In △ABC,OB,OC are angular bisectors and MN∥BC, if AB=12, BC=24 and AC=18, then the perimeter of △AMN is
30
33
36
39
A
30
B
33
C
36
D
39
Open in App
Solution
Verified by Toppr
△AMN⇒?
∠MBO=∠CBO=∠MOB=12∠MBC
AM+MN+NA=AM+MO+NO+NA
=AM+MB+NC+NA
=AB+AC=30
Was this answer helpful?
2
Similar Questions
Q1
In △ABC,OB,OC are angular bisectors and MN∥BC, if AB=12, BC=24 and AC=18, then the perimeter of △AMN is
View Solution
Q2
Right triangle ABC, ¯¯¯¯¯¯¯¯BC = 5, ¯¯¯¯¯¯¯¯AC = 12, and ¯¯¯¯¯¯¯¯¯¯AM = x; MN is perpendicular to AC, NP is perpendicular BC; N is on AB. If y= ¯¯¯¯¯¯¯¯¯¯¯MN+¯¯¯¯¯¯¯¯¯NP, one-half the perimeter of rectangle MCPN, then:
View Solution
Q3
If O is a point within ΔABC then show that : 1) AB+AC=OB+OC 2) AB+BC+CA>OA+OB+OC 3) OA+OB+OC>12(AB+BC+CA).
View Solution
Q4
. O is any point in the interior of a triangle ABC. Prove that
1. AB + AC > OB + OC
2.AB + BC + CA > OA + OB + OC
3.OA + OB + OC > 1/2(AB + AC + BC)
View Solution
Q5
In fig., if OC = 9 cm and OB = 15 cm, then BC + BD is equal to: