The area of the heptagon is 221.65 mm²
Angle subtended by two radii
Since the regular heptagon is inscribed in the circle of radius r = 9 mm, the angle subtended by any two radii at the center of the circle is Ф = 360/n where n = number of sides of heptagon = 7.
So Ф = 360/7
= 51.43°
Now, two radii and a chord between the two radii form a triangle.
Area of triangle between two radii
The area of the triangle A = (1/2)r²sinФ
So, there are 7 of such triangles in the heptagon.
Area of heptagon
So, the area of the heptagon, A' = number of sides of heptagon × area of one triangle
= nA
= (nr²sinФ)/2
So, substituting the values of the variables into the equation, we have
A' = (nr²sinФ)/2
A' = (7 × (9 mm)²sin51.43°)/2
A' = (7 × 81 mm² × 0.7818)/2
A' = 443.3 mm²/2
A' = 221.65 mm²
So, the area of the heptagon is 221.65 mm²
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