# finding the area of the shaded part.

• Oct 19th 2012, 11:56 AM
rcs
finding the area of the shaded part.
there some parts of the solution on the book where i found this problem.

Area of shaded region = A sector + [ A triangle - 2 (A sector 2)]
= 5/6 A circle + [ (s^2 * sqrt of 3)/4 - 2(1/6 A of circle)]
... i understand the 5/6 A circle and the 2(1/6 Area of circle, but i dont understand the (s^2*sqrt of 3)/4 for the Area of triangle.... how was that being done? please help me understand this. Thank you.
• Oct 19th 2012, 12:57 PM
bjhopper
Re: finding the area of the shaded part.
I am not looking at what extra info you provide but try this

area 8 radius circle +equilateral triangle side lenght 8 - 3 times the segment of a sector of 8 radius where sector is 1/6 of circle
• Oct 19th 2012, 12:59 PM
MarkFL
Re: finding the area of the shaded part.
The area $A$ of an equilateral triangle having sides $s$ may be computed as follows:

$A=\frac{1}{2}\cdot s\cdot s\cdot\sin(60^{\circ})=\frac{1}{2}s^2\cdot\frac{ \sqrt{3}}{2}=\frac{s^2\sqrt{3}}{4}$
• Oct 19th 2012, 01:12 PM
HallsofIvy
Re: finding the area of the shaded part.
The area of a circle of radius 8 is, of course $64\pi$. Every angle of an equilateral triangle has measure 60 degrees which is 60/360= 1/6 of the entire circle. The portion of the circle outside the triangle has area $(5/6)(64\pi)$. If you draw a horizontal line where the triangle cuts the circle, you have an equilateral triangle with side length 8 and can use the formulas given above.

The area of the trapezoid below that has bases 8 and 16 and height $4\sqrt{3}$ and the the area of such a trapezoid is (h/2)(b1+ b2). Now you have to remove unshaded area in that trapezoid which is again the area of 1/6 of a circle of radius 8.
• Oct 20th 2012, 06:56 PM
rcs
Re: finding the area of the shaded part.
thank you guys