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ASHUTOSH KUMAR

· started a discussion

· 1 Months ago

Check explanation! (Question: 117234.0 )

Here 'E' and 'F' are not the midpoints of BC and CD respectively so how can you conclude that EF=1/2BD?

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 \(=\cfrac{\sqrt[]{3}}{4}\left ( \cfrac {a}{2} \right )^2=\cfrac{\sqrt[]{3}a^2}{16}\)

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