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RITUPARNA DUTTA

· started a discussion

· 1 Months ago

yes test series are full of errors (Question: 117234.0 )

plz correct it

Question:
What is the area of the inner equilateral triangle if the side of the outermost square is ‘a’? (ABCD is a square).

Options:
A) \(\cfrac{3\sqrt[]{3}a^2}{32}\)
B) \(\cfrac{\sqrt[]{3}a^2}{16}\)
C) \(\cfrac{5\sqrt[]{3}a^2}{32}\)
D) \(\cfrac{5\sqrt[]{3}a^2}{64}\)
Solution:
Ans: (b) \(BD=a,EF=\cfrac{a}{2}\)

\(\therefore\) area of equilateral triangle EFG \(=\cfrac{\sqrt[]{3}}{4}\left ( \cfrac {a}{2} \right )^2=\cfrac{\sqrt[]{3}a^2}{16}\)

Knowledge Expert

· commented

· 1 Months ago

By closely watching the diagram we can clearly locate 3 isosceles triangles in the equilateral triangles
In the triangle equilateral each angle is 60degrees
The radius will divide each angle in halves that is 30degress each
Applying cos(30degree) in one of the smaller triangle
cos30=base/hypotenus
root3/2=base/(a/2*root2)
Correct answer is (a)
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