Question

In the adjoining figure,BD||CA, E is the midpoint of CA and BD=$\frac{1}{2}CA$. The ratio of $\frac{ar\left(△ABC\right)}{ar\left(△DBC\right)}$ is _____.

• A.
  1 : 1
• B.
  1 : 2
• C.
  2 : 1
• D.
  1 : 4

Answer 1

The correct option is C

2 : 1
ANSWER:
E is the midpoint of CA.
So, AE = EC
.....(1)
Also, BD = $\frac{1}{2}$ CA
(Given)
So, BD = AE
.....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of △ABC
so, ar(△BCE) = ar(△ABE) = $\frac{1}{2}$ ar△ABC
.....(1)
ar(△DBC) = ar(△BCE)
.....(2)
(Triangles on the same base and between the same parallels
are equal in area)
From (1) and (2)
ar(∆ABC) = 2ar(∆DBC)