O being the center of the circle of radius unitáio, the area of triangle ABC has the catheter in diameter AC, it:

http://www.tutorbrasil.com.br/forum/...le.php?id=3338

Answer:

$\displaystyle \frac{3\sqrt{3}}{8}$

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- Mar 23rd 2009, 02:37 PMApprentice123Trigonometry on the triangle rectangleO being the center of the circle of radius unitáio, the area of triangle ABC has the catheter in diameter AC, it:

http://www.tutorbrasil.com.br/forum/...le.php?id=3338

Answer:

$\displaystyle \frac{3\sqrt{3}}{8}$

- Mar 25th 2009, 07:09 AMearboth
1. Unfortunately you forgot to tell us that $\displaystyle |\overline{OC}|$ is half of the radius. Since r = 1 you know that $\displaystyle |\overline{AC}| = \dfrac34$

2. The right triangle in question is part of the right triangle ADB (see attachment). The side CB is the height in the triangle ADB.

3. To calculate the length of the height use Euklid's theorem:

$\displaystyle |\overline{AC}| \cdot |\overline{CD}|= h^2~\implies~ h=\sqrt{\dfrac34 \cdot \dfrac14} = \dfrac14 \cdot \sqrt{3}$

4. Calculate the area of the triangle ACB:

$\displaystyle a = \dfrac12 \cdot |\overline{AC}| \cdot |\overline{CB}| = \dfrac12 \cdot \dfrac34 \cdot \dfrac14 \cdot \sqrt{3} = \dfrac3{16} \cdot \sqrt{3}$

According to your headline you maybe want to calculate the area of a rectangle(?). If so you have to double the result from #4. And then you'll get the result which is given by the answer.