KVPY2018MathematicsProperties of Triangles
Let P   Q   R be an acute-angled triangle in which P   Q < Q   R   . From the vertex Q draw the altitude Q Q 1 the angle bisector Q Q 2 and the median Q Q 3 , with Q 1 , Q 2 , Q 3 lying on P   R . Then,
Options
- AP Q 1 < P Q 2 < P Q 3
- BP Q 2 < P Q 1 < P Q 3
- CP Q 1 < P Q 3 < P Q 2
- DP Q 3 < P Q 1 < P Q 2
Correct answer
A. P Q 1 < P Q 2 < P Q 3
Step-by-step solution
Given, P Q R is an acute angle triangle. P Q Q R ∠ Q R P ∠ Q P R P Q 3 = 1 2 P R P Q 2 : Q 2 R = r : p P Q 2 = r r + p P R But r p P Q 2 1 2 P R Comparison between altitude and angle bisector ∠ Q P Q 2 + ∠ P Q 2 Q + ∠ P Q Q 2 = ∠ R Q Q 2 + ∠ Q Q 2 R + ∠ Q R Q 2 ∴ ∠ P Q Q 2 = ∠ R Q Q 2 [since, Q Q 2 is angle bisector of ∠ Q ∠ Q P Q 2 + ∠ P Q 2 Q = ∠ Q Q 2 R + ∠ Q R Q 2 ∴ P Q Q R the ∠ Q P Q 2 ∠ Q R Q 2 Hence, ∠ Q Q 2 P ∠ Q Q 2 R But ∠ Q Q 2 P + ∠ Q Q 2 R = 180 ° Hence, ∠ Q Q 2 P 90 ° and ∠ Q Q 2 R > 90 ° ∵ Foot from