Question

Let PQR be an isosceles triangles. Suppose that the sides PQ and PR are equal and let the length of PQ be K cm. If $\angle PQR=\theta ,cos\theta =\frac{4}{5}$ and area of triangle is M sq. cm, then which of the following is true about M?

• A. $M<\frac{K^2}{4}$
• B. $M>\frac{K^2}{2}$
• C. $\frac{K^2}{4}<M<\frac{K^2}{2}$
• D. $\frac{K^2}{2}<M<K^2$

Answer 1

The correct option is C $\frac{K^2}{4}<M<\frac{K^2}{2}$

Given that PQ= PR = K cm, $\angle PQS=\theta ,cos\theta =\frac{4}{5}$ and area of $\Delta PQR=M$ sq. cm.
Let PS be the perpendicular bisector on QR
$\Rightarrow \angle PSQ=\angle PSR={90}^{\circ }andQS=SR$
Now in $\Delta PQS.$
$sin\theta =\frac{PS}{PQ}$
$\Rightarrow \sqrt{1-co{s}^2\theta }=\frac{PS}{K}(\because si{n}^2\theta +co{s}^2\theta =1)$
$\Rightarrow PS=K\sqrt{1-{\left(\frac{4}{5}\right)}^2}(\because cos\theta =\frac{4}{5})$
$\Rightarrow PS=K\sqrt{\frac{25-16}{25}}=K\sqrt{\frac{9}{25}}$
$\Rightarrow PS=\frac{3K}{5}.......\left(i\right)$
$Now,cos\theta =\frac{QS}{PQ}$
$\Rightarrow \frac{4}{5}=\frac{QS}{K}$
$\Rightarrow QS=\frac{4K}{5}...\left(ii\right)$
Therefore, area of $\Delta PQR=\frac{1}{2}\times PS\times QR$
$=\frac{1}{2}\times PS\times 2QS(\because QR=2QS$
$=PS\times QS$
$\frac{3K}{5}\times \frac{4K}{5}\left[From\right(i\left)and\right(ii\left)\right]$
$=\frac{12K^2}{25}$
$\therefore M=\frac{12}{25}{K}^2$
$\Rightarrow \frac{K^2}{4}<M<\frac{K^2}{2}$
Hence, the correct answer is option (3).