Ex.6.2 Q2 Triangles Solution - NCERT Maths Class 10

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Question

 \(E\) 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) \(PQ = 1.28\,\rm{cm},\; PR = 2.56 \,\rm{cm}, \;PE = 0.18\,\rm{cm}\) and \(PF = 0.36\,\rm{cm}\)

Text 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 \(BPT,\;EF\parallel QR\) \(\)

  
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