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13 Sep 2018, 04:44
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
60% (01:38) correct
40%
(01:55)
wrong
based on 88
sessions
History
Date
Time
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What is the area of the largest isosceles right triangle that can be inscribed in a semicircle of area 18π?
A. 72
B. 36
C. 24
D. 18
E. 6
A. 72
B. 36
C. 24
D. 18
E. 6
13 Sep 2018, 06:27
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Bunuel
Formulae used: Area of semi-circle = \(\frac{1}{2}*\pi*r^2\) where r - radius of the semicircle
Area of a triangle - \(\frac{1}{2}\)* Base of triangle * Height of triangle
Since the area of the semicircle is 18*\pi, the radius has to be 6 (as \(\frac{1}{2} * \pi * r^2 = 18*\pi\))
The diameter of this semicircle(12) is the hypotenuse of the right-angled triangle.
In a 45-45-90 triangle, the sides are in the ratio \(1:1:\sqrt{2}\)
The equal sides which form the base and height of the triangle measure \(\frac{12}{\sqrt{2}}\)
Therefore, the area of the largest isosceles right-angled triangle is \(\frac{1}{2}*\frac{12}{\sqrt{2}}*\frac{12}{\sqrt{2}} = 36\)(Option B)
13 Sep 2018, 21:30
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Bunuel
Area of the triangle = (1/2)*Base*Height
\((1/2)*π*r^2 = 18π\)
i.e. \(r = 6\)
For area to be maximum, Base should be maximum and the biggest base can be the diameter of the semicircle
The height will be maximum when the third vertex is taken on circumference above the centre as shown in figure
Hence, Maximum Area \(= (1/2)*12*6 = 36\)
Answer: Option B
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