B is my answer.

If we draw a circle with center in A, the area will be PIE*(10)^2 =100 pie. So the area of segment ABC will (1/4) of 100 pie=25 pie.
SHADED REGION FROM A SIDE= AREA OF SEGMENT ABC-AREA OF TRIANGLE ABC
=25 pie -50.

Same thing happens from C side.

So, Shaded region = 2*(25pie-50)=50 (pie-2)
B is my answer.

1)Region 1: quarter circle with center A
Area: 10*10*π/4
2)Region 2: quarter circle with center C
Area: 10*10*π/4

Area of Shaded Portion: (area of region 1+region 2)-area of sq
=(100*π/4+100*π/4)-100
=50π-100
=50(π-2)

In a square ABCD the shaded region is the intersection of two circular regions centered at A and C. if AB =10 then what is the area of the shaded region ?

If we draw a circle by considering center as C with radius 10 then area will be
area= π r^2 = 100 π
But for square we have only 1/4th circle
area in square = 100 π /4 = 25 π

Area of triangle BCD will be = 1/2 * 10 * 10 = 50

Area of shaded region = 2 * (25 π - 50) = 50 ( π - 2)

Answer: B

_________________

Area of the square is unshaded region(USD) + quarter of a circle
100= USD+(π/4)*10^2
USD=100-25π

And square is having two unshaded regions


so shaded region = area of square - 2(area of unshaded region)
=100-2(100-25π)
=50π-100

OA:B

Posted from my mobile device

with center as A and C points the area of the region of circle would be 1/4 th of radius 10 circle ; viz. 100pi/4 *2 = 50 pi
and ∆ BCD and ∆ ABD area = 1/2 * 10*10 * 2 ;100
so area of shaded region
50pi-100
or say
50(pi-2)
IMO B

In a square ABCD the shaded region is the intersection of two circular regions centered at A and C. if AB =10 then what is the area of the shaded region ?

A.25(π−2)
B.50(π−2)
C. 25π
D. 50π
E. 50

Area of the sector of a circle ascribed by one corner = \(90/360 * \pi * 10^{2}\) =\(1/4 * \pi * 10^{2}\)

Shaded area = 2*Area ascribed by one corner - Area of the square
\(= 2 * 1/4 * \pi * 10^{2} - 10^{2}
= 10^{2}( \pi /2 -1)
= 50(\pi-2)\)

Area of half of the arc = Area OG sector ABD - Area of triangle ABD
= 90/360*π*10^2 - 1/2*10*10
= 25π - 50
= 25(π - 2)

—> Area of full arc = 2*25(π - 2) = 50(π - 2)

IMO Option A

Posted from my mobile device

In a square ABCD the shaded region is the intersection of two circular regions centered at A and C. if AB =10 then what is the area of the shaded region ?

A. 25(π−2)
B. 50(π−2)
C. 25π
D. 50π
E. 50

Area of two circular regions centered at A and C = Ar. of sector ABD + Ar. of sector CBD
= \(π * \frac{r^2}{4}\) + \(π * \frac{r^2}{4}\) = \(π * 2 * \frac{100}{4}\)
= 50π
Now,

50π = Non - shaded Ar. of sector ABD + Non - shaded Ar. of sector CBD + 2 * (Area for shaded region)
50π = (Non - shaded Ar. of sector ABD + Non - shaded Ar. of sector CBD + Area for shaded region) + Area for shaded region
50π = Ar. of ABCD + Area for shaded region
50π = 10*10 + Area for shaded region
Area for shaded region = 50π - 100
= 50 (π - 2)

ALTERNATIVELY:
As we can see that area of two circular regions is 50π in which area of shaded region is calculated twice. Thus, area of shaded region has to be subtracted once from 50π.

Looking at answer options except 'B' all are out since only has 50π component in it.

Answer B.
_________________

Join the diagonal BD
we get a sector DAB angled 90 and another DAB right angle triangle
so are of half shaded portion towards c is given by
= area of sector DAB - area of triangle DAB
= 90/360 .pi.(100)^2 - 1/2.10.10
= 25pi - 50
=25(pi-2)
area of another half of shaded region will be twice this so= 2. 25(pi-2)= 50(pi-2) (B)