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Re: The hypotenuse of right triangle ABC is the diameter of the circle. If [#permalink]
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Given:
Diameter of the circle = 18
Diameter of the circle = Hypotenuse of the triangle

Need:
Area of shaded region = Area of circle - Area of triangle

Area of Circle:
Diameter = 18, so radius = 9 , Area of circle = \(81\pi\)

Area of Triangle:

Triangle = 30-60-90 , so we know the sides are in the ratio x:x\(\sqrt{3}\): 2x

From the diagram, we know that 2x = 18 , so x = 9
We can now calclulate the other 2 sides , height of triangle = 9 and base is equal to 9\(\sqrt{3}\)

Area of the triangle is equal to \((9 * 9\sqrt{3})/2\) = \((81\sqrt{3})/2\)

Area of shaded region = Area of circle - Area of triangle
Area of shaded region = \(81\pi\) - \((81\sqrt{3})/2\)

Ans: B
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Re: The hypotenuse of right triangle ABC is the diameter of the circle. If [#permalink]
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Just as the area of an equilateral triangle is \(\frac{\sqrt{3}}{4}* a^2\), where a is the side of the equilateral triangle, we can also remember the areas of a 30 - 60 - 90 and a 45 - 45 - 90 triangle in terms of the hypotenuse.

For a 30 - 60 - 90 triangle, the area = \(\frac{\sqrt{3}}{8}* h^2\)

For a 45 - 45 - 90 triangle, the area = \(\frac{1}{4}* h^2\)


Area of the shaded region = Area of the circle (with radius = 9) - Area of the 30-60-90 triangle (with hypotenuse = 18) = \(\pi r^2 - \frac{\sqrt{3}}{8}* h^2\)


Area = \(9^2 \pi - \frac{\sqrt{3}}{8}* 18^2 = 81 \pi - \frac{81\sqrt{3}}{2}\)



Option B

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Re: The hypotenuse of right triangle ABC is the diameter of the circle. If [#permalink]
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Re: The hypotenuse of right triangle ABC is the diameter of the circle. If [#permalink]
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