SAT question: guessed at answer regarding circles

[1questionawayfrom800]

I think I did pretty well on my SAT's. I did well last time, and there was only one question this time around where I wasn't confident of my answer. Maybe I should have left it blank; I guessed that the answer was "10pi". Please tell me if that's right.

I had no idea where to start with this. I know that the area "A" of a triangle with base "b" and height "h" is A = 0.5bh, so I figured that the base was 10 and guessed that the height was something like 7. I picked the answer "10pi" because it seemed the most reasonable option. The other options were also close to (10)(7)/2.

Help me out, please? Thank you!

 

[wjm11]

You have an equilateral triangle with side length 10 inside the shaded region. Calculate its area. Also, calculate the area of the sector with angle = 60 degrees. Can you take it from here?

 

[1questionawayfrom800]

I understand where the equilateral triangle came from but I'm not sure where you got the angle measure from?
I can see three different sectors for every side of the triangle that goes unnaccounted for, I'm not sure how to get the area of these three sections or how the angle 60 got there.
Wait, I get it 60 degrees because the corresponding angle in the triangle, but I'm not sure what the formula for the area of that shape would be? The answer seems (my logic is what has gotten me here but I'll go ahead and try it anyway) to be that it would be the area of the traingle "A" and 3x the area of the section "Asec" to the 60 degrees. so totA = A + 3(Asec) ??

Thanks for the help all I need to know now (I think) is how to get the area of those three arcs.

 

[wjm11]

One of those "arcs" is just the complete sector area (1/6 of a circle) minus the triangle area. Good work.

 

[Denis]

Or the area of a complete sector = 1/6 pi 100;
3 of them = 3(1/6 pi 100) = pi 50

Now the pi 50 contains 3 10by10by10 triangles; take 2 of them away:
pi 50 - 2(100 * sqrt(3) / 4)

Formula for area of equilateral triangle (s = side length):
s^2 * sqrt(3) / 4

 

[galactus]

You could also use the area of a sector plus the area of two segments.

\(\displaystyle \L\\\text{Area of sector:} \;\ \frac{1}{2}r^{2}{\theta} \;\ = \;\ \frac{1}{2}(10)^{2}(\frac{\pi}{3}) \;\ = \;\ \frac{50{\pi}}{3}\)

\(\displaystyle \L\\\text{Area of segment:} \;\ \frac{1}{2}r^{2}({\theta}-sin({\theta})) \;\ = \;\ \frac{1}{2}(10)^{2}(\frac{\pi}{3}-sin(\frac{\pi}{3})) \;\ = \;\ \frac{50{\pi}}{3}-25\sqrt{3}\)

\(\displaystyle \L\\\text{Area of sector plus\\area of 2 segments\\ = area of shaded region:} \;\ \frac{50{\pi}}{3}+(\frac{100{\pi}}{3}-50\sqrt{3}) \;\ = \;\ \fbox{50({\pi}-\sqrt{3})}\)

 

[Denis]

My turn?
Area shaded region = r^2(pi - sqrt(3)) / 2 :roll: