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What is the area of the largest isosceles right triangle that can be

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New post 13 Sep 2018, 04:44
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

48% (01:12) correct 52% (01:50) wrong based on 36 sessions

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What is the area of the largest isosceles right triangle that can be  [#permalink]

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New post 13 Sep 2018, 06:27
Bunuel wrote:
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


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)
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What is the area of the largest isosceles right triangle that can be  [#permalink]

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New post 13 Sep 2018, 21:30
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Bunuel wrote:
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


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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What is the area of the largest isosceles right triangle that can be   [#permalink] 13 Sep 2018, 21:30
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