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
Solution
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`
