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03 Jan 2021, 07:28
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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:
95%
(hard)
Question Stats:
33% (03:27) correct
67%
(02:49)
wrong
based on 39
sessions
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What is the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points of the adjacent sides of a square of side ‘a’ ?
(A) \(\frac{3a^2}{16}\)
(B) \(\frac{3\sqrt{3}a^2 }{16}\)
(C) \(\frac{3a^2(\pi - 12)}{4}\)
(D) \(\frac{\sqrt{3}*a^2}{32}\)
(E) \(\frac{3\sqrt{3}*a^2}{32}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(\frac{3a^2}{16}\)
(B) \(\frac{3\sqrt{3}a^2 }{16}\)
(C) \(\frac{3a^2(\pi - 12)}{4}\)
(D) \(\frac{\sqrt{3}*a^2}{32}\)
(E) \(\frac{3\sqrt{3}*a^2}{32}\)
Are You Up For the Challenge: 700 Level Questions
03 Jan 2021, 08:48
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Bunuel
We would prefer to read the question backwards, a square with 4 midpoints on its lengths creates another square that inscribes a circle that inscribes an equilateral triangle (see attachment). Then the bigger square has a length of a, the smaller square has a side length of \(\frac{\sqrt{2}}{2}a\). The radius of the circle in terms of \(a\) is \(r = \frac{\sqrt{2}}{4}a\). Next we would like to represent the triangle in terms of r, and plug in r so we get our answer in terms of a.
I do this by splitting the triangle in 3, so we have 3 smaller triangles of 120 - 30 - 30. Split each of those in half and we would have 6 triangles of 60-30-90. The radius which is the hypotenuse of the 60-30-90 triangle is r, so the area is \(\frac{1}{2}*\frac{r}{2}*\frac{\sqrt{3}}{2}r\), multilpy that by 6 to get the area is \(\frac{3}{4}\sqrt{3}r^2\).
Finally plug in \(r^2 = \frac{2}{16}a^2 = \frac{a^2}{8}\) and we get:
Area of triangle = \(\frac{3}{4}\sqrt{3}*\frac{1}{8}a^2 = \frac{3}{32}\sqrt{3}*a^2\)
Ans: E
Attachments
Inscribed Circle Graph.png [ 25.33 KiB | Viewed 4382 times ]
11 Nov 2022, 00:25
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Bunuel
Solution:
- an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points of the adjacent sides of a square
Attachment:
tricirclesquare.png [ 34.3 KiB | Viewed 2460 times ]
- Side \(HE=\sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2}=\frac{a}{\sqrt{2}}\)
- Diameter of circle \(=HE=\frac{a}{\sqrt{2}}\). Radius of circle \(=\frac{HE}{2}=\frac{a}{2\sqrt{2}}\)
- The equilateral triangle is then divided into 6 equal triangles with height \(=\frac{a}{2\times 2\sqrt{2}}\) and base \(=\frac{\sqrt{3}a}{2\times 2\sqrt{2}}\)
Attachment:
tricirclesquare2.png [ 8.56 KiB | Viewed 2412 times ]
- Area of each such triangle \(=\frac{1}{2}\times \frac{a}{4\sqrt{2}}\times \frac{\sqrt{3}a}{4\sqrt{2}}\)
- Area of full triangle \(=6\times \frac{1}{2}\times \frac{a}{4\sqrt{2}}\times \frac{\sqrt{3}a}{4\sqrt{2}}=\frac{3\sqrt{3}a^2}{32}\)
Hence the right answer is Option E
