TS EAMCET · Maths · Straight Lines
Let \(\mathrm{ABC}\) be a triangle. Let a point \(\mathrm{P}\) divide \(\mathrm{AB}\) in the ratio \(1: 2\) internally and a point \(\mathrm{Q}\) divide \(\mathrm{BC}\) in the ratio \(1: 2\) internally. Let \(\mathrm{D}\) be the point of intersection of \(\mathrm{AQ}\) and \(\mathrm{CP}\). If the area of the triangle \(\mathrm{ABC}\) is \(\mathrm{k}\) square units then the area of the triangle \(\mathrm{BCD}\) in sq. units is
- A \(\frac{4 k}{7}\)
- B \(\frac{2 k}{7}\)
- C \(\frac{7 k}{2}\)
- D \(\frac{7 k}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 k}{7}\)
Step-by-step Solution
Detailed explanation
\(\frac{AP}{PB} = \frac{1}{2}\) \(\frac{BQ}{QC} = \frac{1}{2} \Rightarrow \frac{BC}{CQ} = \frac{BQ+QC}{QC} = \frac{3}{2}\) By Menelaus' Theorem on \(\triangle ABQ\) with transversal C-D-P: \(\frac{AP}{PB} \cdot \frac{BC}{CQ} \cdot \frac{QD}{DA} = 1\)…
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