1
$\begingroup$

enter image description here

$ABCD$ is a parallelogram. Point $E$ divides $BC$ into two equal lengths. If the area of $BEF$ is 126, what is the area of $ABCD$?

Source: Bangladesh Math Olympiad 2017 Junior Category.

I can not solve this problem. I am confused on the position of $'F'$. There is no information mentioned about it. Is there any lack of information in this question? Can anyone give me a hint?

$\endgroup$
  • 1
    $\begingroup$ The area of $BEF$ doesn't depend on the position of $F\in AD$ because it is equal $\frac{BE\cdot h}{2}$ where $h$ is the height of the parallelogram. $\endgroup$ – SMM Feb 15 at 18:03
2
$\begingroup$

Hint:

  1. Consider the area of $\triangle BFC$, which shares the same height with $\triangle BEF$.
  2. Draw an auxiliary line through $F$ parallel to $AB$ to see that the area found in step (1) is actually half of the area of the parallelogram $ABCD$.
$\endgroup$
2
$\begingroup$

Hint: CEF has the same area as BEF as they are both a half of the same parallelogram.

$\endgroup$
2
$\begingroup$

Let $FK$ be an altitude of $ABCD$.

Thus, $$S_{ABCD}=BC\cdot FK=2BE\cdot FK=4\cdot\frac{BE\cdot FK}{2}=4\cdot126=504.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.