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Solve this: ABCD is a square. Diagonals AC and BD intersect at O. AB = 10.5 cm. A circle with centre O is constructed, such that area is one-fifth the area of the square. Calculate the area and perimeter of the shaded region. [HOT S]

Dear Student,

Please find below the solution to the asked query:

Given : ABCD is a square and side of square =  AB =  10.5 

And we know area of square =  ( Side )2 , So

Area of square ABCD  =  ( 10.5 )2  = 110.25 cm2                  

Also given : Circle with center ' O '  have its area is one fifth of area of square , So

Area of circle with center ' O ' = 110.255 = 22.05

And diagonal of square AC and BD intersect at ' O ' , and we know diagonal of square bisect it to equal parts , So

Area of triangle ACD = 12×110.25 = 55.125 cm2

As diagonal AC passing through center ' o ' of given circle then that also bisect area of given circle , So

Area of half circle = 12×  22.05 = 11.025 cm2

Then,

Area of shaded region  =  Area of triangle ACD  -  Area of half circle  = 55.125 - 11.025 = 44.1 cm2                       ( Ans )

We know :  area of circle  = πr2   and we calculate area of circle with center ' O ' = 22.05 cm2 , So

πr2  = 22.05                     ( Here r  =  Radius of given circle )

227×r2 = 22.05r2 = 22.05×722r2 =154.3522r2 =7.0159r = 2.648 2.65 cm
We know perimeter of half circle ( Without diagonal ) = πr , So

Perimeter of half circle with center ' O '  = 227×2.65 = 58.37 = 8.3285 8.33 cm 

Here , XY  =  Diameter of circle with center ' O '  = 2 ( 2.65 ) = 5.3 cm 

And

We know diagonal of square  = 2 ( Side ) , So 

Length of AC  =  2 ( 10.5 ) =  1.414 × 10.5 = 14.847 14.85 cm 

Therefore ,

Perimeter of shaded region  =  AD +  CD +  AC -  XY + Perimeter of half circle with center ' O ' , Substitute all values we get :

Perimeter of shaded region  = 10.5 + 10.5 + 14.85 - 5.3 + 8.33 = 38.88 cm                                                    ( Ans )
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