IMO C . Explanation as in the attached image

Questions seem easy if you go by options and logic

Square side = 10
Area of square = 100
As the orange region incribed in share its area must be less than 100 .

Hence option D and E gone as they are greater than 100.

Now if u convert option ABC in decimal it will be

A.3.2 B.16 C.32

Now area of quarter circle = (3.14*10*10)÷4 =78.5

If you observe carefully quarter circle ,the orange area compromised around half area of quarter circle ie.38 ...So
Answer C.

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IMO B.
The options are key to find aright answer in PS.Luckily what options we have here are fairly spreaded.
The entire area of the square is 100 so, the orange shade would be less than 100 .
C D E CAN be eliminated.
Now A is far less than expected.
Remaining is B.So we can proceed with B.

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Explanation:

Side of square,a = 10
Radius of Quater circle is equal to side of square, (r=a) = 10

Zone A,B,C,D,E,F,G,H,I are area in figure attached.

Area of B,C,D,E are equal (Symmetry)
Area of F,G,H,I are equal.
(Symmetry)

As shown in figure, zx=zy=xy=a (as, zx & xy are radius of Quater circles & xy is side of square) & xyz is equalateral Triangle.

So area of Zone A,D,H,E is = 2*(60/360)*π*a^2 - (√3/4)*a^2
= (1/3)*π*a^2 - (√3/4)*a^2

Area of Zone C+G = Area of Quater Circle - Area of Zone A,D,E,H
= (1/4)*π*a^2 - (1/3)*π*a^2 + (√3/4)*a^2
= (√3/4)*a^2 - (1/12)*π*a^2

As Area of Zone C+G = Area of Zone D+H = Area of Zone E+I = Area of Zone B+F

So Area of Zone A = Area of Square - 4*(Area of Zone C+G)
= a^2 - 4*{(√3/4)*a^2 - (1/12)*π*a^2}
= a^2 - (√3)*a^2 + (1/3)*π*a^2
= a^2 * (1 + π/3 - √3)
Where a=10 (side of Square)

= 100*(1 + π/3 - √3)

IMO-C


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Let me be honest with you, this question took me more than 4 minutes to solve. It is definitely a hard one but with practice you will be able to figure out the approach to such areas based question.


Let's start with labeling the different regions in the figure:
1) The figures with the smallest area formed with the edges of the square to be named as 'a'
2) The area to be calculate to be named as 'c'
3) Lastly, the triangular regions formed with the vertices of the square and sides of the figure 'c'

The total area of the square = \(10^{2}\) = \(4a + 4b + c\) = 100 ------------------------------------------(1)

The area of a quarter circle = \(2a + 3b + c = \frac{area of square}{4}\) -----------------------------------------(2)

The area of the circular sector = \(pi*10^{2}* \frac{60}{360}\)

The area of the equilateral triangle inscribed in the arc with sides measuring 10 units each = \(\frac{\sqrt{3}}{4}*10^{2}\)


Now, in order to find the area of the circular segment, we subtract the above two areas, i.e area of circular sector minus area of triangle.

Thus, area of segment = \((pi*10^{2}* \frac{60}{360})\) - \(\frac{\sqrt{3}}{4}*10^{2}\)

Thus forming our 3rd equation, \(a + 2b + c\) = \( \frac{pi*10^{2}}{3}\) - \(\frac{\sqrt{3}}{4}*10^{2}\) ------------------------------------------------------------------------------(3)

That's 3 equations and 3 variables, just by substituting the variables we obtain our expression for c which is OPTION C

The figure above shows four identical quarter circles drawn between the vertices of the square. If a side length of the square is 10, what is the area of the orange region?

Area of quarter circle AECDB
Area of AECDB=\(\frac{1}{4} \pi(10^2)=25 \pi cm^2\)

Area of sector AFCGB
Area of sector AFCGB=\(\frac{1}{6} \pi(10^2)=\frac{50}{3} \pi cm^2\)


Area of segment AFCE
Area of segment AFCE=\(\frac{1}{6} \pi(10^2)−\frac{1}{2}(10^2)sin60^{\circ}\)
AFCE=\(\frac{50}{3} \pi−25\sqrt{3 }cm^2\)


Area of BGCD
BGCD=AECDB−AFCGB−AFCE
BGCD=\(25 \pi cm^2\)-\(\frac{50}{3} \pi cm^2\)-\(\frac{50}{3} \pi−25\sqrt{3 }cm^2\)
=\(25 \pi - \frac{100}{3} \pi +25\sqrt{3 }\)

