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A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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29 May 2014, 14:51
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A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle? A. \(\frac{3\pi}{8*\sqrt{3}}\) B. \(\sqrt{3}\pi\) C. \(2\pi\) D. \(\sqrt{6}\pi\) E. \(2\sqrt{3}\pi\) Attachment: File comment: Triangle
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Last edited by Bunuel on 15 Aug 2016, 03:34, edited 2 times in total.
Edited the question.



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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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goodyear2013 wrote: Attachment: The attachment Triangle.png is no longer available A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle? A. \(\frac{3\pi}{8*\sqrt{3}}\) B. \(\sqrt{3}\pi\) C. \(2\pi\) D. \(\sqrt{6}\pi\) E. \(2\sqrt{3}\pi\) Attachment:
Untitled1.png [ 7.49 KiB  Viewed 6517 times ]
The area of equilateral triangle = \(side^2*\frac{\sqrt{3}}{4}=16\) > \(side=\frac{8}{\sqrt[4]{3}}\). Now, the half of the side (\(\frac{4}{\sqrt[4]{3}}\)) is the hypotenuse of 306090 triangle, where the radius is the leg opposite 60° angle. In such triangles the sides are always in the ratio\(1 : \sqrt{3}: 2.\). So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) > \(r=2\sqrt[4]{3}\). Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\). Answer: E.
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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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21 Sep 2014, 13:50
Bunuel wrote: So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) > \(r=2\sqrt[4]{3}\).
Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).
Answer: E. Hi Bunuel, How did you get the value of \(r\) as you did? Something with the root of 3 doesn't make sense to me.... Can you elaborate on this step?



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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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22 Sep 2014, 01:30



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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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14 Aug 2016, 23:53
Bunuel wrote: goodyear2013 wrote: Attachment: Triangle.png A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle? A. \(\frac{3\pi}{8*\sqrt{3}}\) B. \(\sqrt{3}\pi\) C. \(2\pi\) D. \(\sqrt{6}\pi\) E. \(2\sqrt{3}\pi\) Attachment: Untitled1.png The area of equilateral triangle = \(side^2*\frac{\sqrt{3}}{4}=16\) > \(side=\frac{8}{\sqrt[4]{3}}\). Now, the half of the side (\(\frac{4}{\sqrt[4]{3}}\)) is the hypotenuse of 306090 triangle, where the radius is the leg opposite 60° angle. In such triangles the sides are always in the ratio\(1 : \sqrt{3}: 2.\). So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) > \(r=2\sqrt[4]{3}\). Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\). Answer: E. HI Bunuel. How did you get 90 degrees angle there? thanks in advance



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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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15 Aug 2016, 03:35
sree9890 wrote: Bunuel wrote: goodyear2013 wrote: Attachment: Triangle.png A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle? A. \(\frac{3\pi}{8*\sqrt{3}}\) B. \(\sqrt{3}\pi\) C. \(2\pi\) D. \(\sqrt{6}\pi\) E. \(2\sqrt{3}\pi\) Attachment: Untitled1.png The area of equilateral triangle = \(side^2*\frac{\sqrt{3}}{4}=16\) > \(side=\frac{8}{\sqrt[4]{3}}\). Now, the half of the side (\(\frac{4}{\sqrt[4]{3}}\)) is the hypotenuse of 306090 triangle, where the radius is the leg opposite 60° angle. In such triangles the sides are always in the ratio\(1 : \sqrt{3}: 2.\). So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) > \(r=2\sqrt[4]{3}\). Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\). Answer: E. HI Bunuel. How did you get 90 degrees angle there? thanks in advance If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency.
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A semicircle is inscribed in equilateral triangle ABC, as [#permalink]
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16 Aug 2017, 14:58
Bunuel wrote: goodyear2013 wrote: Attachment: Triangle.png A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle? A. \(\frac{3\pi}{8*\sqrt{3}}\) B. \(\sqrt{3}\pi\) C. \(2\pi\) D. \(\sqrt{6}\pi\) E. \(2\sqrt{3}\pi\) Attachment: Untitled1.png The area of equilateral triangle = \(side^2*\frac{\sqrt{3}}{4}=16\) > \(side=\frac{8}{\sqrt[4]{3}}\). Now, the half of the side (\(\frac{4}{\sqrt[4]{3}}\)) is the hypotenuse of 306090 triangle, where the radius is the leg opposite 60° angle. In such triangles the sides are always in the ratio\(1 : \sqrt{3}: 2.\). So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) > \(r=2\sqrt[4]{3}\). Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\). Answer: E. Suppose if i forget formula of area of equilateral triangle and use normal area formula 1/2*base*altitude. \(1/2*base*altitude=16\) Base*Altitude =32 As per triangle ratio \(30:60:90= 1:\sqrt{3} : 2\) If hypotenuse is 2x then base of triangle will be 2x and altitude will be \(\sqrt{3} x\) Now how we can proceed futher using this approach. Please help




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