Connecting the circle centers makes a triangle. The sides of the triangle pass through the points where the circles are tangent, which is also where the shaded region ends (meaning it's totally contained in the triangle). The non-shaded area inside the triangle is three 1/6th circles. So it's triangle area minus half a circle.
Theres a neat little trick my trig teacher taught us for 30-60-90 triangles! If the shortest side is x, the hypotenuse is 2x and the other side is x√3. She drilled that into our heads, so I knew the height of the triangle was 8√3 without going through the hassle of Pythagorean theorem and simplifying the √192. Little helpful hack for the curious viewer!
As soon as you have 32π you're limited to options A and C, but factoring A gives you 32 * (sqrt(3) - π) which is a negative number which eliminates that option.
Good spot! Actually it looks like all the answers except for the real one give negative numbers, so you don't need to do any actual work on the problem at all. 🤣
@@shadow4798 While true, i prefer to try an calculate the answer myself before ever looking at multiple choice answers. Personally I don't see the point in multiple choice math questions.
Let X = (PI * r ** 2)/2 is the area of the interior triangle OQP excluding the shaded area (the 3 interior wedges of the circles) == 8*PI. Let Y be the area of that interior triangle OQP == 0.5 * base * height == 0.5 * 8**2 == 32. So, the area of the shaded region is Y - X == 32 - 8*PI == 6.867 approximately.
I made the circle inside a hexagon, the shaded area is equal to the area of the hexagon (Ah) minus the area of the circle (Ac) divided by 2 . (Ah-Ac)/2 --> (128√3 − 64π)/2 --> 64√3 - 32π
Connecting the circle centers makes a triangle. The sides of the triangle pass through the points where the circles are tangent, which is also where the shaded region ends (meaning it's totally contained in the triangle). The non-shaded area inside the triangle is three 1/6th circles. So it's triangle area minus half a circle.
Theres a neat little trick my trig teacher taught us for 30-60-90 triangles!
If the shortest side is x, the hypotenuse is 2x and the other side is x√3. She drilled that into our heads, so I knew the height of the triangle was 8√3 without going through the hassle of Pythagorean theorem and simplifying the √192.
Little helpful hack for the curious viewer!
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As soon as you have 32π you're limited to options A and C, but factoring A gives you 32 * (sqrt(3) - π) which is a negative number which eliminates that option.
Good spot! Actually it looks like all the answers except for the real one give negative numbers, so you don't need to do any actual work on the problem at all. 🤣
@@shadow4798 yes, that too!
@@shadow4798 While true, i prefer to try an calculate the answer myself before ever looking at multiple choice answers. Personally I don't see the point in multiple choice math questions.
True, but its a very simple matter the reduce √192 to 8√3.
@@sashashadowhive6128 Totally agree, but since SATs are time based, and the options are there, take the quick way.
Let X = (PI * r ** 2)/2 is the area of the interior triangle OQP excluding the shaded area (the 3 interior wedges of the circles) == 8*PI. Let Y be the area of that interior triangle OQP == 0.5 * base * height == 0.5 * 8**2 == 32.
So, the area of the shaded region is Y - X == 32 - 8*PI == 6.867 approximately.
I made the circle inside a hexagon, the shaded area is equal to the area of the hexagon (Ah) minus the area of the circle (Ac) divided by 2 . (Ah-Ac)/2 --> (128√3 − 64π)/2 --> 64√3 - 32π
(0.5)(2r)(rtan60) - (3)(60/360)(pi)(r)^2
thanks, i dont do the SAT since am british but i like the way you explain stuff!!
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That's a pretty easy, question don't you think?
I don't think, so.
Difficulty varies for everyone 👍
The shaded area is a Gay Triangle. Next.