Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d${}^2$ = R(R-2r) where R = circum radius, r= inradius

d${}^2$ = R(R-2r) = 6(6 - 2 x 2) = 12

d=2$\sqrt{3}$

Option (b)

 O is the centre of the circle and the mid-point of BC. DO is parallel to AC. So, $\angle$DOB = 60°

 Area of Δ BDO =$\frac{3}{4}$ * 49

Area of sector OBD = $\frac{49}{6}$

Hence area of the shaded region

 = 2[$\frac{49}{6}$ – $\frac{3}{4}$*49]

= 49[$\frac{1}{3}$ – $\frac{3}{2}$]

 Shortcut

By graphical division, there are 3 equilateral triangles of areas of $\frac{\sqrt{3}}{4}$ * 49

Area of interest = (area of semi circle [$\frac{r^2}{2}$] - area of three triangles) = (49*$\frac{1}{2}$ - 3*3*$\frac{49}{4}$)*($\frac{2}{3}$) = 49*($\frac{1}{3}$ – $\frac{3}{2}$)

Let the radius of the identical circles be r. Hence, Areas of the four circles lying within the square = 4 x $\pi$ x $\frac{r^2}{4}$ = $\pi$ x $r^2$ 

Radius of the smaller circle = $\frac{(diagonalofsquare-2r)}{2}$ = $\frac{(2\sqrt{2}r-2r)}{2}$ = r ($\sqrt{2}$-1) 

Required ratio = $\frac{\pi x{r}^2}{\pi x\left[x^2\right(\sqrt{2}-1{)}^2]}$ = $\frac{1}{(\sqrt{2}-1{)}^2}$

Shortcut : Lateral Thinking

The question is basically the ratio areas of larger to smaller circle because sum of areas within the square of the 4 circles add up to a full large circle.

Approximately compare the radii of small and large circles; the ratio is nearly 3.5/1, square of which is approx. 6/1. Only option c) works. This is a good approach which can be used for a lot of geometry questions.

AB = BC and $\angle$ABC=60°. Therefore, ΔABC is an equilateral triangle

Now see that ABOP is a rectangle.

And $\angle$BAN=60°, Therefore, $\angle$NAP=90${}^{\circ }$ − 60${}^{\circ }$ = 30${}^{\circ }$

And $\angle$ANP =$\frac{1}{2}$ * 90 = 45${}^{\circ }$

Now in ΔANP,

$\angle$NPA=180${}^{\circ }$−45${}^{\circ }$−30${}^{\circ }$=105${}^{\circ }$

And hence $\angle$NPO = $\angle$NPA - $\angle$OPA = 105${}^{\circ }$ - 90${}^{\circ }$ = 15${}^{\circ }$

Shortcut

ABC is an equilateral triangle and ABM is a 30 – 60 -90 triangle (M being the point of intersection of AN and the circle). OMN is also 30${}^{\circ }$. MOP = 90${}^{\circ }$, and MNP = 45${}^{\circ }$; MPO = PMO = 45${}^{\circ }$. NPO = 180${}^{\circ }$ – 75${}^{\circ }$ – 45${}^{\circ }$ – 45${}^{\circ }$ = 15${}^{\circ }$

Interior angle = [(2n-4)x90]/n = 108

2x= 360-(108x2) = 144

angle P = 180-144= 36

Total = 36x5= 180

 Useful to remember: Sum of angles in a star= 180${}^{\circ }$

 Shortcut: Clearly the star point trisects the angle in a pentagon. Hence each angle = 108/3 = 36.

Sum of angles = 36*5 = 180

Alternatively:  Sum of angles of a n point star = (n – 4) x 180${}^{\circ }$ , where n = number of sides of a polygon.

Here, n = 5. Sum of angles of a n point star = (5 – 4) x 180${}^{\circ }$ = 180${}^{\circ }$.

Area of the square: (x + 5)${}^2$

So the answer has to be a perfect square: 36 or 64

If area is 36 then x = 1; so EF = 6, OE = 1, OA = 5 and EA = 4

Angle OFB = ABF = BAE = 90, so angle OEA = 90

OE, EA and OA do not form Pythagoras triplet.

If area is 64 then x = 3, EF = 8, EA = 4, OA = 5

Shortcut 1

Reverse Gear approach-With a radius of 5, we will get a square of side 5%undefined2. area will be 50, but we know the side has to be slightly greater than 5%undefined2. hence, choose an option which is slightly greater than 50; i.e. 64

Shortcut 2

Graphical division:- The square gets divided into 8 parts. 4 parts are each right angled triangles of sides 3,4,5 and hence with area 6. 4 of them in total have an area of 24. The other 4 are triangles of the form 4, 5, $\sqrt{41}$ and whose areas are 10 each. Total area = (6+10)*4 = 64.

Areas of circles follow the sequence 1,4,9,16,…………100${}^2$

Areas of Black & Red annuals – 1,3,5,7,9……………199 – General term 2n – 1

Red annuals – 3, 7, 11, …………….. 199 Sum = $\left(\frac{50}{2}\right)$*(3 + 199)

Largest circle’s area = 100${}^2$

Ratio = $\frac{101}{200}$

Note : Ignore pi as it is a ratio of areas

Shortcut: Reverse gear approach –

We know that the denominator is 100${}^2$ and num is an integer, hence the denominator cannot be 101 as it is in no-way a factor of 10000. Hence option d) has been eliminated.

We need to find $\frac{R}{(R+B)}$. R + B = 10000. Each of the red annual is greater in area by 2 units than the subsequent black annual and there are 50 such cases. Hence R = 5050 and B = 4950. $\frac{R}{(R+B)}$. = $\frac{5050}{10000}$ = $\frac{101}{200}$

Deducing a Pattern:

Taking two circles. The number of points of intersection is 2 at most.

If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together).

Shortcut:- Assumption & Reverse Gear:

Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.

Sum of the interior angles of a polygon = (n-2) 180

$\frac{n}{2}$ * (2a+ (n-1)d)= (n-2)180

$\frac{n}{2}$ * [2*(120) + (n-1)5]=(n-2)180

n[48+(n-1)] = (n-2)72

n${}^2$ -25n + 144=0

n=9 and n=16

when n= 16, the greatest angle will be equal to a+15d = 120 + 15 x 5 = 195 and no interior angle of a

polygon can be equal to or greater than 180. hence answer =b

In an equilateral triangle circum radius = $\frac{2}{3}$ h,

10 = $\frac{2}{3}$ * $\frac{\sqrt{3}}{2}S$             

Side = 10\(\sqrt 3\) units.

Area of Triangle = $\frac{\sqrt{3}}{4}(10\sqrt{3}{)}^2$

By graphical division, we need $\frac{3}{4}$${}^{th}$ of the area of the triangle =$\frac{3}{4}\times \frac{\sqrt{3}}{4}\times (10\sqrt{3}{)}^2$ = $\frac{225\sqrt{3}}{4}$