Question

AM is the median of ΔABC. N is on AC such that AN : NC = 3 : 2. If area of ΔMNC = 12 cm², find the area of ΔABC.

Answer 1

Given:

• AM is the median of triangle ABC (M is midpoint of BC)
• N is a point on AC such that ratio AN : NC = 3 : 2
• Area of triangle MNC = 12 cm²

To find: Area of triangle ABC

Now, ΔAMN and ΔMNC lie on the same base AC and have a common vertex M.

So, heights are same.

`(ar(ΔAMN))/(ar(ΔMNC)) = (1/2 xx AN xx h)/(1/2 xx NC xx h)`

⇒ `(ar(ΔAMN))/12 = (AN)/(NC)`

⇒ `(ar(ΔAMN))/12 = 3/2`

⇒ ar(ΔAMN) = 18 cm²

Since the median of a triangle divides it into two triangles of equal areas.

Therefore, ar(ΔABC) = 2 × ar(ΔAMC)

= 2 × ar(ΔAMN) + ar(ΔMNC)

= 2 × 18 + 12

= 2 × 30

= 60 cm²