Increase in area = $x + y + \frac{xy}{100} = 50 + 50 + \frac{50 \times 50}{100} = 125\%$

The diameter of each circle is $d$, so the total diameter in a row of three circles will be $3d$ which is equal to the side of the square. Area of the square = $(\text{side})^2 = (3d)^2 = 9d^2$.

$\frac{\sqrt{3}}{4} a^2 = 4\sqrt{3} \Rightarrow a^2 = 16 \Rightarrow a = 4$ cm.

Let the diagonal of the first square be $2d$ and the second square be $d$. Area of a square = $\frac{1}{2} \times (\text{diagonal})^2$. Thus, the ratio = $\frac{\frac{1}{2}(2d)^2}{\frac{1}{2}(d)^2} = \frac{4d^2}{d^2} = 4:1$

Outer radius $R=5$, inner $r=3$. Area = $\pi(R^2-r^2) = \pi(25-9) = 16\pi$ sq m.

Diameter of circle = Diagonal of rectangle = $\sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$ cm.
Radius ($r$) = $20/2 = 10$ cm.
Area = $\pi r^2 = \pi(10)^2 = 100\pi$ sq cm.

Radius increased from $r$ to $2r$. Circumference becomes $4\pi r$ (double). Area becomes $4\pi r^2$ (4 times). According to the question asking 'by how many times will it increase', the option 2, 3 is provided.

Let the radius of the circle be $R$ and the side of the square be $a$. $\pi R^2 = a^2 \implies a = R\sqrt{\pi}$. Ratio of perimeters = $\frac{2\pi R}{4a} = \frac{2\pi R}{4R\sqrt{\pi}} = \frac{\pi}{2\sqrt{\pi}} = \frac{\sqrt{\pi}}{2} = \sqrt{\pi} : 2$.

Let width (b) = $x$, length (l) = $3x$.
Diagonal = $\sqrt{l^2 + b^2} = \sqrt{(3x)^2 + x^2} = \sqrt{9x^2 + x^2} = \sqrt{10x^2} = x\sqrt{10}$.
According to question: $x\sqrt{10} = 8\sqrt{10} \implies x = 8$.
Length = $3 \times 8 = 24$, width = $8$.
Perimeter = $2(l+b) = 2(24+8) = 64$ cm.

Semi-perimeter $S = \frac{25+30+35}{2} = 45$. Area = $\sqrt{S(S-a)(S-b)(S-c)} = \sqrt{45(45-25)(45-30)(45-35)} = \sqrt{45 \times 20 \times 15 \times 10} = \sqrt{135000} \approx 367.42$.

Area = $\frac{50000}{125} = 400$ sq m.
Area of isosceles triangle = $\frac{b}{4} \sqrt{4a^2 - b^2} = 400 \frac{40}{4} \sqrt{4a^2 - 40^2} = 400 10 \sqrt{4a^2 - 1600} = 400 \sqrt{4a^2 - 1600} = 40 4a^2 - 1600 = 1600 4a^2 = 3200 a^2 = 800 a = 20\sqrt{2}$ m.

Volume of water flowing in 30 minutes = $6 \times 1.5 \times (10 \times 1000 \times \frac{30}{60}) = 6 \times 1.5 \times 5000 = 45000$ cubic meters. Required water height = $8$ cm $= 0.08$ meters $= \frac{8}{100}$ meters. Area = $\frac{\text{Volume}}{\text{Height}} = \frac{45000}{8/100} = \frac{4500000}{8} = 562500$ sq meters = $56.25$ hectares.

Let the radii be $x$ and $14-x$. $\pi x^2 + \pi (14-x)^2 = 130\pi \Rightarrow x^2 + (196 - 28x + x^2) = 130 \Rightarrow 2x^2 - 28x + 66 = 0 \Rightarrow x^2 - 14x + 33 = 0 \Rightarrow (x-11)(x-3) = 0$. $x=11$ or $3$. Thus the radii are $11$ cm and $3$ cm.

Area = $\frac{\sqrt{3}}{4} \times (\text{Side})^2 = 25\sqrt{3}$
$(\text{Side})^2 = 100 \Rightarrow \text{Side} = 10$ cm
Perimeter = $3 \times 10 = 30$ cm.

Area = $15 \times 11 = 165$ sq m. Length of carpet = $\frac{165}{0.60} = 275$ m. Cost = $275 \times 12 = 3300$ rupees.

Let width be $x$, length is $x+1$. Perimeter $2(L+B) = 14 \Rightarrow 2(x+1+x) = 14 \Rightarrow 2x+1 = 7 \Rightarrow 2x=6 \Rightarrow x=3$. Length = 4, Width = 3. Area = $4 \times 3 = 12$ sq.cm.

Second diagonal = $\frac{2 \times \text{Area}}{d_1} = \frac{2 \times 30}{12} = 5$ cm.
Side of rhombus = $\sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{(\frac{12}{2})^2 + (\frac{5}{2})^2} = \sqrt{6^2 + 2.5^2} = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5 = 6 \frac{1}{2}$ cm.

Total wall area $= 20 \times 5 = 100$ sq m. Area of doors $= 2 \times (2 \times 1.5) = 6$ sq m. Remaining wall $= 100 - 6 = 94$ sq m. Volume of wall $= 94 \text{ m}^2 \times 0.5$ m $= 47$ m$^3$. Volume of brick $= 0.25 \times 0.16 \times 0.1$ m$^3 = 0.004$ m$^3$. Number of bricks $= \frac{47}{0.004} = 11750$.

Area = $(31+4) \times (14+4) - 31 \times 14 = 35 \times 18 - 434 = 630 - 434 = 196$ sq m.

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding perimeters.
Thus, the ratio of perimeters = √25 : √16 = 5 : 4