Question

If AD is median of ΔABC and P is a point on AC such that
ar (ΔADP) : ar (ΔABD) = 2 : 3, then ar (Δ PDC) : ar (Δ ABC)

Options

• 1 : 5

• 1 : 5

• 1 : 6

• 3 : 5

Answer 1

Given: (1) AD is the Median of ΔABC

(2) P is a point on AC such that ar (ΔADP) : ar (ΔABD) = `2/3`

To find: ar (ΔPDC) : ar (ΔABC)

We know that” the medians of the triangle divides the triangle in two two triangles of equal area.” 

Since AD is the median of ΔABC,

ar (ΔABD) = ar (ΔADC) ……(1)

Also it is given that

ar (ΔADP) : ar (ΔABD) = `2/3` ……(2)

Now,

ar (ΔADC) = ar (ΔADP) + ar (ΔPDC)

ar(ΔADB) = `2/3` ar (ΔADB) + ar (ΔPDC )(from 1 and 2)

ar (ΔPDC ) = `1/3` ar (ΔADB)    ..............(3)

ar (ΔABC) = 2ar (ΔADB)          ...................(4)

Therefore,

`(ar(ΔPDC))/(ar(ΔABC))= (1/3 ar(ΔADB))/(2ar(ΔADB)`

`(ar(ΔPDC))/(ar(ΔABC))=1/6`