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A triangle is inscribed in a circle, whose diameter is one of the sides of triangle. If the triangle is an isosceles triangle, what is the area of triangle to that of circle?
A. π:1
B. 2π:1
C. 1:π
D. 2:π
E. 1:2π
_________________
A. π:1
B. 2π:1
C. 1:π
D. 2:π
E. 1:2π
_________________
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With reference to the figure, in \(\triangle ABC \)
\( \angle ABC\) = 90
OA = OC = OB = r
AB = BC (because its give that the triangle is isosceles)
Area of triangle ABC = \(\frac{1}{2}\) * AB * BC
We know that \(OB^2 + OC^2 = BC^2\)
As OB = OC
\( BC^2 \) = \(2 * OB^2 = 2(r)^2\)
\(BC = r \sqrt[]2\)
Area of triangle = \( \frac{1}{2} * r \sqrt[]2 * r \sqrt[]2\)
Area of triangle = A = \( r^2\)
Area of the circle = C = \( \pi * r^2\)
\(\frac{A }{ C }\) = \(\frac{1 }{ \pi }\)
IMO - C
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Bunuel
A \(\Delta\) inscribed in a circle with one of it's sides as the diameter of the circle is a right \(\Delta.\)
Also we know that the Right \(\Delta \) is Isoceles.
Attachment:
Triangle GMAT.png [ 3.32 KiB | Viewed 2777 times ]
Let each of the equal legs be \(r,\) then Hypotenuse :
\(r^2 +r^2 = 2r^2 = h^2 \)
h= \(r \sqrt{2}\) =D
Hence the diameter is \(r \sqrt{2}\)
radius \(= \frac{r\sqrt{2}}{2}\)
area of \(\Delta \) \(= \frac{1}{2}* r^2\)
area of Circle \(= \pi( \frac{r \sqrt{2}}{2})^2 = \pi \frac{2r^2}{4} = \pi \frac{r^2}{2}\)
Ratio = area of \(\Delta\) : area of circle \(= \frac{1}{\pi}\)
Ans =C
Hope it's clear.
* The word ratio seems to be missing from the question.
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- Stne
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