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# Find The Area (in Terms Of Ac) Of A Square Inscribed Within A Circle Of Area `ac`

*Here is an exemplar solution for the above question. *

**Date **: 25/04/2019

**VISUALISE**: We must begin by visualizing a Square inscribed inside a Circle. Note that owing to the symmetry of the problem the Diagonals of the Square also act as the Diameter of the Circle. This is a key aspect that will help relate the two areas and get to the ultimate answer.

SideNote: Additionally by another line of logic we can arrive at the equivalence between the Diagonal and the Diameter - Because the Diagonals subtend right angles onto the circle the must be Diameters.

**SOLVE**: Let the side of the square be `s` and the diameter of the circle be `d`.

Thus, using the information in the problem statement the area of the circle, we have `A

_{c}` = (pi)*((d/2)

^{2}). This is just by invoking the expression for the area of a circle which is pi*(radius^2) where we have substituted the radius as half of the diametrical length i.e `radius` = d/2.

Or, A

_{c}= (pi*d*d)/4 ............

*Equation.1*

Now, the diagonal `d` of the square is related to the side `s` by the Pythagorean Theorem. This gives:

d

^{2}= s

^{2}+ s

^{2}, because the sum of the squares of the base `s` and the perpendicular `s` is equal to the square fo the hypotenuse `d`. This eventually yields,

2*s*s = d*d ............

*Equation.2*

As the area of the square is nothing but the length into its height which in this case are the same as `s` we have, Area of Square (say, A

_{s}) = s*s. Substituting this value of A

_{s}in

*Equation.2*we get,

2*A

_{s}= d*d ............

*Equation.3*

Now expressing `d*d` from Equation.1 and substituting in

*Equation.3*we get,

A

_{c}= (pi*(2*A

_{s}))/4. Or,

A

_{c}= pi*A

_{s}/2. Now, making As the subject of the equation we get,

A

_{s}= (2/pi)*A

_{c}. & ............

*Equation.4*

**ANSWER:**

*Equation.4*is the required expression relating the area of a circle to that of an inscribed square.

:)

This resource was uploaded by: Priyash

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