# 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) - Mathematics

MCQ

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

• 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(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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Q 10 | Page 61
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