SATHEE: Properties of Triangles 3 Question 11

Properties of Triangles 3 Question 11

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

$(2008, 4 M)$

(a) $\frac{1}{P S} + \frac{1}{S T} < \frac{2}{\sqrt{Q S \times S R}}$

(b) $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}}$

(d) $\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}$

Fill in the Blanks

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Answer:

Correct Answer: 11. (b, d)

Solution:

Let a straight line through the vertex $P$ of a given $△ P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of $△ P Q R$ at the point $T$ .

Points $P, Q, R, T$ are concyclic, then $P S \cdot S T = Q S \cdot S R$

Now, $\frac{P S + S T}{2} > \sqrt{P S \cdot S T} [∵ A M > G M]$

and $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{P S \cdot S T}} = \frac{2}{\sqrt{Q S \cdot S R}}$

Also, $\frac{S Q + Q R}{2} > \sqrt{S Q \cdot S R}$

$\Rightarrow \frac{Q R}{2} > \sqrt{S Q \cdot S R}$

$\Rightarrow \frac{1}{\sqrt{S Q \cdot S R}} > \frac{2}{Q R}$

$\Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} > \frac{4}{Q R}$

$∴ \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \cdot S R}} > \frac{4}{Q R}$

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Properties of Triangles 3 Question 11

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

Fill in the Blanks

Answer:

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Properties of Triangles 3 Question 11

11. A straight line through the vertex P of a △PQR intersects the side QR at the point S and the circumcircle of the △PQR at the point T. If S is not the centre of the circumcircle, then

Fill in the Blanks

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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