What is the area of a regular hexagon inscribed in a circle of radius 8 meters?

Mathematics

1 answer:

MAXImum [283]2 years ago

4 0

Answer:

it's 166.277

Step-by-step explanation:

i took the test & got it right

You might be interested in

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$