Ex.9.4 Q7 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

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Question

\(P\) and \(Q\) are respectively the mid-points of sides \(AB\) and \(BC\) of a triangle \(ABC\) and \(R\) is the mid-point of \(AP\), show that

(i) \(\begin{align}{\rm{ar}} (PRQ)  = \,\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{ ar}} (ARC)\end{align}\)

(ii) \(\begin{align}{\rm{ar }} (RQC) =\,\,\frac{3}{8}{\rm{ ar}}(ABC)\end{align}\)

(iii) \(\begin{align}{\rm{ar }} (PBQ) = {\rm{ar }} (ARC)\end{align}\)

 Video Solution
Areas Of Parallelograms And Triangles
Ex 9.4 | Question 7

Text Solution

Steps:

(i) \(PC\) is the median of \(\Delta  ABC. \)

\(\begin{align}\therefore \operatorname{ar}(\Delta BPC)=\operatorname{ar}(\Delta APC) \ldots .(\mathrm{i})\end{align}\)

\(RC\) is the median of \(\Delta  APC. \)

\(\begin{align}\therefore \operatorname{ar}(\Delta ARC)= \!\frac{1}{2} \!\operatorname{ar}(\Delta APC)\!  \ldots (\mathrm{ii})\end{align}\)

[Median divides the triangle into two triangles of equal area]

\(PQ\) is the median of \(\Delta  BPC. \)

\(\begin{align}\therefore \operatorname{ar}(\Delta PQC) \!\!=\! \frac{1}{2}\operatorname{ar}(\Delta BPC) \!\!\ldots (\mathrm{iii})\end{align}\)

From eq. (i) and (iii), we get,

\[\begin{align} \text{ar}(\Delta PQC)= \frac{1}{2}\operatorname{ar}(\Delta APC) \ldots .(\mathrm{iv})\end{align}\]

From eq. (ii) and (iv), we get,

\[\begin{align} \text{ar}(\Delta PQC)= \operatorname{ar}(\Delta ARC)  \ldots .(\mathrm{v})\end{align}\]

We are given that \(P\) and \(Q\) are the mid-points of \(AB\) and \(BC\) respectively.

\[\begin{align} \therefore PQ \| AC \text { and } PQ=\frac{1}{2} AC \end{align}\]

\[\begin{align} \Rightarrow & \operatorname{ar}(\Delta APQ) \\ &= \operatorname{ar}(\Delta PQC) \ldots .(\mathrm{vi})\end{align}\]

[triangles between the same parallel lines are equal in area]

From eq. (v) and (vi), we get

\(\operatorname{ar}(\Delta APQ)= \operatorname{ar}(\Delta ARC)  \ldots (\mathrm{vii})\)

\(R\) is the mid-point of \(AP\). Therefore, \(RQ\) is the median of \( \Delta  APQ. \)

\[\begin{align}\therefore\;  &\operatorname{ar}(\Delta PRQ) \\ &=\frac{1}{2} \operatorname{ar}(\Delta APQ) \ldots .(\mathrm{viii}) \end{align}\]

From (vii) and (viii), we get,

\[\begin{align}\therefore \operatorname{ar}(\Delta PRQ)=\frac{1}{2} \operatorname{ar}(\Delta ARC)\end{align}\]

(ii) \(PQ\) is the median of \(\Delta  BPC. \)

\(\begin{align}\therefore \; & {\rm{ar}}\,\left( {\Delta PQC} \right)  \\ & = \frac{1}{2}\,{\rm{ar}} (\Delta BPC) \\ &= \frac{1}{2} \times \frac{1}{2} {\rm{ar }}\left( {\Delta ABC} \right)....\rm(ix)\end{align}\)

Also,

\({\text{ar}}\left( {\Delta PRC} \right) = \begin{pmatrix} \frac{1}{2} {\rm{ar}} (\Delta APC) \\ {\text{ [Using (iv)]}} \end{pmatrix}\)

\(\begin{align} \Rightarrow \, & {\rm{ar}}\left( {\Delta PRC} \right) \\ &= \frac{1}{2} \times \frac{1}{2}{\rm{ar}}\,(\Delta ABC) \\ &= \frac{1}{4}{\rm{ar}}\,(\Delta ABC)...\rm(x) \end{align} \)

Adding eq. (ix) and (x), we get,

\(\begin{align} & \operatorname{ar}(\Delta PQC)+\operatorname{ar}(\Delta PRC) \\ & =  \left(\frac{1}{4}+\frac{1}{4}\right) \operatorname{ar}(\Delta ABC) \end{align}\)

\[ \begin{align}\Rightarrow \, & {\rm{ar }}\left( PQCR \right) \\ &= \frac{1}{2}{\rm{ar}}\left( {\Delta ABC} \right)...\left( {{\rm{xi}}} \right) \end{align} \]

