=1.28 \mathrm{~cm}, \mathrm{PR}=2.56 \mathrm{~cm}, \mathrm{PE}=0.18 \mathrm{~cm}$ and $\mathrm{PF}=0.36 \mathrm{~cm}$
$\Rightarrow \mathrm{EQ}=\mathrm{PQ}-\mathrm{PE}=1.28-0.18=1.10 \mathrm{~cm}$
And $\mathrm{ER}=\mathrm{PR}-\mathrm{PF}=2.56-0.36=2.20 \mathrm{~cm}$
Now, $\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55} \mathrm{~cm}$
And $\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55} \mathrm{~cm}$
$\because \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}$
Therefore, EF divides the sides $\mathrm{PQ}$ and $\mathrm{PR}$ of $\triangle \mathrm{PQR}$ in the same ratio.
$\therefore$ EF is parallel to $\mathrm{QR}$.
Ex 6.2 Question 3.

In figure, if $L M \| C B$ and $L N \| C D$, prove that $\frac{A M}{A B}=\frac{A N}{A D}$.

Answer.

In $\triangle \mathrm{ABC}, \mathrm{LM} \| \mathrm{CB}$
$\therefore \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}}$ [Basic Proportionality theorem]
And in $\triangle \mathrm{ACD}, \mathrm{LN} \| \mathrm{CD}$
$\therefore \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AN}}{\mathrm{AD}}$ [Basic Proportionality theorem]
From eq. (i) and (ii), we have
$
\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}
$
Ex 6.2 Question 4.

In the given figure, $\mathrm{DE} \| \mathrm{AC}$ and $\mathrm{DF} \| \mathrm{AE}$. Prove that $\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$.

Answer.

In $\triangle \mathrm{BCA}, \mathrm{DE} \| \mathrm{AC}$
$\therefore \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BD}}{\mathrm{DA}}$ [Basic Proportionality theorem]
And in $\triangle \mathrm{BEA}, \mathrm{DF} \| \mathrm{AE}$

$
\therefore \frac{\mathrm{BE}}{\mathrm{FE}}=\frac{\mathrm{BD}}{\mathrm{DA}}[\text { Basic Proportionality theorem] }
$

From eq. (i) and (ii), we have
$
\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}
$
Ex 6.2 Question 5.

In the given figure, DE $\| O Q$ and DF $\|$ OR. Show that EF $\| Q R$.

Answer.

In $\triangle \mathrm{PQO}, \mathrm{DE} \| \mathrm{OQ}$
$\therefore \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}}[$ Basic Proportionality theorem $]$
And in $\triangle \mathrm{POR}, \mathrm{DF} \| \mathrm{OR}$
$\therefore \frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}}[$ Basic Proportionality theorem]
From eq. (i) and (ii), we have
$
\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}
$
$\therefore \mathrm{EF} \| \mathrm{QR}[$ By the converse of BPT]
Ex 6.2 Question 6.

In the given figure, $A, B$, and $C$ are points on $O P, O Q$ and $O R$ respectively such that $A B$ $\| P Q$ and $A C \| P R$. Show that $B C \|$ QR.

Answer.

Given: $\mathrm{O}$ is any point in $\triangle \mathrm{PQR}$, in which $\mathrm{AB} \| \mathrm{PQ}$ and $\mathrm{AC} \| \mathrm{PR}$.
To prove: $B C \| Q R$
Construction: Join BC.
Proof: In $\triangle \mathrm{OPQ}, \mathrm{AB} \| \mathrm{PQ}$
$\therefore \frac{O A}{A P}=\frac{O B}{B Q}$ [Basic Proportionality theorem]
And in $\triangle \mathrm{OPR}, \mathrm{AC} \| \mathrm{PR}$
$\therefore \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}$ [Basic Proportionality theorem].
From eq. (i) and (ii), we have
$\frac{O B}{B Q}=\frac{O C}{C R}$
$\because$ In $\triangle \mathrm{OQR}, \mathrm{B}$ and $\mathrm{C}$ are points dividing the sides $\mathrm{OQ}$ and $\mathrm{OR}$ in the same ratio.
$\therefore$ By the converse of Basic Proportionality theorem,
$\Rightarrow \mathrm{BC} \| \mathrm{QR}$
Ex 6.2 Question 7.

Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer.

Given: A triangle $A B C$, in which $D$ is the midpoint of side $A B$ and the line $\mathrm{DE}$ is drawn parallel to $\mathrm{BC}$, meeting $\mathrm{AC}$ at $\mathrm{E}$.

