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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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Carcass wrote:
Attachment:
The attachment #greprepclub In the figure above, an equilateral triangle is inscribed in a circle..jpg is no longer available


In the figure above, an equilateral triangle is inscribed in a circle. How many times greater is the area of the circle than the area of the triangle

A. \(\frac{\pi}{\sqrt{3}}\)

B. \(\frac{3\pi}{4}\)

C. \(\frac{4\pi}{3 \sqrt{3}}\)

D. \(3\)

E. \(\frac{2\pi}{\sqrt{3}}\)


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Here,

For the equilateral \(\triangle\) ADM,

let the sides of the \(\triangle\) = S

Since, the sides are equal, hence DB divides the side AM into two equal parts i.e = \(\frac{S}{2}\)

Now in the \(\triangle\) ACB

\(\angle\) CAB = 30

\(\angle\) CBA = 90

therefore, \(\angle\) ACB = 60

hence \(\triangle\) ACB is : 30 : 60 : 90 triangle and the sides are in the ratio

1 : \(\sqrt{3}\) : 2

or\( \frac{S}{{2\{\sqrt{3}}}\) : \(\frac{S}{2}\) :\(\frac{S}{{\sqrt{3}}}\)( since AB = \(\frac{S}{2}\))

Now radius of the circle = AC = \(\frac{S}{{\sqrt{3}}}\)

Area of the circle = \(\pi * {radius}^2 = \frac{{S^2}}{3}\pi\)

and Area of equilateral \(\triangle\) = \(\frac{{\sqrt3*S^2}}{4}\)
the area of the circle is greater than the area of the triangle = \(\frac{{S^2}}{3}\pi\) * \(\frac{4}{{\sqrt3*S^2}}\) = \(\frac{4\pi}{3 \sqrt{3}}\)
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#greprepclub In the figure above, an equilateral triangle is inscribed in a circle..jpg
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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GREAT explanation Sir.

Thank you.

Regards
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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to calculate how many times greater the area of the circle is, why did you multiply the area of the circle by the inverse of the area of the triangle?

to calculate how many times A is greater than B = A * 1/B. Does this method work every time?
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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FunkyMonk0 wrote:
to calculate how many times greater the area of the circle is, why did you multiply the area of the circle by the inverse of the area of the triangle?

to calculate how many times A is greater than B = A * 1/B. Does this method work every time?



It is actually = \(\frac{{Area of circle}}{{ Area of triangle}} \)

When u end up with some tricky statement, put some nice number,

Let me rephrase: how many times is number 6 greater than number 3?

2 times i.e \(\frac{6}{3}\)

*** just a trick (check for yourself): any ques in GRE that ask you how many times is A greater THAN B = the quantity after THAN will be in denominator
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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Now area of triangle = root 3/4a^2
we need sides of triangle. From centre of circle, we drop perpendicular to any one side, we get 30 - 60 - 90 triangle. Thus hypo is 2, (this is also radius), and side opp to 60 degrees is root 3. Thus as this is the half of one side, full side is 2 root 3.

Thus area = pie r^2/root 3/4a^2 = 4 pie/3root 3. Which is C

Kudos if you like this post!
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
Expert Reply
Dear Sir.
We much appreciate the effort of your explanation. However, if it is possible, the solution should be posted as text.

Of course, if you do have an image to upload is perfectly fine but the explanation would be better as text.

This way is easiest to find in the search box.

Many Thanks
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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FUN FACT:

If an equilateral triangle is inscribed in a circle, then area of circle to the area of that triangle is always equal (4*pi)/(3\sqrt3). Remember this relation to get answer for similar type of problem to avoid unnecessary calculation and it helps to save a time a lot
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Re: In the figure above, an equilateral triangle is inscribed in [#permalink]
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