Answer: Option C. -> 101200
:
C
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. $\frac{R}{(R+B)}$. = $\frac{5050}{10000}$ = $\frac{101}{200}$

Answer: Option B. -> 17.32 m
:
B
Option (b)
Perimeter = 3a
Area = $\frac{\sqrt{3}}{4a^2}$
10x$\frac{\sqrt{3}}{4a^2}$= 25x3a
a=$\frac{30}{\sqrt{3}}$= 17.32 m

Answer: Option A. -> 100∘
:
A

Option (a)
Take DBF = a${}^{\circ }$ and ECF = b${}^{\circ }$.
40 + a${}^{\circ }$ + b${}^{\circ }$ = 180 $\Rightarrow$a${}^{\circ }$ + b${}^{\circ }$= 140
DFB + EFC + DFE = 180 $\Rightarrow$ 180 – 2a${}^{\circ }$ + 180 – 2b${}^{\circ }$ + DFE = 180 $\Rightarrow$ 180${}^{\circ }$ – 2(a + b) = - DFE
DFE = 280 – 180 = 100${}^{\circ }$

Answer: Option C. -> √7:√3
:
C

Let the side of the equilateral triangle ΔABC be 3 units. Clearly, BD = 2, CD = 1 and CE = 2.
CD/CE = $\frac{1}{2}$ So $\angle$DEC=30${}^{\circ }$
Hence, ΔCDE is a right angled triangle.
Similarly, ΔBDE is also a right angled triangle.
BE = $\sqrt{7}$ and ED = $\sqrt{3}$ hence BE:ED = $\sqrt{7}$ : $\sqrt{3}$

Answer: Option A. -> 45∘
:
A

$\angle$PTQ= 90 (angle in a semi circle)
, hence $\angle$RTS = 45.
$\angle$ROS = 90 ( Inscribed angle= 2 *Angle at circumference)
OR=OS=OR (radius)
$\angle$RSO = $\angle$SOQ= 45${}^{\circ }$ (interior alternate angles)
Option (a)

Answer: Option D. -> ≥ 45∘
:
D
Consider any triangle ABC, AM is a line connecting A to mid point of BC. Drop a perpendicular from A to BC touching BC at D. The angle ADM is a right angled triangle with 45 – 45 – 90 angles. Hence DAM = 45 and BAM > 45. BAM= 45${}^{\circ }$If $\angle$ABC =90${}^{\circ }$. hence, answer is option (d)

Answer: Option A. -> 64 sq units
:
A
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.

Answer: Option B. -> 1
:
B
Option (b)
Assume a 45-45-90 triangle
sin $\alpha$= P = $\frac{1}{\sqrt{2}}$ . (Sin= opposite side/ hypotenuse)
tan $\alpha$= Q = 1 (Tan = opposite side/ adjacent side)
$\frac{1}{P^2}-\frac{1}{Q^2}$ = 2-1 =1

Shortcut
$cose{c}^2\alpha$ – $co{t}^2\alpha$ = 1 is a trigonometric identity
If you are not aware of this identity, substitute $\alpha$ = 45.

Answer: Option C. -> 75
:
C
Option c
In a quadrilateral, product of area of diagonally opposite triangles is same.
So, the product of other set of opposite triangles will be 12*27
Now, we have to find minimum a and b such that a*b=12*27 and a+b is minimum.
So a = b = 18 (as 12*27 = 324)
Hence the area of the quadrilateral = 18 + 18 + 12 + 27 = 75

Answer: Option A. -> 18 cm2
:
A

Draw perpendicular from D to GF meeting at X.
Also, let GN = GP = k
Since, DE = XF = 3 cm
NX = 3 – 2 = 1 = DM = DP
Therefore, GX = k – 1
$\Delta$DXG is right angled triangle.
Hence, (k + 1)${}^2$ – (k – 1)${}^2$ = 42
Hence, k = 4 and so GF = 6 cm
Area of trapezium DEFG = $\frac{(3+6)}{2}$ * 4 = 18 cm${}^2$