Try this beautiful problem from Geometry based on hexagon and Triangle.
In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK $ is equilateral and FE=BC. What is the area of $\triangle KBC$?
Geometry
Triangle
hexagon
Answer:$12$
AMC-8, 2015 problem 21
Pre College Mathematics
Clearly FE=BC
Can you now finish the problem ..........
$\triangle KBC$ is a Right Triangle
can you finish the problem........
Clearly ,since FE is a side of square with area 32
Therefore FE=$\sqrt 32$=$4\sqrt2$
Now since FE=BC,We have BC=$4\sqrt2$
Now JB is a side of a square with area 18
so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$
Lastly $\triangle KBC$ is a right triangle ,we see that
$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$
i.e$ 90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$
i.e $\angle CBK=90^\circ $
So $\triangle KBC $ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$
Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12
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