The area of a regular polygon of 7 sides whose each side measures 4 cm and the circumradius is 3cm is: [Take $\sqrt{5} = 2.2365$ ]

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31.204 cm2

31.304 cm2

31.404 cm2

31.504 cm2

answer is A.

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Detailed Solution

Concept- Let us consider a regular polygon of 7 sides with its center of circumcircle at “O” and any two adjacent vertices being A and B, as shown in the figure below.
Question Image

Now after joining OA and OB, which are two radii of the circumcircle of this regular polygon, we get an isosceles triangle AOB whose length of all sides are known from data provided in the question statement.
Now if we join all the other 5 vertices with O, we will have a total of 7 equal, isosceles triangles. Thus, we will calculate the area of one triangle and multiply it by the number of sides of the polygon, to obtain the total area of the polygon.
And, formula to calculate area of a Triangle of base “b” and altitude or perpendicular “h” =

And, to calculate the altitude of a triangle, we will have to first draw a perpendicular from the vertex opposite to the base side, on the base side. This joining line will be the altitude of the triangle and in case of an equilateral triangle or an isosceles triangle, this altitude will perpendicularly bisect the base.
Then, we can use the pythagorean theorem to calculate the altitude, i.e.,
As per the pythagorean theorem,
“In case of a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides”
Question Image

i.e., According to pythagoras theorem,
Question Image

Given, regular polygon of 7 sides, each side measuring 4 cm and circumradius measuring 3cm.
Therefore, as per the figure above, (OA = OB = 3) and (AB = 4).
Considering triangle AOB with altitude OM,

.
And, OM divides triangle AOB into two equal, right-angled triangles, OAM and OBM.
Considering triangle OAM, (perpendicular = OM), (base = AM) and (hypotenuse = OA).
Using the pythagorean theorem on triangle OAM, we get,
Question Image

Hence, the correct answer is 1) 31.204 cm2.

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