A regular hexagon is inscribed in a circle as shown determine the measure of FE

Mathematics

2 answers:

iVinArrow [24]3 years ago

8 0

A regular hexagon is inscribed in a circle as shown. Determine the measure of FE.

FE= 6 CM.

Ostrovityanka [42]3 years ago

3 0

The complete question in the attached figure

we know that
1) The triangles that are formed in the hexagon by joining all the vertices with the center of the hexagon are all equilateral and are equal in size
therefore
the radius of the circle is equals to the length side of the regular hexagon
FE=BP--------> FE=6 cm

the answer is
FE=6 cm

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Find all real solutions of the equation by completing the square. (Enter your answers as a comma-separated list.) x2 − 4x − 8 =

Nat2105 [25]

Answer:

After completing the square the expression is (x-2)^2 -12 =0 and real solution are x = 5.46 and x = -1.46

Step-by-step explanation:

We need to complete the square:

x^2 − 4x − 8 = 0

x^2 -2(x)(2)+(2)^2 -8 -4 =0

(x^2 -4x +4) - 12 = 0

(x-2)^2 -12 =0

Now, finding the value of x

(x-2)^2 -12 =0

(x-2)^2 = 12

taking square root on both sides

$\sqrt{(x-2)^2}= \sqrt{12} \\(x-2) = \pm\sqrt{12} \\x = \pm\sqrt{12} +2\\x = \sqrt{12}+2 \,\,and\,\,x=-\sqrt{12}+2\\x= 3.46+2 \,\,and\,\,x=-3.46+2\\x =5.46\,\,and\,\,x=-1.46$