Carcass wrote:
Attachment:
ex.jpg
What is the area of the hexagonal region shown above ?
A. \(54 \sqrt{3}\)
B. \(108\)
C. \(108 \sqrt{3}\)
D. \(216\)
E.
It cannot be determined from the information givenThere are no. of ways to solve it:
From the diagram we know it is a regular hexagon
and formula of the area of a regular hexagon = \((3\sqrt{3} * side^2)\) * \(\frac{1}{2}\)
= \((3\sqrt{3} * 6^2)\) * \(\frac{1}{2}\) = \(54\sqrt{3}\).
Now if we donot know the formula then, please refer to the diagram below:Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.
Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.
Now area of Equilateral triangle = \(\sqrt{3} * (side^2)\) * \(\frac{1}{4}\) = \(\sqrt{3} * (6^2)\) * \(\frac{1}{4}\)
=\(9\sqrt{3}\)
Therefore for 6 triangles we have = 6* \(9\sqrt{3}\) = \(54\sqrt{3}\).
Now if we donot know the area of equilateral triangle then we can always find out:Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = \(\frac{1}{2} * base * altitude\)
Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)
So \(altitude^2 = 6^2 - 3^2\) = 27
or altitude = \(3\sqrt{3}\)
Now area = \(\frac{1}{2} * 6 * 3\sqrt{3}\) = \(9\sqrt{3}\)
Now area of all the six triangle = 6 * \(9\sqrt{3}\) = \(54\sqrt{3}\) = Area of the hexagon.
** Please note this applies only for a regular hexagonHexagon has 720 degrees, not 360... 720 / 6 = 120