3 years ago

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

Mathematics

1 answer:

Maksim231197 [3]3 years ago

4 0

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

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

$A=259.8\ m^2$

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

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