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16 Oct 2017, 03:24
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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:
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Question Stats:
74% (01:23) correct
26%
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The area of an equilateral triangle is \(\sqrt{3}\) times that of a square. What is the ratio of a side of the triangle to a side of the
square?
A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{3}}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\sqrt{3}\)
E. \(2\)
https://www.expertsglobal.com
square?
A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{3}}\)
C. \(\frac{1}{\sqrt{2}}\)
D. \(\sqrt{3}\)
E. \(2\)
https://www.expertsglobal.com
16 Oct 2017, 04:33
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The area of an equilateral triangle is 3√3 times that of a square. What is the ratio of a side of the triangle to a side of the
square?
A. 1/2
B. 1/√3
C. 1/√2
D. √3
E. 2
When dealing with theoretical geometry, first write out the requisite formulas, in this case Area Triangle = 1/2 base x height and Area Square = side^2. From the ratio in the problem it can be determined that √3 = 1/2 base x height and 1 = side^2. Therefore, a side of the square = 1, and 2√3 = base x height. Because the ratio of the base to the height in an equilateral triangle is always 2:√3, divide each side of that equation by √3 to find that the base of the triangle is 2 and that the ratio of the length of a base of the triangle is 2 times that of a side of the square. The correct answer is choice E.
Alternatively, using simple logic and drawing a figure it should be possible to determine that the side of the triangle must be larger for the triangle to have an area √3 times that of a square, so eliminate choices A, B, and C which indicate that the side of the triangle is less than the side of the square. The ratio of the sides and that of the areas also would logically not be the same, so eliminate choice D. Choice E is the only logical option, and can be determined in relatively short order without any calculation. Consider these types of shortcuts when facing theoretical geometry problems.
square?
A. 1/2
B. 1/√3
C. 1/√2
D. √3
E. 2
When dealing with theoretical geometry, first write out the requisite formulas, in this case Area Triangle = 1/2 base x height and Area Square = side^2. From the ratio in the problem it can be determined that √3 = 1/2 base x height and 1 = side^2. Therefore, a side of the square = 1, and 2√3 = base x height. Because the ratio of the base to the height in an equilateral triangle is always 2:√3, divide each side of that equation by √3 to find that the base of the triangle is 2 and that the ratio of the length of a base of the triangle is 2 times that of a side of the square. The correct answer is choice E.
Alternatively, using simple logic and drawing a figure it should be possible to determine that the side of the triangle must be larger for the triangle to have an area √3 times that of a square, so eliminate choices A, B, and C which indicate that the side of the triangle is less than the side of the square. The ratio of the sides and that of the areas also would logically not be the same, so eliminate choice D. Choice E is the only logical option, and can be determined in relatively short order without any calculation. Consider these types of shortcuts when facing theoretical geometry problems.
16 Oct 2017, 11:47
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nkmungila
If you know the formula for area of an equilateral triangle, the answer turns on translation and algebra.*
Area of an equilateral triangle, \(a\) = side:
\(\frac{a^2\sqrt{3}}{4}\)
"The area of an equilateral triangle [with side \(a\)] is \(\sqrt{3}\) times that of a square," with side \(s\). Translate:
(Area of square)(\(\sqrt{3}\)) = Area of triangle
\(s^2\sqrt{3}\) = \(\frac{a^2\sqrt{3}}{4}\)
Factor out \(\sqrt{3}\)
\(s^2 = \frac{a^2}{4}\)
\(\sqrt{s^2} = \sqrt{\frac{a^2}{4}}\)
\(s = \frac{a}{2}\)
Ratio of a side of the triangle to a side of the square? Rearrange the expression immediately above.
\(\frac{a}{s} = \frac{2}{1} = 2\)
Answer E
*If you don't know the formula, although deriving it is painstaking, it is completely possible. You drop an altitude and find the area of two 30-60-90 right triangles, described here and here
