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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Well, the angles of the smaller triangle are 52π,52π,5π.
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
If there is a regular polygon of n sides(n≥3),then its interior angle is equal to (n2n−4×90).Now substitute n=5.
You get each interior angle to be equal to 108∘.
Now take △ABC,
Using Cosine Rule,
k2=a2+b2−2bc.cosB⇒k2=1+1−2.cos108∘⇒k=2.62⇒k=1.62 units.
By using heron's formula,we get the area of △ABC=0.47 units2.△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 △ACD,
Using heron's formula,we get the area of △ACD=0.77 units2.
Therefore,the area of the pentagon ABCDE=area of △ABC+area of △AED+area of △ACD =0.47+0.47+0.77
=1.71units2.
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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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