Answer: Option C. -> 75%
Let original radius be R cm
Then, original circumference = $$2\pi r$$ cm
New radius :
$$\eqalign{
& = \left( {175\% {\text{ of }}R} \right)cm \cr
& = \left( {\frac{{175}}{{100}} \times R} \right)cm \cr
& = \frac{{7R}}{4}\,cm \cr} $$
New circumference :
$$\eqalign{
& = \left( {2\pi \times \frac{{7R}}{4}} \right)cm \cr
& = \frac{{7\pi R}}{2}\,cm \cr} $$
Increase in circumference :
$$\eqalign{
& = \left( {\frac{{7\pi R}}{2} - 2\pi R} \right)cm \cr
& = \frac{{3\pi R}}{2}\,cm \cr} $$
Increase % :
$$\eqalign{
& = \left( {\frac{{3\pi R}}{2} \times \frac{1}{{2\pi R}} \times 100} \right)\% \cr
& = 75\% \cr} $$

Answer: Option D. -> 12 cm
Ratio of sides = $$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$$   = 6 : 4 : 3
Perimeter = 52 cm
So, sides are :
$$\eqalign{
& \left( {52 \times \frac{6}{{13}}} \right)cm \cr
& \left( {52 \times \frac{4}{{13}}} \right)cm\,\& \cr
& \left( {52 \times \frac{3}{{13}}} \right)cm \cr} $$
So, a = 24 cm, b = 16 cm, and c = 12 cm
∴ Length of smallest side = 12 cm

Answer: Option B. -> Equal to 1
A square and a rhombus on the same base are equal in area.

Answer: Option A. -> 27
Distance covered in 4 sec :
$$\eqalign{
& = \left( {\frac{{30}}{7} \times 7} \right)m \cr
& = 30\,m \cr} $$
Distance covered in 1 sec = $$\frac{{30}}{4}$$
Distance covered in 1 revolution :
$$\eqalign{
& = \left( {\frac{{30}}{4}} \right)m \cr
& = \frac{{15}}{2}m \cr} $$
∴ Required speed :$$\eqalign{
& = \left( {\frac{{15}}{2}} \right)m/s \cr
& = \left( {\frac{{15}}{2} \times \frac{{18}}{5}} \right)km/hr \cr
& = 27\,km/hr \cr} $$

Answer: Option A. -> 10 m
Let the radius of the new park be R m
Then,
$$\eqalign{
& \pi {R^2} = \pi \times {8^2} + \pi \times {6^2} \cr
& \Rightarrow \pi {R^2} = 100\pi \cr
& \Rightarrow {R^2} = 100 \cr
& \Rightarrow R = 10 \cr} $$

Answer: Option D. -> $$\frac{{\sqrt 3 }}{3}{x^2}$$
Let the side of the triangle be a
Then,
$$\eqalign{
& {a^2} = {\left( {\frac{a}{2}} \right)^2} + {x^2} \cr
& \Leftrightarrow \frac{{3{a^2}}}{4} = {x^2} \cr
& \Leftrightarrow {a^2} = \frac{{4{x^2}}}{3} \cr} $$
∴ Area :
$$\eqalign{
& = \frac{{\sqrt 3 }}{4}{a^2} \cr
& = \frac{{\sqrt 3 }}{4} \times \frac{{4{x^2}}}{3} \cr
& = \frac{{{x^2}}}{{\sqrt 3 }} \cr
& = \frac{{\sqrt 3 {x^2}}}{3} \cr} $$

Answer: Option B. -> Rs. 1011.50
Area of 4 walls :
$$\eqalign{
& = 2\left( {l + b} \right) \times h \cr
& = \left[ {2\left( {12.5 + 9} \right) \times 7} \right]{m^2} \cr
& = 301\,{m^2} \cr} $$
Area of 2 doors and 4 windows :
$$\eqalign{
& = \left[ {2\left( {2.5 \times 1.2} \right) + 4\left( {1.5 \times 1} \right)} \right]{m^2} \cr
& = 12\,{m^2} \cr} $$
∴ Area to be painted :
$$\eqalign{
& = \left( {301 - 12} \right){m^2} \cr
& = 289\,{m^2} \cr} $$
Cost of painting :
$$\eqalign{
& = {\text{Rs}}{\text{.}}\left( {289 \times 3.50} \right) \cr
& = {\text{Rs}}{\text{. }}1011.50 \cr} $$

Answer: Option D. -> 3 : 8
Let the length of side of the square be a units
Then,
$$BE = EC = DF = FC = \frac{a}{2}$$
$$\eqalign{
& AE = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BE} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {{a^2} + {{\left( {\frac{a}{2}} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {{a^2} + \frac{{{a^2}}}{4}} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{5{a^2}}}{4}} \cr
& \,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 5 a}}{2} \cr} $$
Similarly, $$AF = \frac{{\sqrt 5 a}}{2}$$
$$\eqalign{
& EF = \sqrt {{{\left( {CE} \right)}^2} + {{\left( {CF} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{a}{2}} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{2{a^2}}}{4}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \frac{a}{{\sqrt 2 }} \cr
& EX = \frac{1}{2}EF = \frac{a}{{2\sqrt 2 }} \cr
& AX = \sqrt {{{\left( {AE} \right)}^2} - {{\left( {EX} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {\frac{{\sqrt 5 a}}{2}} \right)}^2} - {{\left( {\frac{a}{{2\sqrt 2 }}} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{5{a^2}}}{4} - \frac{{{a^2}}}{8}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{9{a^2}}}{8}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \frac{{3a}}{{2\sqrt 2 }} \cr} $$
$$\eqalign{
& \therefore \,{\text{Area of }}\left( {\vartriangle AEF} \right): \cr
& = \frac{1}{2} \times EF \times AX \cr
& = \frac{1}{2} \times \frac{a}{{\sqrt 2 }} \times \frac{{3a}}{{2\sqrt 2 }} \cr
& = \frac{{3{a^2}}}{8} \cr} $$
$$\eqalign{
& {\text{Required ratio :}} \cr
& = \frac{{3{a^2}}}{8}:{a^2} \cr
& = 3:8 \cr} $$

Answer: Option C. -> 41 cm
Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively.
Then,
$$\eqalign{
& x + \left( {x + 1} \right) + \left( {x + 2} \right) = 120 \cr
& \Rightarrow 3x + 3 = 120 \cr
& \Rightarrow 3x = 117 \cr
& \Rightarrow x = 39 \cr} $$
∴ Length of greatest side :
= (39 + 2) cm
= 41 cm

Answer: Option A. -> 64 cm
Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively.
Then,
$$\eqalign{
& 9x \times 7x = 252 \cr
& \Rightarrow 63{x^2} = 252 \cr
& \Rightarrow {x^2} = 4 \cr
& \Rightarrow x = 2 \cr} $$
So, length = 18 cm, breadth = 14 cm
∴ Perimeter :
= 2(18 + 14) cm
= 64 cm