NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2
Chapter 6 Ex.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, \(\Delta ABC \sim \Delta ADE\)
\[\begin{align} \frac{AD}{DB}&=\frac{AE}{EC} \\ \frac{AD}{7.2} &=\frac{1.8}{5.4} \\ AD&=\frac{7.2\times 1.8}{5.4} \\ AD&=2.4\,\text{cm} \\\end{align}\]
Chapter 6 Ex.6.2 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 \(BPT,\;EF\parallel QR\) \(\)
Chapter 6 Ex.6.2 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
\[\begin{align} \frac{MB}{AM}+1&=\frac{DN}{AN}+1 \\ \frac{MB+AM}{AM}&=\frac{DN+AN}{AN} \\ \frac{AB}{AM}&=\frac{AD}{AN} \\ \frac{AM}{AB}&=\frac{AN}{AD} \\ \end{align}\]
Chapter 6 Ex.6.2 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 \(\rm (ii)\)
\[\begin{align} & \frac{BD}{AD}=\frac{BE}{EC}=\frac{BF}{FE} \\ & \frac{BE}{EC}=\frac{BF}{FE} \\ \end{align}\]
Chapter 6 Ex.6.2 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 \(BPT\))
Chapter 6 Ex.6.2 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 \(\Delta OQR\)
\[\begin{align}\frac{OB}{BQ}=\frac{OC}{CR} \end{align}\]
\(BC\,||\,QR \;\; [ \because \text{ Converse of BPT} ]\)
Chapter 6 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).
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 \(AC.\)
Chapter 6 Ex.6.2 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 \(BPT\))
\[DE||BC\]
Chapter 6 Ex.6.2 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 \(\rm (ii)\)
\[\begin{align} \,\,\,\,\,\,\frac{OA}{OC}&=\frac{OB}{OD}\,\,\,\\ \Rightarrow\quad \frac{OA}{OB}&=\frac{OC}{OD} \\ \end{align}\]
Chapter 6 Ex.6.2 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 \(\Delta BCD\)
\[\begin{align}\rm \frac{OB}{OD}&=\frac{BE}{CE} \\ \rm OE&\,\,||CD \\ \rm OE& \,\,||AB\,\,\text{and}\,\,\rm OE||CD \\ \Rightarrow \rm AB& \,\,||CD \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow \rm ABCD\,\,\text{is a trapezium} \end{align}\]