NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2

Question 1

In Figure, (i) and (ii), \(DE || BC.\) Find \(EC\) in (i) and \(AD\) in (ii)

Solution

Video Solution

Reasoning:

As we all know the Basic Proportionality Theorem (B.P.T) or (Thales Theorem)

Two triangles are similar if :

(i) Their corresponding angles are equal

(ii) Their corresponding sides are in the same ratio (or proportion)

Steps:

(i) In\(\,\,\Delta ABC\) 

\[\begin{align} BC||DE \\ \end{align}\]

In\(\begin{align}\Delta ABC\,\,&\And \,\,\Delta ADE \end{align}\)

\[\begin{align} & \angle ABC=\angle ADE \\ & \left( \because \text{Corresponding}\,\text{angles} \right) \\\\ & \angle ACB=\angle AED \\ & \left( \because \text{Corresponding}\,\text{angles} \right) \\\\ &\quad \;\;\angle A=\angle A \\ &\text{(Common)} \\\\ & \Rightarrow \Delta ABC \sim \Delta \text{ADE} \\ \end{align}\]

\[\begin{align} \frac{AD}{DB}&=\frac{AE}{EC} \\  \frac{1.5}{3}&=\frac{1}{EC} \\ EC&=\frac{3\times 1}{1.5} \\  EC&=2\,\text{cm}\end{align}\]

(ii) Similarly, \(\Delta ABC \sim \Delta ADE\)

\[\begin{align} \frac{AD}{DB}&=\frac{AE}{EC} \\ \frac{AD}{7.2} &=\frac{1.8}{5.4} \\ AD&=\frac{7.2\times 1.8}{5.4} \\ AD&=2.4\,\text{cm} \\\end{align}\]

Question 2

\(E\) and \(F\) are points on the sides \(PQ\) and \(PR\) respectively of a \(\Delta \mathrm{PQR}\). For each of the following cases, state whether \(EF || QR:\)

(i)

\(PE = 3.9\,\rm{cm},\; EQ = 3\,\rm{cm},\; PF = 3.6 \,\rm{cm}\) and \(FR = 2.4\,\rm{cm}\)

(ii)

\(PE = 4\,\rm{cm},\; QE = 4.5\,\rm{cm},\; PF = 8\,\rm{cm}\) and \(RF = 9\,\rm{cm}\)

(iii)

\(\begin{align}PQ &= 1.28\,\rm{cm},\;\\ PR &= 2.56 \,\rm{cm}, \\\;PE &= 0.18\,\rm{cm} \\PF &= 0.36\,\rm{cm}\end{align}\)

Solution

Video Solution

(i) Reasoning:

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of \(BPT\))

Steps:

Here,

\[\begin{align} \frac{PE}{EQ}&=\frac{3.9}{3} =1.3\,\text{cm} \\\\ & \rm{and}\\\\ \frac{PF}{FR}&=\frac{3.6}{2.4}= 1.5 \end{align}\]

Hence,

\(\begin{align} \frac {{PE}}{EQ} \not= \frac {{PF}}{FR} \end{align}\)

According to converse of \(BPT,\;EF\) is not parallel to \(QR\).

(ii) Reasoning:

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side(converse of \(BPT\))

Steps:

Here,

\(\begin{align} \frac{{PE}}{EQ}= \frac{{4}}{4.5}=\frac{{8}}{9} \end{align}\)

and

\(\begin{align} \frac{{PF}}{FR}= \frac{{8}}{9} \end{align}\)

Hence,

\(\begin{align} \frac{{PE}}{EQ}=\frac{{PF}}{FR} \end{align}\)

According to converse of \(BPT\), \(EF \parallel QR\)

(iii) Reasoning:

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side(converse of \(BPT\))

Steps:

Here,

\(PQ=1.28 \,\rm{cm}\) \(\) and \(PE=0.18\,\rm{cm}\) \(\) 

\[\begin{align}  {EQ} &={PQ}-{PE} \\ &=(1.28-0.18)\,\rm{cm} \\ &=1.10\, \mathrm{cm} \end{align}\]

\(PR=2.56\,\rm{cm}\) and \(PF=0.36\,\rm{cm}\)

\[\begin{align}  {FR} &={PR}-{PF}\\ &=(2.56-0.36)\,\rm{}cm \\ &=2.20\; \mathrm{cm} \end{align}\]

Now,

\[\begin{align} \frac{PE}{EQ}&=\frac{0.18\text{cm}}{1.10\text{cm}} =\frac{18}{110}  =\frac{9}{55} \\ \frac{PF}{FR} &=\frac{0.36\text{cm}}{2.20\text{cm}}  =\frac{36}{220}=\frac{9}{55} \end{align}\]

\[\begin{align}\Rightarrow\frac{PE}{EQ}=\frac{PF}{FR} \end{align}\]

