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\(\mathbf{A B}=\mathbf{a}\) and \(\mathbf{A C}=\mathbf{b}\) are the sides of \(\mathbf{a} \triangle A B C . P\) is a point on \(\mathbf{A B}\) and \(Q\) is a point on \(\mathbf{B C}\) such that \(\frac{A P}{P B}=\frac{1}{2}\) and \(\frac{B Q}{Q C}=\frac{1}{2}\). If the point of intersection of \(\mathbf{A Q}\) and \(\mathbf{C P}\) is \(D\) and the area of \(\triangle B C D\) is 7 square units, then the area of the \(\triangle A B C\) (in the same sq units) is
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Verified Answer
The correct answer is:
\(\frac{49}{4}\)
According to given informations,
\(\mathbf{A P}=\frac{\mathbf{a}}{3} \text { and } \mathbf{A Q}=\frac{2 \mathbf{a}+\mathbf{b}}{3}\)
Let \(D\) divides the line \(\mathbf{A Q}\) in ratio \(\lambda: 1\) and \(\mathbf{C P}\) in \(\mu: 1\).
So, \(\quad=\frac{\lambda(2 \mathbf{a}+\mathbf{b})}{3(\lambda+1)}=\frac{\mu \frac{\mathbf{a}}{3}+\mathbf{b}}{\mu+1}\)
On comparing
\(\begin{array}{rlrl}
& & \frac{\mu}{3(\mu+1)} & =\frac{2 \lambda}{3(\lambda+1)} \text { and } \frac{1}{\mu+1}=\frac{\lambda}{3(\lambda+1)} \\
\Rightarrow & \frac{\mu}{3(\mu+1)} & =\frac{2}{\mu+1} \Rightarrow \mu=6 \\
\text {So, } & \frac{1}{7} & =\frac{\lambda}{3(\lambda+1)} \Rightarrow 3 \lambda+3=7 \lambda \Rightarrow \lambda=\frac{3}{4} \\
\Rightarrow & & \text { AD } & =\frac{2 a+\mathbf{b}}{7}
\end{array}\)
So, \(\frac{1}{7}=\frac{\lambda}{3(\lambda+1)} \Rightarrow 3 \lambda+3=7 \lambda \Rightarrow \lambda=\frac{3}{4}\)
\(\Rightarrow \quad \mathbf{A D}=\frac{2 \mathbf{a}+\mathbf{b}}{7}\)
Now area of \(\triangle B C D=\frac{1}{2}|\mathbf{B C} \times \mathbf{B D}|\)
\(\begin{aligned}
& =\frac{1}{2}\left|(\mathbf{b}-\mathbf{a}) \times\left(\frac{\mathbf{b}-5 \mathbf{a}}{7}\right)\right| \\
& =\frac{1}{14}|(-\mathbf{b} \times \mathbf{a})-(\mathbf{a} \times \mathbf{b})| \\
& =\frac{4}{14}|\mathbf{a} \times \mathbf{b}|=7 \text { (given) }
\end{aligned}\)
\(\Rightarrow \quad \frac{1}{2}|\mathbf{a} \times \mathbf{b}|=\frac{49}{4}\)
\(\Rightarrow\) Area of \(\triangle A B C=\frac{49}{4}\)
Hence, option (a) is correct.
\(\mathbf{A P}=\frac{\mathbf{a}}{3} \text { and } \mathbf{A Q}=\frac{2 \mathbf{a}+\mathbf{b}}{3}\)
Let \(D\) divides the line \(\mathbf{A Q}\) in ratio \(\lambda: 1\) and \(\mathbf{C P}\) in \(\mu: 1\).
So, \(\quad=\frac{\lambda(2 \mathbf{a}+\mathbf{b})}{3(\lambda+1)}=\frac{\mu \frac{\mathbf{a}}{3}+\mathbf{b}}{\mu+1}\)
On comparing
\(\begin{array}{rlrl}
& & \frac{\mu}{3(\mu+1)} & =\frac{2 \lambda}{3(\lambda+1)} \text { and } \frac{1}{\mu+1}=\frac{\lambda}{3(\lambda+1)} \\
\Rightarrow & \frac{\mu}{3(\mu+1)} & =\frac{2}{\mu+1} \Rightarrow \mu=6 \\
\text {So, } & \frac{1}{7} & =\frac{\lambda}{3(\lambda+1)} \Rightarrow 3 \lambda+3=7 \lambda \Rightarrow \lambda=\frac{3}{4} \\
\Rightarrow & & \text { AD } & =\frac{2 a+\mathbf{b}}{7}
\end{array}\)
So, \(\frac{1}{7}=\frac{\lambda}{3(\lambda+1)} \Rightarrow 3 \lambda+3=7 \lambda \Rightarrow \lambda=\frac{3}{4}\)
\(\Rightarrow \quad \mathbf{A D}=\frac{2 \mathbf{a}+\mathbf{b}}{7}\)
Now area of \(\triangle B C D=\frac{1}{2}|\mathbf{B C} \times \mathbf{B D}|\)
\(\begin{aligned}
& =\frac{1}{2}\left|(\mathbf{b}-\mathbf{a}) \times\left(\frac{\mathbf{b}-5 \mathbf{a}}{7}\right)\right| \\
& =\frac{1}{14}|(-\mathbf{b} \times \mathbf{a})-(\mathbf{a} \times \mathbf{b})| \\
& =\frac{4}{14}|\mathbf{a} \times \mathbf{b}|=7 \text { (given) }
\end{aligned}\)
\(\Rightarrow \quad \frac{1}{2}|\mathbf{a} \times \mathbf{b}|=\frac{49}{4}\)
\(\Rightarrow\) Area of \(\triangle A B C=\frac{49}{4}\)
Hence, option (a) is correct.
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