Question

What radius of a circle is required to inscribe a regular hexagon with an area of 210.44 cm2 and an apothem of 7.794 cm?

Answer 1

$\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap\quad \begin{cases} a=apothem\\ p=perimeter\\ -------\\ a=7.794\\ A=210.44 \end{cases}\implies 210.44=\cfrac{1}{2}(7.794)p \\\\\\ 420.88=7.794p\implies \cfrac{420.88}{7.794}=p\implies 54\approx p \\\\\\ \textit{a \underline{hexa}gon has 6 sides, so }\cfrac{54}{6}\implies \stackrel{\textit{length of one side}}{9}$

so.. now, if we know one side is of 9 units long... ok, bear in mind that a regular hexagon, splits the center of the circumscribing circle in 6 even angles, that'd be 60° each, if you run a bisector from that angle, it'd split into two 30° angles, check the picture below.

So it ends with a 30-60-90 triangle with one side being just half of the 9 units, and we can just use the 30-60-90 rule to get the radius itself, which is just one of the sides of the 30-60-90 triangle.

Answer 2

Answer:

The answer is D) 9cm

6 equilateral triangles in a regular hexagon

210.44

6

= 35.073 cm2

The apothem cuts each equilateral triangle into two right triangles.

35.073

2

= 17.5367 cm2

Base of each right triangle.

area =  

1

2

bh

17.5367 =  

1

2

b(7.794)

b = 4.5

Thus, the sides of each equilateral triangle is 2b = 2(4.5) = 9; which is also the radius of the circle needed to construct the inscribed regular hexagon.

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