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# Triangle ABC is inscribed in a semicircle centered at

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Triangle ABC is inscribed in a semicircle centered at [#permalink]  04 Aug 2017, 07:55
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#GREpracticequestion Triangle ABC is inscribed in a semicircle centered.jpg [ 20.18 KiB | Viewed 619 times ]

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. $$18 \sqrt{3}$$
[Reveal] Spoiler: OA

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Re: Triangle ABC is inscribed in a semicircle centered at [#permalink]  01 Oct 2017, 00:28
1/2* (2√3)*3 = 6√3
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Re: Triangle ABC is inscribed in a semicircle centered at [#permalink]  01 Oct 2017, 08:09
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Carcass wrote:
Attachment:
The attachment half_circle.jpg is no longer available

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

now in triangle BDC

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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Re: Triangle ABC is inscribed in a semicircle centered at [#permalink]  01 Oct 2017, 08:28
Since the triangle ABC is built on the diameter, the angle in B is equal to 90°.

Then, moving to triangle BCD, it is easy to notice that sides BD and DC are equal since they are both radii of the circumference, thus the triangle is at least isosceles.

Since the angle in D is 60°, the other two angles (which are equal) are (180°-60)/2 = 60°. Thus the triangle is actually equilateral.

Now, moving back to triangle ABC, we have now two angles, 90° in B and 60° in C. Thus, the last angle must be 30° (180°-60°-30°). Thus, triangle ABC is a 30-60-90.

Then, we know that the longer leg is computed as $$l\sqrt(3)$$, thus equating this expression to 6, we get $$l = 2\sqrt(3)$$.

Finally, we can compute the area as $$\frac{6*2sqrt(3)}{2} = 6\sqrt(3)$$

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Re: Triangle ABC is inscribed in a semicircle centered at [#permalink]  09 Jan 2019, 12:54
thanks
Re: Triangle ABC is inscribed in a semicircle centered at   [#permalink] 09 Jan 2019, 12:54
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