# Another geo problem. I just need to make sure if i am on the right track

1. Jun 19, 2006

### sssddd

http://img47.imageshack.us/img47/4519/geo23mo.jpg [Broken]

It's an equliteral triangle

I'm suppose to find the area of the shaded region. first of all are the chords of each those regions half 5cm? Should I find the areas of the sectors and then subtract the areas of the triangles?

Please help me get this started and put me on the right track

Last edited by a moderator: May 2, 2017
2. Jun 19, 2006

### HallsofIvy

Staff Emeritus
No, that's not true. The lengths of the chords are not 5 cm (which is what I think you meant). The point of intersection of the sides of the triangle and the circle are not the midpoints of the sides. If you call the point where the line AC intersects the circle "D", the the length of AD is $\sqrt{15}$.

I don't know what you are "allowed" to use, but I wrote the circle as
$x^2+ y^2= 25$ and then found the equation of the line from A to C to be $y= \sqrt{3}(x+ 5)$. The points of intersection of the line and circle are A= (-5, 0) (of course) and $D= (-\frac{5}{4},\frac{15\sqrt{3}}{4})$. You can use those to find the area of the circular sector "AOD" and the triangle "AOD" and subtract.

3. Jun 21, 2006

### Tide

Halls,

You were a little hasty with your quadratic formula! :)

A different approach is to draw a radius from D (the intersection point) to the center of the circle. That radius, the radius from the center to A and the chord AD form a triangle which is clearly isosceles. The angle at A is 60 deg as is the angle at D so we conclude the triangle is also equilateral!

Therefore the length of the chord is equal to the radius of the circle. Of course, solving the equations you specified will give the same length.