Let M be apoint in the triangle ABC such that area ABM =2 area ACM ,then locus of M A)must be a line B)must be a parabola C)cannot be a circle D)cannot be a line
Answers
Answered by
4
Answer:
Correct option is
A
5
Let AM be x and MC be y.
MN is parallel to BC.
∠ANM=∠ABC
∠AMN=∠ACB
MP is parallel to NB.
∠ANM=∠MPC
Therefore,
ΔANM∼ΔMPC∼ΔABC
By theorem, ratio of areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.
areaΔABC
areaΔANM
=
(AC
2
)
(AM
2
)
=
(x+y)
2
x
2
−−−(1)
areaΔABC
areaΔMPC
=
(AC
2
)
(MC
2
)
=
(x+y)
2
y
2
−−−(2)
areaΔANC+areaΔMPC=area ΔABC−area □NMCB=area ΔABC−
18
5
area ΔABC=
18
13
area ΔABC
Adding equation (1) and (2):
18
13
=
(x+y)
2
x
2
+y
2
5x
2
−26xy+5y
2
=0
5x
2
−25xy−xy+5y
2
=0
(5x−1)(x−5y)=0
y
x
=5 OR
y
x
=
5
1
But x>y
Therefore answer is 5.
Similar questions