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AP EAMCET · Maths · Three Dimensional Geometry

\(\mathbf{A B}=\mathbf{a}\) 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

  1. A \(\frac{49}{4}\)
  2. B \(\frac{49}{2}\)
  3. C \(\frac{7}{2}\)
  4. D \(\frac{7}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{49}{4}\)

Step-by-step Solution

Detailed explanation

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,…
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