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# Area of hexagon through a star

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

Note by Prince Loomba
1 year, 2 months ago

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If there is a regular polygon of n sides($$n\geq 3)$$,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 $$\triangle ACD$$ $$=0.47+0.47+0.77$$
$$=\boxed{1.71 {units}^2}.$$ · 1 year ago

But I asked to calculate area through a star · 1 year ago

i know it bro!! · 1 year ago

But if people read this then they may find it correct · 1 year ago

okkkk!!! · 1 year ago

Well, 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. · 1 year, 2 months ago