Hello,
Can you please assist with this:
In the figure above, circle O has its center on the intersection of the two
diagonals of square ABCD. What is the area of circle O if y is the measure of the
diagonal of each of the small squares in the figure and PQ = RS = TU = VW?
(1) y = 3√2
(2) PQ = 5
OA: B
To find the area of the circle
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Here's a (crude) attempt to draw and label the figure
The basic idea:
* If the diagonal of the little squares is y, the sides of those squares are y/√2, or (y√2)/2, in proper form.
* Since the diagonal of the big square is (2r + 2y), the sides of that square are (2r + 2y)/√2, or √2(r+y), in proper form.
* Since the sides of the smaller squares sum to y√2, the distance between those squares on the sides of the big square is r√2 (i.e. PQ = r√2).
* Once you know r√2, you can find r, the radius of the circle, and solve.
* Since S2 gives you r√2, it's sufficient. Since S1 doesn't, it's not.
The basic idea:
* If the diagonal of the little squares is y, the sides of those squares are y/√2, or (y√2)/2, in proper form.
* Since the diagonal of the big square is (2r + 2y), the sides of that square are (2r + 2y)/√2, or √2(r+y), in proper form.
* Since the sides of the smaller squares sum to y√2, the distance between those squares on the sides of the big square is r√2 (i.e. PQ = r√2).
* Once you know r√2, you can find r, the radius of the circle, and solve.
* Since S2 gives you r√2, it's sufficient. Since S1 doesn't, it's not.