According to converse of \(BPT,\;EF\parallel QR\) \(\)

Question 3

In Figure if \( LM || CB\) and \(LN || CD\), prove that

\[\begin{align}\frac{AM}{AB}=\frac{AN}{AD}\end{align}\]

Solution

Video Solution

Reasoning:

As we know if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Steps:

In \(\begin{align} \Delta ABC\end{align}\)

\[\begin{align} &LM||CB \\ &\frac{AM}{MB}=\frac{AL}{LC}\;\dots(1) \\ \end{align}\]

In \(\Delta ACD\) 

\[\begin{align}&LN||CD \\ &\frac{AN}{DN}=\frac{AL}{LC}\;\dots(2) \\ \end{align}\]

From equations \((1)\) and \((2)\)

\[\begin{align}\frac{AM}{MB}=\frac{AN}{DN}\end{align}\]

\[\begin{align}\Rightarrow \frac{MB}{AM}=\frac{DN}{AN}\end{align}\]

Adding \(1\) on both sides

\[\begin{align} \frac{MB}{AM}+1&=\frac{DN}{AN}+1 \\ \frac{MB+AM}{AM}&=\frac{DN+AN}{AN} \\ \frac{AB}{AM}&=\frac{AD}{AN} \\ \frac{AM}{AB}&=\frac{AN}{AD} \\ \end{align}\]

Question 4

In Figure, \(DE || AC\) and \(DF || AE.\) Prove that

\[\begin{align}\frac{BF}{FE}=\frac{BE}{EC}\end{align}\]

Solution

Video Solution

Reasoning:

As we know if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Steps:

In \(\Delta ABC\)

\[\begin{align} & DE||AC \end{align}\]

\[\begin{align}\frac{BD}{AD}=\frac{BE}{EC}\;\dots\text{(i)} \\ \end{align}\]

In\(\begin{align}\Delta \,ABE \end{align}\)

\[\begin{align}&DF||AE \\ &\frac{BD}{AD}=\frac{BF}{FE}\;\dots\text{(ii)} \end{align}\]

From \(\rm (i)\) and \(\rm (ii)\)

\[\begin{align} & \frac{BD}{AD}=\frac{BE}{EC}=\frac{BF}{FE} \\ & \frac{BE}{EC}=\frac{BF}{FE} \\ \end{align}\]

Question 5

In Figure, \(DE || OQ\) and \(DF || OR.\) Show that \(EF || QR.\)

Solution

Video Solution

Reasoning:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Steps:

In \(\Delta P O Q\)

\[\begin{align}D E \;&|| O Q \;\text { (Given) } \\ \frac{P E}{E Q}&=\frac{P D}{O D} \;\dots(1)\end{align}\]

In \(\Delta POR\)

\[\begin{align}& DF||OR\;\;\left( \text{Given} \right) \\ & \frac{PF}{FR}=\frac{PD}{DO} \;\dots (2) \end{align}\]

From \((1)\) & \((2)\)

\[\begin{align} \frac{PE}{EQ}&=\frac{PF}{FR}=\frac{PD}{DO} \\ \frac{PE}{EQ}&=\frac{PF}{FR} \\ \end{align}\]

In \(\Delta PQR\)

\[\begin{align}\frac{PE}{EQ}=\frac{PF}{FR} \end{align}\]

\(\therefore\;\;QR||EF\)  (Converse of \(BPT\))

Question 6

In Figure, \(A, \;B\) and \(C\) are points on \(OP, \;OQ \) and \(OR\) respectively such that \(AB || PQ\) and \(AC || PR.\) Show that \(BC || QR.\)

Solution

Video Solution

Reasoning:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Steps:

In \(\Delta OPQ\)

\[\begin{align}AB\,&||\,PQ \;\;\text{(Given)}\\ \frac{OA}{AP}&=\frac{OB}{BQ}\;\dots\rm{(i)} \end{align}\]

\([\because\)  Thales Theorem (BPT)]

In \(\Delta OPR \) 

\[\begin{align} AC\,&||\,PQ \;\;\text{(Given)} \\ \frac{OA}{AP}&=\frac{OC}{CR}\;\dots\rm{(ii)} \end{align}\]

[ \(\because\)Thales Theorem (\(BPT\))]

From \(\rm (i)\) & \(\rm (ii)\)

\[\begin{align}\frac{OA}{AP}=\frac{OB}{BR}=\frac{OC}{CR}\end{align}\]

\[\begin{align}\frac{OB}{BQ}=\frac{OC}{CR}\end{align}\]

Now In \(\Delta OQR\)

\[\begin{align}\frac{OB}{BQ}=\frac{OC}{CR} \end{align}\]

\(BC\,||\,QR \;\; [ \because \text{ Converse of BPT} ]\)

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).

