Carcass wrote:

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Triangle ABC is inscribed in a semicircle centered at D. What is the area of triangle ABC ?

A. \(\frac{12}{\sqrt{3}}\)

B. \(6 \sqrt{3}\)

C. 12

D. \(12 \sqrt{3}\)

E. \(18 \sqrt{3}\)

Now consider triangle ABC

we have AD=DC=AD (AD is also radius since it is drawn from the centre of the circle.)

now in triangle BDC

BD=DC (both are radius)

and their angles are also equal.

Since one angle in triangle is 60 degree therefore the other two angles will be equal to 60 degree and triangle BDC is equilateral triangle.i.e BD=DC=BC

Now consider ABC

we have a 30=60-90 triangle

i.e \(1:\sqrt{3}:2\)

or \(2\sqrt{3} :6 : 4\sqrt{3}\) (since the sides opposite to 60 degree =6)

therefore area of triangle = \(\frac{1}{2} * base * altitude\)

=\(1/2 * 6 * 2\sqrt{3}\) (here base of the triangle ABC is 6, and height is the radius of the circle)

\(= 6\sqrt{3}\)

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