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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.

 

 Jan 19, 2021
 #1
avatar+128475 
+1

 

 

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

 

 

cool cool cool

 Jan 19, 2021
edited by CPhill  Jan 19, 2021
edited by CPhill  Jan 19, 2021
 #2
avatar+1639 
+3

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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

r = 4

CO = sqrt(2r2)

AC = CO + r

AB = sqrt(AC2/2)

[ABCD] = AB2 ≈ 46.63 u2

 

 Jan 20, 2021

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