I tried several options, but certainly do not know anything about this figure, since I can not solve it.

a) The length of one side of given regular hexagon is $4cm$. What is the are of the shaded region: enter image description here

b) In the figure, $ABCD$ is a parallelogram. $|AF|=8cm,|BF|=4cm, |CE|=9cm.$ What is $|EF|.$ (I find DE, but nothing helps me.) enter image description here

Please help me. Thank you very much.

  • $\begingroup$ For a), you maybe know that the side of a regular hexagon has the same length that the radius of the circumscribed circle. $\endgroup$ – N74 Dec 8 '16 at 15:51
  • $\begingroup$ if you can help me to solve the example you then, please $\endgroup$ – brain host Dec 8 '16 at 15:52
  • $\begingroup$ For b), FBE and CDE are similar so you can find a proportion for the lengths of EF and CE. $\endgroup$ – N74 Dec 8 '16 at 15:58
  • $\begingroup$ I understand no more, for this reason if you can please tell me detail solution $\endgroup$ – brain host Dec 8 '16 at 16:02

(a) As we know that the side of a regular hexagon is equal to the radius of circumscribed circle.

and we also know that the hypotenuse is the diameter the circle.

Hence, The hypotenuse of the given triangle $H=2\times4=8cm$ .

therefore the perpendicular $P=\sqrt{8^2-4^2}$ ...(Pythagoras theorem)

$P=4\sqrt{3} $


Area of shaded region $=\frac{1}{2}\times4\sqrt{3}\times4$

$=8\sqrt{3}cm^2 $

(b) $\angle CDB=\angle DBF$ ...(Alternate interior Angles)

$\angle DCF=\angle CFB$ ...(Alternate interior Angles)

Therefore, $\triangle DEC\sim \triangle BEF$ ...(AA-Similarity)

As you said you had found $|DE|$ you can use it to find $|EF|$


For (a):

Since the given regular hexagon each internal angle is $120$ degrees and every side is equal. So by the given data, we have each angle=$120$ and each side= $4cm$.

The height in your black shaded triangle can be easily calculated by applying cosine rule.

$H^2=4^2+4^2-2\times 4\times 4\times \cos120$.

Which gives $H=4\sqrt{3} $. So, area of right triangle $=\frac{1}{2}\times4\times 4\sqrt{3}=8\sqrt{3}$.

For (b):

Notice that $\triangle DEC$ is similar to $\triangle BEF $.

So, $\frac{DC}{BF}=\frac{EC}{EF}$ $\implies \frac{12}{4}=\frac{9}{EF}\implies EF=3$


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