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The area of a regular polygon with n sides inscribed in a circle with radius r is given by `A = (n*r^2/2)*sin((2*pi)/n)` and the perimeter of the polygon is `P = (2*n*r)*sin(pi/n)`
A pentagon has five sides and it is inscribed in a circle with radius 8 m.
The area of the pentagon is `((5*64)/2)*sin 72` = 152.17 m^2
The perimeter of the pentagon is `2*5*8*sin 36` = 47.02 m
The area of the pentagon is 152.17 m^2 and the perimeter is 47.02 m
This question can be solved using trigonometry and the Pythagorean theorem if you recognize that a pentagon is made up of 5 identical triangles, arranged in a fan shape, where each of the triangles is isosceles. The lengths of the two sides of each triangle radiating out from the centre are equal to the radius of the circle, which is 8.
Since the interior angles of all 5 triangles (the ones at the centre of the circle) must total 360 degrees (they form a circle!) each of the 5 angles must be 360/5 or 72 degrees. You now have the length of two sides and one contained angle for each of the triangles which means you can use the cosine law to get the third side:
This is the length of one of the sides of the pentagon. So the perimeter is 5x9.4=47.02.
For the area we can divide one of the triangles into two smaller right angled triangles, where the hypotenuse has length 8 (its one of the radii of the circle) and the length of the base is half of 9.4 or 4.7, since the height or altitude of an isosceles triangle bisects the side it intersects.
Use the Pythagorean theorem to find the height: c^2=a^2+b^2 where c is the length of the hypotenuse, or 8.
So the length of the height is a=sqrt(8^2-4.7^2) which is 6.47
You now have everything you need to calculate the area of one of the right angled triangles, which are half the area of the isosceles triangles:
Area=1/2 x base x height
Area=.5 x 6.47 x 4.7=15.205
So the area of each of the isosceles triangles is twice this or 30.41
Since the pentagon is made up of 5 triangles just like this, its total area is 30.41 x 5 or 152.05
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