Mensuration : Area of Triangle , AMC 8 2015 Problem 21

Can you find the lengths of one side of the squares $ABJI$ and $FEHG$ ??

If you can, then try to use the given conditions to find out the lengths of two sides of $\triangle KBC$

Area of a square =$(\textbf{Side of the square})^2$

Then Side of a square =$\sqrt{(\textbf{Area of the square})}$

Then you can easily find the lengths of a side of the squares $ABJI$ and $FEHG$ which is $3\sqrt{2}$ and $4\sqrt{2}$ respectively.

Then by the given condition $\overline{FE}=\overline{BC}=4\sqrt{2}$

Since $\triangle JBK$ is an equilateral triangle then all of its sides are equal.

and $\overline{JB}=\overline{BK}=3\sqrt{2}$

Now as you know that $\triangle JBK$ is equilateral and the hexagon $ABCDEF$ is equiangular. Can you find out the measure of $\angle KBC$ ??

From the figure we can clearly see $\angle JBA + \angle ABC + \angle KBC + \angle KBJ = 360^{\circ}$

$\angle KBC = 360^{\circ}-90^{\circ}-120^{\circ}-60^{\circ}$ [Since $\angle JBA=90^{\circ}$ (an angle of a square) $\angle ABC=120^{\circ}$ (an angle of an equiangular hexagon) and $\angle JBK= 60^{\circ} $(an angle of an equilateral triangle)]

i.e., $\angle KBC= 90^{\circ}$

Then $\triangle KBC$ is a right angle triangle

Then the base and height of $\triangle KBC$ are $BC$ and $KB$

So the area of $\triangle KBC=\frac12 \times KB \times BC = \frac12 \times 4\sqrt{2} \times 3\sqrt{2} = 12$

Can you find the lengths of one side of the squares $ABJI$ and $FEHG$ ??

If you can, then try to use the given conditions to find out the lengths of two sides of $\triangle KBC$

Area of a square =$(\textbf{Side of the square})^2$

Then Side of a square =$\sqrt{(\textbf{Area of the square})}$

Then you can easily find the lengths of a side of the squares $ABJI$ and $FEHG$ which is $3\sqrt{2}$ and $4\sqrt{2}$ respectively.

Then by the given condition $\overline{FE}=\overline{BC}=4\sqrt{2}$

Since $\triangle JBK$ is an equilateral triangle then all of its sides are equal.

and $\overline{JB}=\overline{BK}=3\sqrt{2}$

Now as you know that $\triangle JBK$ is equilateral and the hexagon $ABCDEF$ is equiangular. Can you find out the measure of $\angle KBC$ ??

From the figure we can clearly see $\angle JBA + \angle ABC + \angle KBC + \angle KBJ = 360^{\circ}$

$\angle KBC = 360^{\circ}-90^{\circ}-120^{\circ}-60^{\circ}$ [Since $\angle JBA=90^{\circ}$ (an angle of a square) $\angle ABC=120^{\circ}$ (an angle of an equiangular hexagon) and $\angle JBK= 60^{\circ} $(an angle of an equilateral triangle)]

i.e., $\angle KBC= 90^{\circ}$

Then $\triangle KBC$ is a right angle triangle

Then the base and height of $\triangle KBC$ are $BC$ and $KB$

So the area of $\triangle KBC=\frac12 \times KB \times BC = \frac12 \times 4\sqrt{2} \times 3\sqrt{2} = 12$