Solution

Video Solution

Reasoning:

We know that theorem \(6.1\) states that “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (\(BPT\))”.

Steps:

In \(\Delta ABC\)\(, D\) is the midpoint of \(AB\)

Therefore,

\[AD = BD\]

\[ \frac{AD}{BD}=1\]

Now,

\[\begin{align}DE&||BC \\ \Rightarrow\frac{AE}{EC}&=\frac{AD}{BD}\,\,\,\text{[Theorem} \,6.1] \\ \Rightarrow \frac{AE}{EC}&=1 \end{align}\]

Hence \(,E\) is the midpoint of \(AC.\)

Question 8

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

Solution

Video Solution

Reasoning:

We know that theorem \(6.2\) tells us if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of \(BPT\))

Steps:

In \(\Delta ABC\)

\(D\) is the midpoint of \(AB\)

\[\begin{align}AD &= BD\\  \frac{AD}{BD} &= 1\;\dots\rm{(i)}\end{align}\]

\(E\) is the midpoint of \(AC\)

\[\begin{align} AE &= CE\\ \frac{AE}{BE} &= 1\;\dots\rm{(ii)}\end{align}\]

From (i) and (ii)

\[\begin{align} \frac{AD}{BD}&=\frac{AE}{BE} = 1 \\ \frac{AD}{BD}&=\frac{AE}{BE} \\\end{align}\]

According to theorem \(6.2,\) (Converse of \(BPT\))

\[DE||BC\]

Question 9

\(ABCD\) is a trapezium in which \(AB || DC\) and its diagonals intersect each other at the point \(O.\) Show that \(\begin{align}\frac{AO}{BO}=\frac{CO}{DO} \end{align}\)

Solution

Video Solution

Reasoning:

As we know if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Steps:

In trapezium \(ABCD\)

\( AB||CD,\)  \(AC\) and \(BD\) intersect at \(‘O’\)

Construct \(XY\) parallel to \(AB \) and \(CD\) \((XY\parallel AB,\,(XY \parallel CD)\) through \(‘O’\)

In \(\begin{align}\Delta ABC\end{align}\)

\[OY\,||\,AB\;\;(\because \text{Construction})\]

According to theorem \(6.1\) \((BPT)\)

\[\begin{align}\frac{BY}{CY}=\frac{OA}{OC}\;\dots\rm{(i)}\end{align}\]

In \(\Delta BCD\)

\(\rm OY||CD\;\; (\because\, \rm{Construction})\)

According to theorem \(6.1\) (\(BPT\))

\[\begin{align}\frac{BY}{CY}=\frac{OB}{OD}\;\dots\left( \text{ii} \right) \\ \end{align}\]

From \(\rm (i)\) and \(\rm (ii)\)

\[\begin{align} \,\,\,\,\,\,\frac{OA}{OC}&=\frac{OB}{OD}\,\,\,\\ \Rightarrow\quad \frac{OA}{OB}&=\frac{OC}{OD} \\ \end{align}\]

Question 10

The diagonals of a quadrilateral \(ABCD\) intersect each other at the point \(‘O’\) such that \(\begin{align}\frac{AO}{BO}=\frac{CO}{DO}\end{align}\). Show that \(ABCD\) is a trapezium.

Solution

Video Solution

Reasoning:

As we know if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.

Steps:

In quadrilateral \(ABCD\)

Diagonals \(AC, BD\) intersect at \(‘O’\)

Draw \( OE||AB\)

In \(\Delta ABC\)

\(\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OE||AB \\ \end{align}\)

\[\Rightarrow \frac{{OA}}{OC} = \frac{{BE}}{CE} \,\,(BPT)\; \cdots(1)\]

But \(\begin{align}\frac{OA}{OB}=\frac{OC}{OD}\,\,\left( \rm{Given} \right)\end{align}\)

\[\begin{align}\Rightarrow \frac{{OA}}{OC}=\frac{{OB}}{{OD}}\;\dots(2)\end{align}\]

From \((1)\) and \((2)\)

\[\begin{align}\frac{OB}{OD}=\frac{BE}{CE}\,\,\,\end{align}\]

In \(\Delta BCD\)

\[\begin{align}\rm \frac{OB}{OD}&=\frac{BE}{CE} \\ \rm OE&\,\,||CD \\ \rm OE& \,\,||AB\,\,\text{and}\,\,\rm OE||CD \\ \Rightarrow \rm AB& \,\,||CD \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow \rm ABCD\,\,\text{is a trapezium} \end{align}\]