In triangle ABC, AD is the bisector of ∠A. If AB = 5 cm, AC= 7.5 cm and BC = 10 cm, then what is the distance of D from the mid-point of BC (in cm)?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is Option 2 i.e 1.
In the above diagram AB = 5 cm, AC = 7.5 cm, BC = 10 cm. AD is the angle biasector and P is the midpoint of BC.
AB/AC = BD/DC
⇒ 5/7.5 = BD/DC
⇒ 2x : 3x = BD : DC
BC = BD + DC
⇒ 10 = 2x + 3x
x= 2
BD = 4 cm
DC = 6 cm
Midpoint BP = CP = BD/2 = 10/2 = 5 cm
So, the distance between midpoint and D = DP - BD = 5 - 4 = 1 cm