# Area enclosed by chord and another similar problem

• Jun 19th 2011, 11:34 PM
skyd171
Area enclosed by chord and another similar problem

Both the problems on page 15, especially #2.
Thanks.
• Jun 19th 2011, 11:56 PM
Prove It
Re: Area enclosed by chord and another similar problem
In the first question, you need to note that the two triangles are similar. So the length of the sides in the second triangle are the same constant multiple of the length of the sides in the first triangle.

In the second question, you need to create a triangle using the chord and two radii. Evaluate the area of the sector and subtract the area of the triangle.
• Jun 20th 2011, 08:09 AM
skyd171
Re: Area enclosed by chord and another similar problem
Quote:

In the second question, you need to create a triangle using the chord and two radii. Evaluate the area of the sector and subtract the area of the triangle.
I understand how to create the triangle (I tried isosceles with side length 6). I believe the area of the triangle is 9 root 3.
The circle is of area 36pi, hence each quadrant of the circle should be of area 9pi

I don't know how to get the area of the blackened sector, though.
• Jun 20th 2011, 08:23 AM
Prove It
Re: Area enclosed by chord and another similar problem
The sector is actually $\displaystyle \frac{1}{6}$ of the circle, so its area is actually $\displaystyle 6\pi$.

So surely the blackened area is $\displaystyle 6\pi - 9\sqrt{3}$...
• Jun 20th 2011, 08:25 AM
Plato
Re: Area enclosed by chord and another similar problem
Quote:

Originally Posted by skyd171
I understand how to create the triangle (I tried isosceles with side length 6). I believe the area of the triangle is 9 root 3.
The circle is of area 36pi, hence each quadrant of the circle should be of area 9pi

The $\Delta OAB$ is equilateral. So that central angle is $60^o$.
Thus the area of the whole sector is $\left( {\frac{1}{6}} \right)\left( {\pi \cdot 6^2 } \right)$.
To find the shaded area, subtract the area of the triangle from the area of the sector.