# Thread: Area enclosed by chord and another similar problem

1. ## Area enclosed by chord and another similar problem

Both the problems on page 15, especially #2.
Thanks.

2. ## 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.

3. ## Re: Area enclosed by chord and another similar problem

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.

4. ## Re: Area enclosed by chord and another similar problem

The sector is actually $\displaystyle \displaystyle \frac{1}{6}$ of the circle, so its area is actually $\displaystyle \displaystyle 6\pi$.

So surely the blackened area is $\displaystyle \displaystyle 6\pi - 9\sqrt{3}$...

5. ## Re: Area enclosed by chord and another similar problem

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