KVPY SA Exam 2018 Question Paper

Question : 14 of 80

Marks: +1, -0

Let PQR be an acute-angled triangle in which PQ < QR. From the vertex Q draw the altitude Q

Q_1

, the angle bisector Q Q

_2

and the medium Q Q

_3

with

Q_1, Q_2, Q_3

lying on PR. Then

Solution:

PQ_3 = Q_3R (∴ QQ_3

ie median)

PQ_3 = 1/2PR

PQ_2 : Q_2 R = r : p

(By property of angle bisector)

PQ_2 =(r/{r+P})PR

But r < P (Given)

PQ _2 <1/2PR

Comparison between Altitude and angle bisector
⇒

∠ QPQ_2 + ∠ PQ_2Q + ∠ PQQ_2

= ∠ RQQ_2 +∠ QQ_2R + ∠QRQ_2

∴

∠PQQ_2 = ∠RQQ_2

{Since angle bisector}

∠QPQ_2 + ∠PQ_2Q = ∠QQ_2R + ∠QRQ_2

∴ $PQ < QR \then ∠ QRQ_2$
Hence

∠Q\ Q_2P < ∠QQ_2\ R

But

∠QQ_2P + ∠QQ_2R = 180°

Hence

∠QQ_2P < 90°\ &\ ∠QQ_2R > 90°

⇒ Foot from Q to side PR lies inside

∆\PQQ_2

⇒

PQ_1 < PQ_2 < PQ_3

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