To prove: $\mathrm{AE}=\mathrm{EC}$
Proof: Since DE $\|$ BC
$\therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ [Basic Proportionality theorem] $\qquad$
But $\mathrm{AD}=\mathrm{DB}[$ Given $]$
$
\begin{aligned}
& \Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=1 \\
& \Rightarrow \frac{\mathrm{AE}}{\mathrm{EC}}=1 \text { [From eq. (i)] } \\
& \Rightarrow \mathrm{AE}=\mathrm{EC}
\end{aligned}
$

Hence, $\mathrm{E}$ is the midpoint of the third side $\mathrm{AC}$.
Ex 6.2 Question 8.

Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer.

Given: A triangle $\mathrm{ABC}$, in which $\mathrm{D}$ and $\mathrm{E}$ are the midpoints of sides $A B$ and $A C$ respectively.

To Prove: DE \| $\mathrm{BC}$
Proof: Since D and E are the midpoints of $A B$ and $A C$
respectively.
$\therefore \mathrm{AD}=\mathrm{DB}$ and $\mathrm{AE}=\mathrm{EC}$
Now, $\mathrm{AD}=\mathrm{DB}$
$
\begin{aligned}
& \Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=1 \text { and } \mathrm{AE}=\mathrm{EC} \\
& \Rightarrow \frac{\mathrm{AE}}{\mathrm{EC}}=1 \\
& \therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=1 \\
& \Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}
\end{aligned}
$

Thus, in triangle $\mathrm{ABC}, \mathrm{D}$ and $\mathrm{E}$ are points dividing the sides $\mathrm{AB}$ and $\mathrm{AC}$ in the same ratio.
Therefore, by the converse of Basic Proportionality theorem, we have
$
\mathrm{DE} \| \mathrm{BC}
$
Ex 6.2 Question 9.

$A B C D$ is a trapezium in which $A B \| D C$ and its diagonals intersect each other at the point 0 . Show that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$.

Answer.

Given: A trapezium $\mathrm{ABCD}$, in which $\mathrm{AB} \| \mathrm{DC}$ and its diagonals
$\mathrm{AC}$ and $\mathrm{BD}$ intersect each other at $\mathrm{O}$.

To Prove: $\frac{A O}{B O}=\frac{C O}{D O}$
Construction: Through O, draw OE $\| \mathrm{AB}$, i.e. OE $\|$ DC.
Proof: In $\triangle \mathrm{ADC}$, we have $\mathrm{OE} \| \mathrm{DC}$
$\therefore \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{CO}}$ [By Basic Proportionality theorem]
Again, in $\triangle \mathrm{ABD}$, we have $\mathrm{OE} \| \mathrm{AB}[$ Construction $]$
$\therefore \frac{\mathrm{ED}}{\mathrm{AE}}=\frac{\mathrm{DO}}{\mathrm{BO}}$ [By Basic Proportionality theorem]
$
\Rightarrow \frac{A E}{E D}=\frac{B O}{D O}
$

From eq. (i) and (ii), we get
$
\begin{aligned}
& \frac{A O}{C O}=\frac{B O}{D O} \\
& \Rightarrow \frac{A O}{B O}=\frac{C O}{D O}
\end{aligned}
$
Ex 6.2 Question 10.

The diagonals of a quadrilateral $A B C D$ intersect each other at the point $O$ such that $\frac{A O}{B O}=\frac{C O}{D O}$. Show that ABCD is a trapezium.

Answer.

Given: A quadrilateral $\mathrm{ABCD}$, in which its diagonals $\mathrm{AC}$ and $B D$ intersect each other at $O$ such that $\frac{A O}{B O}=\frac{C O}{D O}$, i.e.

$
\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{DO}}
$

To Prove: Quadrilateral $\mathrm{ABCD}$ is a trapezium.
Construction: Through O, draw $\mathrm{OE} \| \mathrm{AB}$ meeting $\mathrm{AD}$ at $\mathrm{E}$.
Proof: In $\triangle \mathrm{ADB}$, we have $\mathrm{OE} \| \mathrm{AB}$ [By construction]
$\therefore \frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{OD}}{\mathrm{BO}}[$ By Basic Proportionality theorem]
$\Rightarrow \frac{E A}{D E}=\frac{B O}{D O}$
$\Rightarrow \frac{E A}{D E}=\frac{B O}{D O}=\frac{A O}{C O}$
$\left[\because \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{DO}}\right]$
$\Rightarrow \frac{\mathrm{EA}}{\mathrm{DE}}=\frac{\mathrm{AO}}{\mathrm{CO}}$
Thus in $\triangle A D C, E$ and $\mathrm{O}$ are points dividing the sides $\mathrm{AD}$ and $\mathrm{AC}$ in the same ratio. Therefore by the converse of Basic Proportionality theorem, we have
$\mathrm{EO} \| \mathrm{DC}$
But EO \| AB[By construction]
$\therefore A B \| D$
$\therefore$ Quadrilateral ABCD is a trapezium