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# A semicircle is inscribed in equilateral triangle ABC, as

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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. $$2\sqrt{3}\pi$$

[Reveal] Spoiler:
Attachment:
File comment: Triangle

Triangle.png [ 13.47 KiB | Viewed 6531 times ]
[Reveal] Spoiler: OA

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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29 May 2014, 15:34
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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 5679 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 30-60-90 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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

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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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

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?

Kudos [?]: 81 [0], given: 58

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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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ronr34 wrote:
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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

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?

$$\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}$$;

$$r = \frac{\sqrt{3}}{2}*(\frac{4}{\sqrt[4]{3}})$$;

Reduce by 2 and write $$\sqrt{3}$$ as $$(\sqrt[4]{3})^2$$: $$r = (\sqrt[4]{3})^2*(\frac{2}{\sqrt[4]{3}})$$;

Reduce by $$\sqrt[4]{3}$$: $$r=2\sqrt[4]{3}$$;
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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 30-60-90 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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

HI Bunuel. How did you get 90 degrees angle there?

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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 30-60-90 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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

HI Bunuel. How did you get 90 degrees angle there?

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 30-60-90 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 $$\frac{\pi{r^2}}{2}=2\sqrt{3}\pi$$.

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.

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A semicircle is inscribed in equilateral triangle ABC, as   [#permalink] 16 Aug 2017, 14:58
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