A square is partly within a circle and partly outside it, with two sides tangent to the circle as shown in the figure.
If the radius of the circle is 4 units,
Find the area of the square correct to two decimal places.
Call the side of the square, S
BH = sqrt (2) S - 8 + 4 = sqrt (2)S - 4
BC = 4
EH = sqrt (2)S
EF = S
Using similar triangles
EH/ EF = BH / BC
sqrt (2) S / S = (sqrt (2)S - 4) / 4
sqrt (2) * 4 = sqrt (2)S - 4
4sqrt (2) + 4 = sqrt (2) S
S = 4 ( sqrt (2) + 1) / sqrt (2)
S = 2sqrt(2) ( sqrt (2) + 1) = 4 + 2sqrt (2) = 4 + sqrt (8)
Area of square = (4 + sqrt (8) )^2 = 16 + 8sqrt (8) + 8 = 24 + 8sqrt(8) =
24 + 16sqrt (2) ≈ 46.63
A square is partly within a circle and partly outside it, with two sides tangent to the circle as shown in the figure.
If the radius of the circle is 4 units. Find the area of the square correct to two decimal places.
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r = 4
CO = sqrt(2r2)
AC = CO + r
AB = sqrt(AC2/2)
[ABCD] = AB2 ≈ 46.63 u2