http://readyforgmat.com/math/documen...s-version2.pdf

Both the problems on page 15, especially #2.

Thanks.

Printable View

- Jun 19th 2011, 11:34 PMskyd171Area enclosed by chord and another similar problem
http://readyforgmat.com/math/documen...s-version2.pdf

Both the problems on page 15, especially #2.

Thanks. - Jun 19th 2011, 11:56 PMProve ItRe: 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 AMskyd171Re: Area enclosed by chord and another similar problemQuote:

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.

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 AMProve ItRe: 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 $\displaystyle \displaystyle 6\pi - 9\sqrt{3}$... - Jun 20th 2011, 08:25 AMPlatoRe: Area enclosed by chord and another similar problem
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.