Carcass wrote:
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. \(12 \sqrt{3}\)
This is a great question - there are many different geometry properties tested at once here.
I'd refer to pranab's picture in conjuction with this explanation.First off, notice that triangle \(ABC\) is inscribed in a semicircle with a side as its diameter.
This must mean that angle \(ABC\) is 90 degrees..Also notice that angle \(ADB\) is 180-60 = 120. \(AD\) and \(AB\) are both radius' of the semicircle, and so triangle \(ADB\) is an isosceles triangle.
We'll denote the radius of the semicircle \(r\). So we can let \(AD\) and \(AB\) be length \(r\).
Since triangle \(ADB\) is isosceles, this means that angle \(DAB\) and \(ABD\) are the same. Let their angles b \(x\).
So:
\(2x + 120 = 180\)
\(2x = 60\)
\(x = 30\)
Now recall from above in purple, that angle \(ABC\) is 90 degrees. Since angle \(ABD\) = 30, angle \(DBC\) = 60.
This must mean that angle \(CAB\) = 60, and triangle \(DBC\) is an equilateral triangle.We already know that \(DB\) is the radius \(r\). It's clear that \(DC\) is also the radius \(r\). But since triangle \(DBC\) is an equilateral triangle, we know that \(BC\) is also the length of the radius \(r\).
To continue, we'll have to break this equilateral triangle into two 30-60-90 triangles, so bring a vertical line down the middle of triangle \(DBC\).
The base of this triangle is now \(\frac{r}{2}\). Using the \(x:x\sqrt{3}:2x\) ratio, we know that the length of the vertical line we drew must be \(r\sqrt{3} * \frac{1}{2}\).
So we have the dimensions of this new small right triangle. From the beginning, in purple, we know that triangle \(ABC\) is a right triangle as well.
We can use the similar triangle property to proceed.So we have the height of triangle \(ABC\) = 6, and its base is \(r\). The small right triangle has a height of \(r\sqrt{3} * \frac{1}{2}\) and a base of \(\frac{r}{2}\).
So we can set up the following proportion:
\(\frac{height}{base}\) of triangle \(ABC\) = \(\frac{height}{base}\) of new small right triangle.
\(\frac{6}{r}\) = \((r\sqrt{3} * \frac{1}{2})\) / \((\frac{r}{2})\)
\(\frac{6}{r}\) = \(r\sqrt{3} * \frac{1}{2} * \frac{2}{r}\)
\(\frac{6}{r}\) = \(\sqrt{3}\)
\(\frac{6}{\sqrt{3}}\) = \(r\)
Simplifying:
\(2\sqrt{3} = r\)
So we've found the length of the radius!Going back to right triangle \(ABC\), we see that the base is the length \(BC\), which we know is \(2\sqrt{3}\), and the height which is 6.
To find the area:
\(\frac{bh}{2} = Area\)
\(6*2\sqrt{3} * \frac{1}{2} = Area\)
\(6\sqrt{3} = Area\)
So the answer is B.