Question

In ΔABC, M is the mid-point of BC, L is a point on AB such that AL = 2LB. Find the area of ΔALM if area of ΔABC = 72 cm².

Answer 1

Given:

• △ABC
• M = midpoint of BC
• L is a point on AB such that AL = 2LB = (AL : LB = 2 : 1),
• Area(△ABC) = 72 cm²
• We need to find Area(△ALM).

Given that, ar(ΔABC) = 72 cm², AL = 2LB

Median is a line from any one vertex of a triangle to the mid-point of the opposite side.

Since, M is the mid-point of BC.

Therefore, AM is the median of ΔABC.

Median of a triangle divides it into two triangles of equal areas.

∴ ar(ΔABM) = `1/2` × ar(ΔABC) = `1/2` × 72 = 36 cm²

Now, ΔABM and ΔALM lie on the same base AB and have a common vertex M.

So, heights are same.

`(ar(ΔABM))/(ar(ΔALM)) = (1/2 xx AB xx h)/(1/2 xx AL xx h)`

⇒ `36/(ar(ΔALM)) = (AB)/(AL)`

= `(AL + LB)/(AL)`

= `(2LB + LB)/(2LB)`

= `(3LB)/(2LB)`

= `3/2`

⇒ ar(ΔALM) = `36 xx 2/3`

= 24 cm²