Carcass wrote:
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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. \(\frac{2\pi}{\sqrt{3}}\)
Kudos for the right answer and explanation
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 [ 19.05 KiB | Viewed 4688 times ]