Answer: Option A. -> 20           
:
A
Ans. (a) A, B, C will lie on a circle with centre at D (as the angle subtended by the arc at the centre i.e. 260${}^{\circ }$ is twice subtended at the circle i.e. 130${}^{\circ }$)
In triangle DAC, $\angle$DAC = $\angle$DCA = 40${}^{\circ }$.Let $\angle$ADB = 2x $\Rightarrow$$\angle$ACB = x, and let $\angle$BDC = $\angle$CBD = y $\Rightarrow$ 2x+y = 100${}^{\circ }$, and 2y+x = 140${}^{\circ }$.
Hence (a) is the right answer

Answer: Option C. -> 1(√2−1)2
:
C
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.

Answer: Option D. -> none of these
:
D
option (d)

From the figure, 0.338xy = O.2ba $\therefore$ab = 1.69xy ..... (1)Also, since ¨ABCD and ¨PQRS are similar,$\frac{b}{a}-\frac{y}{x}\therefore \frac{b}{y}-\frac{a}{x}$..... (2)From (1) and (2), we get $\frac{a}{x}$ = 1.69 $\frac{x}{a}$a${}^2$ = 1.69x${}^2$ a = 1.3x $\frac{a}{x}$= 1.3 answer = 1.3.

Answer: Option B. -> 49[π2−√32] cm2
:
B
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*($\frac{1}{3}$ – $\frac{3}{2}$)

Answer: Option B. -> 122sq cm
:
B
Altitude of the traingular faces = $\sqrt{\frac{a^2}{4}+9}$
Area of faces =2 x a x$\sqrt{\frac{a^2}{4}+9}$
The total surface area = area of base + lateral surface area
= a${}^2$ + 2 x a x$\sqrt{\frac{a^2}{4}+9}$
Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 $\times$ 8 $\times$ 5 =144)
= 12${}^2$

Answer: Option C. -> 4:1
:
C
Graphical Division: -
Assume the square to have a side 2a. Hence, area of square = 4a${}^2$
Using graphical division, we can divide the figure into 8 parts as follows. The triangle in question is the shaded part



Thus the triangle is $\frac{2}{8}$${}^{th}$ of the square area
Hence ratio = 1:4
Shortcut:- Assumption method. Assume a simple case as the isosceles triangle and square in this case. Just substitute and solve

Answer: Option B. -> 15∘
:
B

AB = BC and$\angle$ABC=60°. Therefore, ΔABC is an equilateral triangle
Now see thatABOP 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${}^{\circ }$

Answer: Option B. -> 12.5
:
B
Option b
It’s important that we note that it is a Pythagoras triplet
7${}^2$ = 25${}^2$+24${}^2$
Now,
In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle.
As the hypotenuse is 25, the circum radius is 12.5

Answer: Option B. -> 225√34
:
B
In an equilateral triangle circum radius = $\frac{2}{3}$ h,
10 = $\frac{2}{3}$ * $\frac{\sqrt{3}}{2}S$
Side = 10\(\sqrt3\) 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 =$\frac{3}{4}\times \frac{\sqrt{3}}{4}\times (10\sqrt{3}{)}^2$ = $\frac{225\sqrt{3}}{4}$

Answer: Option B. -> 9
:
B
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