Question

In the given figure, $AC||FD.IfBC=\frac{1}{4}ACandED=4BC,$ then the ratio of ar(∆ABE) to ar(∆BED) is equal to

• A.
  $1:2$
• B.
  $1:4$
• C.
  $3:4$
• D.
  $3:2$

Answer 1

The correct option is C

$3:4$
Let the height of the $\Delta ABE$ be h.

$AC||ED\left(Given\right)BC=\frac{1}{4}AC\left(Given\right)\Rightarrow 4BC=AC\dots ..\left(i\right)Also,ED=4BC\left(Given\right)\Rightarrow AC=ED$

Now, we know that triangles on the same base and between the same parallels are equal in area.

$\Delta ACEand\Delta BED$ lie on the same base AC = DE and between the same parallels AC and ED.

$\therefore Ar\left(\Delta ACE\right)=Ar\left(\Delta BED\right)Now,\frac{Ar\left(\Delta ABE\right)}{Ar\left(\Delta BED\right)}=\frac{\frac{1}{2}\times AB\times h}{Ar\left(\Delta ACE\right)}=\frac{\frac{1}{2}\times (AC-BC)\times h}{\frac{1}{2}\times AC\times h}=\frac{4BC-BC}{4BC}\left[using\right(i\left)\right]=\frac{3}{4}$

Hence, the correct answer is option (c).