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01 Nov 2009, 13:14
Kudos
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N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Please explain
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01 Nov 2009, 17:49
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Clearly we have two similar triangles.
TIP for similar triangles:
\(Ratio of length=\frac{a}{b}\)
\(Ratio of area= \frac{a^2}{b^2}\)
We know ratio of area \(=2=\frac{S^2}{s^2}\) --> \(S^2=2*s^2\) --> \(S=\sqrt{2}*s\)
Answer: C.
TIP for similar triangles:
\(Ratio of length=\frac{a}{b}\)
\(Ratio of area= \frac{a^2}{b^2}\)
We know ratio of area \(=2=\frac{S^2}{s^2}\) --> \(S^2=2*s^2\) --> \(S=\sqrt{2}*s\)
Answer: C.
01 Nov 2009, 19:08
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Is C saying the "the square root of 2s" or "the square root of 2 times s"? I can't tell from the graphic but I keep getting the same answer of C if it says "the square root of 2 times s"
Assume both triangles are equilateral and the measurement of s = 4. That would make the area of the smaller triangle 1/2(b)(h) = 1/2(4)(2sqrt(3)) = 4sqrt(3)
Twice that is 8sqrt(3)
Big triangle = 1/2(b)(h)=8 sqrt(3)
since the big triangle is also equilateral base = x and height = (1/2) (x)( sqrt(3))
substitute and you get (1/2)(x)(x sqrt(3)) = sqrt 32 = 4 sqrt(2)
makes answer c because s = 4 (which is the base of the smaller triangle)
Assume both triangles are equilateral and the measurement of s = 4. That would make the area of the smaller triangle 1/2(b)(h) = 1/2(4)(2sqrt(3)) = 4sqrt(3)
Twice that is 8sqrt(3)
Big triangle = 1/2(b)(h)=8 sqrt(3)
since the big triangle is also equilateral base = x and height = (1/2) (x)( sqrt(3))
substitute and you get (1/2)(x)(x sqrt(3)) = sqrt 32 = 4 sqrt(2)
makes answer c because s = 4 (which is the base of the smaller triangle)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
