Question

Find the area of a regular hexagon with side length 10 m. Round to the nearest tenth.​

Answer 1

Answer:

$A=259.8\ m^2$

Step-by-step explanation:

we know that

The area of a regular hexagon is the same that the area of 6 equilateral triangles

The area of 6 equilateral triangles applying the law of sines is equal to

$A=6[\frac{1}{2}b^2sin(60^o)]$

where

b is the length side of the regular hexagon

we have

$b=10\ m$

substitute

$A=6[\frac{1}{2}(10)^2sin(60^o)]$

$A=259.8\ m^2$