We can find area of a regular hexagon by creating a star with two equilateral triangles. Let the side of them be 3 cm. Add their area and subtract from it the area of 6 small triangles. Divide by 2 to get the area of the regular hexagon of length 1 cm.

Here area of big triangles=2X(9/4X3^(1/2))

area of small triangles=6X(1/4X3^(1/2))

area of hexagon= (9X3^(1/2)/2-3X3^(1/2)/2)/2= 6X3^(1/2)/4=3X3^(1/2)/2
Also, can someone find area of a pentagon through a pentagonal star

Hint: Consider each small line segment of the star to be 1

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TopNewestWell, the angles of the smaller triangle are \( \frac{2\pi}{5}, \frac{2\pi}{5}, \frac{\pi}{5}\).

You would need to derive the area of such a triangle, and then use the exact same argument that you made.

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I used herons formula for (5 big triangles of side (2,2,3)-2 X (area of 5 small triangles(1,1,1)))/5 but the answer is wrong

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If there is a regular polygon of

nsides(\(n\geq 3)\),then its interior angle is equal to (\(\frac{2n-4}{n}\times 90).\)Now substitute \(n=5.\)You get each interior angle to be equal to \(108^\circ.\)

Now take \(\triangle ABC,\)

Using Cosine Rule,

\(k^2=a^2+b^2-2bc.cosB\Rightarrow k^2=1+1-2.cos108^\circ\Rightarrow k=\sqrt{2.62}\Rightarrow k=1.62\) \(units.\)

By using heron's formula,we get the area of \(\triangle ABC=0.47\) \({units}^2.\triangle AED\) has the area as that of \triangle ABC since both triangles are congruent.So,the length of AD is \(1.62\) \(units.\)

Now take \(\triangle ACD,\)

Using heron's formula,we get the area of \(\triangle ACD=0.77\) \({units}^2.\)

Therefore,the area of the pentagon ABCDE\(=\)area of \(\triangle ABC+\)area of \(\triangle AED+\)area of \(\triangle ACD\) \(=0.47+0.47+0.77\)

\(=\boxed{1.71 {units}^2}.\)

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But I asked to calculate area through a star

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i know it bro!!

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