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23 Jun 2013, 00:26
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
84% (01:25) correct
16%
(02:09)
wrong
based on 19
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
Triangle.png
Four equilateral triangles are inscribed within each other as pictured here. If the perimeter of the outer triangle is 24, what is the area of the innermost triangle?A. \(\frac{1}{6}\)
B. \(\sqrt{3}/8\)
C. \(\frac{1}{3}\)
D. \(\sqrt{2}/4\)
E. \(\sqrt{3}/4\)
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23 Jun 2013, 01:12
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Four equilateral triangles are inscribed within each other as pictured here. If the perimeter of the outer triangle is 24, what is the area of the innermost triangle?
A. \(\frac{1}{6}\)
B. \(\sqrt{3}/8\)
C. \(\frac{1}{3}\)
D. \(\sqrt{2}/4\)
E. \(\sqrt{3}/4\)
The length of a side of the first inscribed triangle is half of that of outer triangle:
The same way:
The length of a side of the second inscribed triangle is 4/2=2;
The length of a side of the third inscribed triangle (the smallest one) is 2/2=1.
The area of an equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\). Therefore, the are of the smallest triangle is \(\frac{\sqrt{3}}{4}\).
Answer: E.
For more check Triangles chapter of Math Book: math-triangles-87197.html
Hope it helps.
A. \(\frac{1}{6}\)
B. \(\sqrt{3}/8\)
C. \(\frac{1}{3}\)
D. \(\sqrt{2}/4\)
E. \(\sqrt{3}/4\)
The length of a side of the first inscribed triangle is half of that of outer triangle:
Attachment:
Untitled2.png
Since the length of the side of bigger triangle is 24/3=8, then the length of a side of the first inscribed triangle is 8/2=4.The same way:
The length of a side of the second inscribed triangle is 4/2=2;
The length of a side of the third inscribed triangle (the smallest one) is 2/2=1.
The area of an equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\). Therefore, the are of the smallest triangle is \(\frac{\sqrt{3}}{4}\).
Answer: E.
For more check Triangles chapter of Math Book: math-triangles-87197.html
Hope it helps.
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!
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