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27 Oct 2021, 04:45
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A
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Difficulty:
25%
(medium)
Question Stats:
85% (02:27) correct
15%
(02:45)
wrong
based on 93
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Triangle ABC is inscribed in a circle, such that AC is a diameter of the circle and angle BAC is 45°. If the area of triangle ABC is 72 square units, how much larger is the area of the circle than the area of triangle ABC?
A. \(72\pi - 72\)
B. \(72\pi - 36\)
C. \(72\pi - 18\)
D. \(72\pi - 1\)
E. \(72\pi\)
A. \(72\pi - 72\)
B. \(72\pi - 36\)
C. \(72\pi - 18\)
D. \(72\pi - 1\)
E. \(72\pi\)
27 Oct 2021, 21:33
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Bunuel
Any triangle that has a side equal to the diameter, the triangle will be right angled where the diameter is the hypotenuse.
As one angle of the right angled triangle is 45, the other angle apart from 90 angle will also be 45. Hence we have an isosceles right angled triangle.
The isosceles right angled triangle will have each of the smaller side AB=BC=\(\frac{dia \ \ AC}{\sqrt{2}}\)
Area of triangle = \(\frac{1}{2}* (\frac{dia \ \ AC}{\sqrt{2}})^2=72\)
\(AC^2=288…….AC=12\sqrt{2}\)
Thus the area of circle = \(\pi r^2=\pi (6\sqrt{2})^2=72\pi\)
Difference in the area = \(72\pi - 72\)
