Question

Find the area of a regular hexagon ABCDE with the given consecutive vertices A(-4,2) and B (0,5).

Answer 1

Answer:

Option C. $65.0\ units^{2}$

Step-by-step explanation:

we know that

The area of a regular hexagon can be divided into six equilateral triangles

Applying the law of sines

The area is equal to

$A=6[\frac{1}{2}b^{2} sin(60\°)]$

where

b is the length side of the regular hexagon

The length side of the regular hexagon is equal to the distance from consecutive vertices A(-4,2) and B (0,5)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$b=\sqrt{(5-2)^{2}+(0+4)^{2}}$

$b=\sqrt{(3)^{2}+(4)^{2}}$

$b=\sqrt{25}$

$b=5\ units$

Find the area

$A=6[\frac{1}{2}(5)^{2} sin(60\°)]=65.0\ units^{2}$