Area required=Area of square−4* Area of BGCD
=\(10^2-4*(25 \pi - \frac{100}{3} \pi +25\sqrt{3 })\)

= \(100(1+ \frac{\pi}{3}−\sqrt{3})\)

Answer is C

A. \(10(1+ \frac{\pi}{3}−\sqrt{3})\)

B. \(50(1+ \frac{\pi}{3}−\sqrt{3})\)

C. \(100(1+ \frac{\pi}{3}−\sqrt{3})\)

D. \(100(1+ \frac{\pi}{3}+\sqrt{3})\)

E. \(100(1+ \frac{\pi}{3}+2\sqrt{3})\)

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The answer isC

Given that:
Side of square = 10
Area of square = 100

Radius of quadrant = side of square = 10
Area of 4 quads = 100 × 22/7
Area of 1 quAd = 25 × 22/7


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The figure above shows four identical quarter circles drawn between the vertices of the square. If a side length of the square is 10, what is the area of the orange region?


A. 10(1+π/3−√3)

B. 50(1+π/3−√3)

C. 100(1+π/3−√3)

D. 100(1+π/3+√3)

E. 100(1+π/3+2√3)

Area of square = 100 = 4x+4y+z - eq 1
Area of each quarter = (90/360) 100π = 25π = 2x+3y+z – eq 2
3 variable need 3 equations
Area of segment AC = Area of the sector BAC – Area of the equilateral triangle = (60/360) 100 π - (100√3)/4 = (100 π)/6-25 √3
Area of the sector BAC + Area of segment AC = x+2y+z = (60/360) 100 π + ((100 π)/6-25 √3) = ((200 π)/6-25 √3) - eq 3

Solving the 3 equations, we get option C as the answer.

Very difficult question. If it comes in the exam, Im pretty sure I'll take more than 3-4 minutes under pressure.

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The figure above shows four identical quarter circles drawn between the vertices of the square. If a side length of the square is 10, what is the area of the orange region?


A. \(10(1 + \frac{\pi}{3} - \sqrt{3} )\)

B. \(50(1 + \frac{\pi}{3} - \sqrt{3} )\)

C. \(100(1 + \frac{\pi}{3} - \sqrt{3} )\) --> correct

D. \(100(1 + \frac{\pi}{3} + \sqrt{3} )\)

E. \(100(1 + \frac{\pi}{3} + 2\sqrt{3} )\)

Solution:
AEB is an equilateral triangle w/ side a (a is side length of the square & a= 10) --> each angle of AEB = 60 degree
So area of AEB = \(\frac{\sqrt{3}}{4}{a^2}\)
Area of circular region-AEFBA = \(\frac{60}{360}*\pi*a^2 = \frac{{\pi}a^2}{6}\), as each angle of AEB is 60 degree so angle EAB=60 degree
So the area of the region-BEFB(blue region) = \( \frac{{\pi}a^2}{6} - \frac{\sqrt{3}}{4}{a^2}\)
Area of circular region-BECB= \(\frac{30}{360}*\pi*a^2 = \frac{{\pi}a^2}{12}\) , as each angle of AEB is 60 degree so angle EBA=60 degree & Angle ABC=90 degree, so angle EBC = 90-60=30 degree
So the area of the region-BFECB(blue vertical strike region) = \( \frac{{\pi}a^2}{12} - (\frac{{\pi}a^2}{6} - \frac{\sqrt{3}}{4}{a^2}) = \frac{\sqrt{3}}{4}{a^2} - \frac{{\pi}a^2}{12} \)

So the area of the orange region = (area of the square) - 4 * [the area of the region-BFECB(blue vertical strike region)] = \( a^2 - 4*(\frac{\sqrt{3}}{4}{a^2} - \frac{{\pi}a^2}{12}) = a^2 - \sqrt{3}{a^2} + \frac{{\pi}a^2}{3}= a^2(1 + \frac{\pi}{3} - \sqrt{3})= 100(1 + \frac{\pi}{3} - \sqrt{3}) \)
Answer: C


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Ans looks like C.
I tried making a guess by calculating the value in the bracket and it looks like the shaded area has to be < 100. Hence 100 * (less than 1) should be the answer. Thus C.
_________________

Here,
Side of square = Radius of the (quarter) circle = 10.

In this kind of figure, the area of the central part(orange region) can be determined by following formula:

Area= r²[1+(pie/3)-Root 3]

[Please understand the formula; the symbol are not written cause they are not in my mobile.]

