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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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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:
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[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Aug 2016, 03:34, edited 2 times in total.
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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]

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New post 29 May 2014, 15:34
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goodyear2013 wrote:
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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:
Untitled1.png
Untitled1.png [ 7.49 KiB | Viewed 5066 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 \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).

Answer: E.
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Re: A semicircle is inscribed in equilateral triangle ABC, as [#permalink]

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New post 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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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 \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).

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?


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

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

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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New post 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 \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).

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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New post 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 \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).

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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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A semicircle is inscribed in equilateral triangle ABC, as [#permalink]

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New post 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 \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).

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