A square with sides of is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the square?
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Hence, (3sqrt(2))^2 + (3sqrt(2))^2 = d^2
Hence, d=6 (you can do the math following the equation above)
so divide this by 2 in order to get the radius of the circle. therefore, r = 3
Next, you will notice that the coloured area we are looking for is only a quarter of the entire circle minus the area of a quarter of the entire square.
So knowing that, first, you must the area of the circle using the equation: pi*r^2 = pi*(3)^2 = 9pi
you need a quarter of that so divide it by 4 to get 9pi/4.
Next, find the area of the square. length x width. So, 2*(3sqrt(2)) = 6sqrt(2)
but similarly, you only need a quarter of that, so divide that by 4 to get 3sqrt(2)/2
Therefore, your final answer of the area of the coloured region will be the subtraction of the two areas:
[(9pi/4)-(3sqrt(2)/2)] = (9pi-6sqrt(2))/4