Subtracting ar \( (\Delta  PRQ) \) from the both sides,

\[\begin{align} &{\rm{ ar}}(\Delta PQCR )-\operatorname{ar}(\Delta PRQ) \\ & \quad =\frac{1}{2} \operatorname{ar}(\Delta ABC)-\operatorname{ar}(\Delta PRQ) \end{align}\]

\[\begin{align}&\Rightarrow {\rm{ar}}\,\Delta RQC) \\ & \quad = \begin{pmatrix} \frac{1}{2} {\rm{ar}}( \Delta ABC) \\ - \frac{1}{2}{\rm{ar}}(\Delta ARC) \\ \text{[Using result (i)]} \end{pmatrix}\end{align} \]

\[\begin{align} &\Rightarrow {\rm{ar}}( \Delta ARC) \\ & \quad=\begin{pmatrix} \frac{1}{2}{\rm{ ar}} (\Delta ABC) \\ - \frac{1}{2} \times \frac{1}{2}{\rm{ar}}(\Delta APC) \end{pmatrix} \\ &\Rightarrow {\rm{ar}}(\Delta RQC) \\ & \quad =\begin{pmatrix} \frac{1}{2}{\rm{ ar}}(\Delta ABC) \\ - \frac{1}{4} {\rm{ar}}(\Delta APC)\end{pmatrix} \end{align}\]

\[\begin{align} \Rightarrow \;& \text{ar}(\Delta RQC) \\ &= \begin{pmatrix}\frac{1}{2} \operatorname{ar}(\Delta ABC)  - \\ \frac{1}{4} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) \\ \begin{bmatrix} PC \text { is median of }\\  \Delta ABC\end{bmatrix} \end{pmatrix}  \end{align}\]

\[\begin{align}\Rightarrow   \,&\text{ar}(\Delta RQC) \\ &= \begin{pmatrix} \frac{1}{2} \operatorname{ar}(\Delta ABC) \\ -\frac{1}{8} \operatorname{ar}(\Delta ABC) \end{pmatrix} \end{align}\]

\[\begin{align}\Rightarrow \, & \text{ar}(\Delta RQC) \\ &= \begin{pmatrix} \left(\frac{1}{2}-\frac{1}{8}\right) \\ \times \operatorname{ar}(\Delta ABC) \end{pmatrix} \end{align}\]

\[\begin{align}\Rightarrow \, &\text{ar}(\Delta RQC) \\ &=\frac{3}{8} \times \operatorname{ar}(\Delta ABC) \end{align}\]

(iii)

\[\begin{align}  {\rm{ar }}\left( {\Delta PRQ} \right) &= \frac{1}{2}{\rm{ar }}\left( {\Delta ARC} \right)\, \\ & \quad [\rm Using\,{\rm{result (i)}}]  \end{align}\]
\[ \begin{align} \Rightarrow \,  & 2{\rm{ar}}\left( {\Delta PRQ} \right) \\ &=  {\rm{ar}}\left( {\Delta ARC} \right)...\rm(xii)\end{align}\]

\[\begin{align}\text{ar}(\Delta PRQ) &=\frac{1}{2} \text{ ar }(\Delta APQ) \\ & \;\; \begin{bmatrix} RQ \text{ is the}\\ \text{ median of} \\ \Delta APQ \end{bmatrix} \!\!\dots \text{(xiii)}\end{align}\]

But,

\[ \begin{align} & \text{ ar } \left( {\Delta APQ} \right) \\  & = {\rm{ ar }}\left( {\Delta PQC} \right) \\ & \quad \begin{bmatrix} \text{Using reason } \\ \text{of eq}.( {\rm{vi}} ) \end{bmatrix}...\left( {\rm{xiv}} \right) \end{align} \]

From eq. (\(\rm{xiii}\)) and (\(\rm{xiv}\)), we get,

\[\begin{align}\operatorname{ar}(\Delta PRQ)=\frac{1}{2} \operatorname{ar}(\Delta PQC)  \ldots(\mathrm{xv})\end{align}\]

But,

\[\begin{align}\text{ ar} \left( {\Delta BPQ} \right) &= {\rm{ ar}}\left( {\Delta PQC} \right) \\ & \begin{bmatrix} PQ \text{ is the} \\ \text{ median of }  \\ \Delta BPC \end{bmatrix}\!\!...\left( {\rm{xvi}} \right) \end{align} \]

From eq. (\(\rm{xv}\)) and (\(\rm{xvi}\)), we get,

 \[\begin{align} &\text{ar} (\Delta PRQ) \\ &=\frac{1}{2} \operatorname{ar}(\Delta BPQ) \!\! \ldots\rm (xvii)\end{align}\]

Now from (\(\rm{xii}\)) and (\(\rm{xvii}\)), we get,

\[\begin{align} 2  \times \frac{1}{2} \text{ar} (\Delta BPQ) & = {\rm{ar}}(ARC) \\ \Rightarrow \quad \quad {\rm{ar}} (\Delta BPQ)  & = {\rm{ar(}}\Delta ARC) \end{align} \]

 Video Solution
Areas Of Parallelograms And Triangles
Ex 9.4 | Question 7