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A puzzle board is in the form of an equilateral triangle that has an a
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29 Jan 2025, 13:12
29 Jan 2025, 13:12
Expert Reply
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Question Stats:
71% (01:58) correct
28% (01:48) wrong based on 7 sessions
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A puzzle board is in the form of an equilateral triangle that has an area of $\(7 \sqrt{3}\)$ square inches. If the puzzle board is placed on a circular table, what should be the minimum area of the table so that whole of the board fits inside the table?
(A) \(30\)
(B) $\(\frac{88}{3}\)$
(C) \(22\)
(D) $\(\frac{44}{3}\)$
(E) \(11\)
(A) \(30\)
(B) $\(\frac{88}{3}\)$
(C) \(22\)
(D) $\(\frac{44}{3}\)$
(E) \(11\)
A puzzle board is in the form of an equilateral triangle that has an a
[#permalink]
05 Mar 2025, 13:30
05 Mar 2025, 13:30
Expert Reply
Area of the equilateral triangle is $\(\frac{\sqrt{3{4} \times\)$ Side $\(^2=7 \sqrt{3} \Rightarrow\)$ Side $\(=\sqrt{28}=2 \sqrt{7}\)$
The radius of the circle which circumscribes an equilateral triangle of side ' $\(a\)$ ' units is $\(\frac{a}{\sqrt{3\)$ (= circum-radius), so, the radius of the circle which would circumscribe an equilateral triangle of side $\(2 \sqrt{3}\)$ would be $\(\frac{2 \sqrt{7}}{\sqrt{3}}=2 \sqrt{\frac{7}{3}}\)$
So, the area of the circular table would be $\(\pi r^2=\frac{22}{7} \times\left(2 \sqrt{\frac{7}{3\right)^2=\frac{88}{3}\)$
Hence the answer is (B).
The radius of the circle which circumscribes an equilateral triangle of side ' $\(a\)$ ' units is $\(\frac{a}{\sqrt{3\)$ (= circum-radius), so, the radius of the circle which would circumscribe an equilateral triangle of side $\(2 \sqrt{3}\)$ would be $\(\frac{2 \sqrt{7}}{\sqrt{3}}=2 \sqrt{\frac{7}{3}}\)$
So, the area of the circular table would be $\(\pi r^2=\frac{22}{7} \times\left(2 \sqrt{\frac{7}{3\right)^2=\frac{88}{3}\)$
Hence the answer is (B).
gmatclubot
A puzzle board is in the form of an equilateral triangle that has an a [#permalink]
05 Mar 2025, 13:30
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