In a square ABCD the shaded region is the intersection of two circular

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)

1/4(π r^2)+1/4(π r^2)-area of square(10^2)

r=10

therefore 50(π-2) is the answer

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 ?

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

IMO it is B

Area of quarter circle=\((π* 10^2)/4=25π\)
Are of triangle in the quarter circle =\(1/2 *10*10= 50\)

Area of 1/2 of the shaded arc= \(25π-50\)
Area of full shaded arc = \(2*(25π-50)= 50*(π-2)\)

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

Approach:

- Divide the Square in two half diagonally.
- area of triangle BDC = \(\frac{1}{2}*10*10 = 50\)
- area of circular region BDC(it is exactly 1/4th of full circle) = \(\frac{1}{4}*π*r^2\)
radius is 10 => \(\frac{1}{4}*π*10^2 = 25π\)
- to get the area of shaded region we'll minus the area of triangle from entire circular region and multiple by 2 to cover full shaded region area. -> \(2(25π - 50) = 50(π-2)\)

IMO Answer is Option B!

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

Please, find the attached file.
The answer is B.

Posted from my mobile device

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
\(= 2 * 1/4 * \pi * 10^{2} - 10^{2}\\
= 10^{2}( \pi /2 -1)\\
= 50(\pi-2)\)

In the figure we can see 2 quarter circles defined by arc ABD and arc DBC.

By visualization we can deduce the following -

Area (arc. ABD + arc. DBC) = Area. Shaded Region + Area Square ABCD
=> Ar. Shaded Region = Area (arc. ABD + arc. DBC) - Area Square ABCD

=> Required Area = Pi/4 (10^2 + 10^2) - 10^2
=> Required Area = 100 [ Pi/2 - 1]
=> Required Area = 50 [Pi - 2]

Answer is (B).

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

Pir²/4 - 10×10/2=25pi-50, the area of half the shaded region.
So, 2×(25pi-50)
Option B

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.

we can derived the shaded region by drawing a quarter of circle minus with the triangle

Finding area of a quarter of circle = 1/4 * π r^2 = 1/4 * π * 10*10 = 25π
Area of a triangle = 1/2*10*10 = 50

Thus, the shaded region is the double of the above mentioned drawing.
So, the shaded region = 2* (25π - 50) = 50π - 100 = 50 (π-2) - 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)

Really good question, thanks for posting, Bunuel!

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.