Now,
In the question,

Area of shadded region
= 10²[1+(pie/3)-Root3]
= 100[1+(pie/3)-Root3]

So, the answer is C.

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The answer is C), see attached image

_________________

The answer is (C).


What formulas will be used in this question?
- area of quarter circle: \(\frac{πr^2}{4}\)
- ratio of 30:60:90 triangle is \(1:\sqrt{3}:2\)
- area of a triangle: \(\frac{base*hight}{2}\)

Now let’s say A, B, C, and D are vertices of the square, and E, F, G and ,H are vertices of orange region \(x\).

Here, we can remove area of figure AFED, DEHC, CHGB, and BGFA from area of the square in order to get the area of \(x\), which is the orange region.

Step 1
The area of a quarter circle DAC is \(25π\)

Step 2
Here, we can make a segment DF and CF to make a 60-60-60 equilateral triangle. Why it’s a equilateral triangle? Because the segment DF is a radius of the quarter circle DAC, as well as the segment CF. Thus segment DF, CF and DC have same length.

Step 3
When we draw a perpendicular foot FI, which half the angle DFC, the 30-60-90 right triangle forms. Therefore, the length of segment FI is \(5\sqrt{3}\), and the area of triangle DFI is \(\frac{25\sqrt{3}}{2}\). Thus, the area of equilateral triangle DFC is \(25\sqrt{3}\).

Step 4
Since the angle CDF is 60, we can also calculate the area of sector DFC. -> \(\frac{100π}{6}\)=\(\frac{50π}{3}\)
In the same way, the angle ADF is 30, therefore sector ADF is \(\frac{25π}{3}\)
Also, we can calculate the area of sector CFG, which is area of sector DFC - area of the equilateral triangle DFC = \(\frac{50π}{3}-25\sqrt{3}\)

Step 5
Finally, we can get the value of area of figure AFED, DEHC, CHGB, and BGFA.
-> area of figure AFED, DEHC, CHGB, and BGFA = \(4*(\frac{25π}{3}-(\frac{50π}{3}-25\sqrt{3})\)=\(4*(25\sqrt{3}-\frac{25π}{3}\)

Step 6
At last, let’s get the area of \(x\)
\(x=100-4*(25\sqrt{3}-\frac{25π}{3})=100+\frac{100π}{3}-100\sqrt{3}=100(1+\frac{π}{3}-\sqrt{3})\)

I attach a phto to help you understand the equations


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D) and E) are out since the area cannot be more than 100 (10 to the power of 2)

A) and B) are also out because your base is 100 and you need to substract some sectors to get the answer, so the answer needs the number 100

Only possible answer C)

Area for A+B+C= Area of Square- 2* (Area of yellow region) = 100- 2* (100- Pie*10^2/4)= 100- 2*(100- 25Pie)
= 50pie- 100~ 57 units.(Approx.) and it will greater than 19 (by pic we are assuming it, more than 1/3 of 57)

So, Area of B will be more than Less than 57. So, option D and E goes out.

Please note- 1+ pie/3- sqrt(3) = 0.3
A- No (less than 10), B- No (between 15-17), C- Yes (value 32 approx.)

So, I think ANS, C.

C is the answer by plugging in each answer. Except C the rest of the answers are too small for it be the area of the shaded region.

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Ans will Be C

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Figure is attached for reference.
From figure I,let a,b,and c be the area of depicted portion.And let r be the length of side of square.
we have to find AREA of c=?
From figure II,Symmetrical arc's radius is two sides of triangle,which is equal to side of square.Hence triangle formed is equilateral as all sides are equal.Also from drawing it is clear ARC of 60 degree is formed.
Hence we can find area of shaded part(in figure II)= x =Area of arc of 60 degree-area of equilateral triangle=60/360*pi r^2- square root 3*r^2/4
Now from figure III,
Area of shaded part in figure III=2*x+area of equilateral triangle.=2b+c+a from figure I
2b+c+a=2*(pi r^2/6-square root 3 r^2/4)+square root 3 r^2/4
2b+c+a=pi r^2/3- square root 3 r^2/4
Also from figure I, 2b+2c+a+b+a= pi r^2/4 (quarter of circle)
hence putting value of 2b+c+a, we get b+a=pi r^2/4-(pi r^2/3-square root 3 *r^2/4)
b+a=square root 3*r^2/4- pi r^2/12
Area of portion c= area of square - 4*(b+a)= 100 -4*(square root 3*r^2/4-pi r^2/12)
given r=10, Area of c= 100-100* suare root 3+ 100*pi/3
HENCE ANSWER IS OPTION C.