HOT Competition 31 Aug/8PM: The figure above shows four identical quar

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

Since the area of shaded region is less than area of the square (100), and approx 50 (half the area of the square as per visual estimation), let us calculate the areas given in the options.

\(A. 10(1+\pi/3-\sqrt{3}=3.1514\)
Incorrect


\(B. 50(1+\pi/3-\sqrt{3}=15.75\)
Incorrect

\(C. 100(1+\pi/3-\sqrt{3}=31.51\)
Closer. Keep it.

\(D. 100(1+\pi/3+\sqrt{3}=377.92>100\)
Incorrect

\(E. 100(1+\pi/3+2\sqrt{3}=551.13>100\)
Incorrect

C is the most close to 50 and is the right answer

IMO C

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The answer is C

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So I used the Ball-parking method here, I am sure there must be a standard way to solve this.

If we take one quarter of the circle, with radius = 10 (given as side of square)
Area = ( pi*10^2 )/4 = ( 3*100 )/4 = 75

We can be sure that the area of orange region is less than 75 -- (between 1/3 and 1/2 of the sector area)

Plugging in pi = 3 in answer choices:

A. 10(1 + π/3 − √3) = 10(0.27) = 2.7 = TOO LESS

B. 50(1 + π/3 − √3) = 50(0.27) = 13.5 = TOO LESS

C. 100(1 + π/3 − √3) = 100(0.27) = 27 = PERFECT

D. 100(1 + π/3 + √3) = 100(3.73) = 373 = TOO BIG

E. 100(1 + π/3 + 2√3) = 100(5.4) = 540 = TOO BIG


Correct answer is C

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Answer choice C.
We will divide the square in 9 parts:
A= 4 parts will be the shapes that arrives to the vertices of the square
B= 4 parts will be the shapes between A.
O= The orange region.

None of this 9 parts overlap.

The formula for the orange region would be:

O = Area of square - 4A - 4B..... (1)

so, first we will estimate the area of 1 part of type A.

To do this, we create an equilateral triangle and two parts of a circle.

A = Area of square - Equilateral triangle - 2 (Parts of circle)
A= 100 - 25*3^(1/2) - 2 *(PI * 25/3)
A= 25 *(4 - 3^(1/2) - 2* PI /3)

We need 4 times A.

Second, we will estimate the area of 1 part of B.

B = Area of square - 1/4 of circle - 2A
B = 100 - 25 PI -2A
B = 25 *(-4 + PI/3 + 2*3^(1/2) )

We need 4 times B.

So, replacing in (1).

O = 100 - 100 * (4 - 3^(1/2) - 2* PI /3) - 100 *(-4 + PI/3 + 2*3^(1/2) )
O = 100 (1 - 4 + 3^(1/2) + 2* PI /3) +4 - PI/3 - 2*3^(1/2) )
O = 100 (1 + PI/3 - 3^(1/2) )

Answer choice C.

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IMO C

Refer to the image below:

Let ABCD be the triangle & EFGH be the point of intersection of the arcs.

Join AEF, BGF, CHG,DHE

Now, lets take region ABEFDA- a quarter of a circle with center at A
Here, Triangle ABE, AEF, AFD are concurrent isosceles triangle, So angle (BAE)=EAF=FAD=30
AE=AB=AF=AD= radius of quarter circle=10
Area of traingle AEF= 1/2 AE AF Sin30= 100/4
Area of sector AEFA= pi (10)^2 * 30/360 = 100*pi/12

So, Area of region bounded between Side of triangle EF and arc EF = 100*pi/12- 100/4 ............(I)

From Cosine rule, EF^2= 10^2+10^2- 2*10*10 cos 30
EF^2= 200-200(sqrt3)/2

So Area of shaded region= Area of Square EFGH + 4* Area between Side of triangle EF and arc EF
Area of shaded region = EF^2 + 4* (100*pi/12- 100/4) = 200-200(sqrt3)/2 + 100 *pi/3- 100 = 100 +100*pi/3 -100(sqrt3)= 100(1+pi/3- sqrt3)

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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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Since the figure is symmetric, we can classify the various areas as a, b, and c (The vertices of various points are also marked as shown in the figure)
Our objective is to find c,
If we take the quarter circle ABC,
Area of circle ABC = 25 π ( ratio of area of circle to square = π : 4)
So, 3a + 2b+ c = 25 π;
Area of the square ABCD = 100;
4a + 4b + c = 100;
Area ABCD – ABC = a + 2b = 100 – 25 π ---1

Now let’s find the area of the segment AEB;
AEB = 2a + b + c ;
Area of the equilateral triangle AEC = √ (3)/4 *100
Area of the circular segment AEB = (60/360) * π * 100
Area of circular portion EB = π/6*100 - √ (3)/4 *100
So, total area of area of the segment AEB = Area of the circular segment AEB + Area of circular portion EB = π/6*100 + π/6*100 - √ (3)/4 *100
= π/3 *100 - √ (3)/4 *100
Area of circle ABC - area of the segment AEB = a + b = 25 π – π/3 *100 + √ (3)/4 *100 --- 2
From these equation 1 & 2
b = eqn 1 – 2 = 100 – 25 π – ( 25 π – π/3 *100 + √ (3)/4 *100)
= 100 – 50 π + π /3*100 – √ (3)/4 *100 ---- 3
a = eqn 2 – 3 = 25 π – π/3 *100 + √ (3)/4 *100 – (100 – 50 π + π /3*100 – √ (3)/4 *100)
= 75 π – 100
c = AEB – 2* a – b
= (π/3 *100 - √ (3)/4 *100) – 150 π + 200 – 100 + 50 π - π /3*100 + √ (3)/4 *100
= 100(1+π/3−√3)

Let us first used logic to eliminated some of the obvious choices. Since,

Area (square) = 10*10 = 100 sq. units

Option D and E > 100 sq. units. Since this is greater than the area of square itself, these two options cannot be the area of the shaded region which clearly lies inside the square.

Eliminate option D, E

Now area of one LEAF = 50 pie – 100 (refer to screenshot attached to see how the area is calculated)

= 50(3.14) – 100

~= 57 sq. units

If the three regions of the LEAF - R1, R2, R3 (refer diagram 2, screenshot) were symmetrical, then

Area of each region = 1/3*57

= 19 sq. units

However, region R2 > R1 and R3, so

Area (SHADED ORANGE Region = R2) > 19 sq. units

Now, let us look at the three remaining options – A, B, C

Value of the common term = [1 + pie/3 - sq. root (3)]

~= 0.3

Option A: 10 * 0.3 = ~3 sq. units (< 19 sq. units) ----- REJECT

Option B: 50 * 0.3 = ~15 sq. units (< 19 sq. units) ----- REJECT


Option C: 100 * 0.3 = ~30 sq. units (>19 sq. units) ----- POSSIBLE

Therefore, Option C

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This looks difficult but use logic to figure out an answer.

Side of the Square = 10. Area = 100
The highlighted portion is definitely less than 100.

Option D & E are > 100 . Eliminate.

Remaining options have the term 1+pi/3 - 3^1/2 common. Calculate. It becomes equal to approx 2-1.73 = 0.27

A) approx 2.7 , B) approx 13.5 , C) approx 27

From the diagram , split the square into 2 halves. The shaded region occupies approx 1/3rd the area.
So in the square also , the shaded portion occupies approx 33% of the area =100X1/3 = 33.33

Match the options with this area. Option C comes the closest. Therefore Best answer.

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