INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

March 14, 2020

Hexagon and Triangle |AMC 8- 2015 -|Problem 21

Try this beautiful problem from Geometry based on hexagon and Triangle.

Area of Triangle | AMC-8, 2015 |Problem 21


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 $\triangle KBC$?

hexagon and triangle

  • 9
  • 12
  • 32

Key Concepts


Geometry

Triangle

hexagon

Check the Answer


Answer:$12$

AMC-8, 2015 problem 21

Pre College Mathematics

Try with Hints


Clearly FE=BC

Can you now finish the problem ..........

$\triangle KBC$ is a Right Triangle

can you finish the problem........

hexagon and triangle

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

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com