A straight line through the vertex $P$ of a triangle $P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the triangle $P Q R$ at the point $T$ . If $S$ is not the centre of the circumcircle then

\frac{1}{P S} + \frac{1}{S T} < \frac{2}{\sqrt{Q S \times S R}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{P S} + \frac{1}{S T} < \frac{4}{Q R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}}$
C $\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}$
Since $S$ is not the centre of the circumcircle $P S \neq S T, Q S \neq S R$
Since $A . M . > G . M .$
$\frac{1}{2} (\frac{1}{P S} + \frac{1}{S T}) > \frac{1}{\sqrt{P S \times S T}}$
$\Rightarrow \frac{1}{P S} \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}}$ ... $[∵ P S \times S T = Q S \times S R]$
We know that if $x + y = k$ a constant, then $x y$ is maximum if $x = y = k / 2$ .
Thus $Q S \times S R < \frac{1}{4} . (Q R)^{2}$
$\Rightarrow \frac{1}{Q S \times S R} > \frac{4}{(Q R)^{2}}$
$\Rightarrow \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}} > \frac{4}{Q R}$

Suggest Corrections

Similar questions

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

The diagonal PR of a quadrilateral PQRS bisects the angles P and R, then

△ P Q R,

P S

P

Q R

Q R

S .

P S : Q S : R S = 2 : 4 : 1,

?

Join BYJU'S Learning Program