Answer: Option B. -> 5 : 2
Let the radius of each of the circle and the semi-circle be r units
Diagonal of the first square = 2r units
Let the side of the second be a units
Then,
$$\eqalign{
& {r^2} = {a^2} + {\left( {\frac{a}{2}} \right)^2} \cr
& \Rightarrow {r^2} = \frac{{5{a^2}}}{4} \cr
& \Rightarrow {a^2} = \frac{{4{r^2}}}{5} \cr} $$
∴ Ratio of the areas of the two squares :
$$\eqalign{
& = \frac{{\frac{1}{2} \times {{\left( {2r} \right)}^2}}}{{{a^2}}} \cr
& = \frac{{2{r^2}}}{{\left( {\frac{{4{r^2}}}{5}} \right)}} = \frac{5}{2} \cr
& = \frac{5}{2} \cr
& = 5:2 \cr} $$

Answer: Option C. -> $$40\sqrt 2 \,$$ cm
$$\eqalign{
& d = \sqrt 2 \times l \cr
& \Rightarrow l = \frac{{20}}{{\sqrt 2 }} \cr} $$
∴ Perimeter :
$$\eqalign{
& = \left( {4l} \right)cm \cr
& = \left( {\frac{{4 \times 20}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}} \right)cm \cr
& = 40\sqrt 2 \,cm \cr} $$

Answer: Option E. -> None of these
$$\eqalign{
& QR = \sqrt {{{\left( {PR} \right)}^2} - {{\left( {PQ} \right)}^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{25}^2} + {3^2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {625 - 9} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {616} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {154} \,cm \cr} $$

Answer: Option A. -> $$2\sqrt 3 $$ cm
Let the length and breadth of the rectangle be $$l$$ cm and n cm respectively
Then,
$$\eqalign{
& 2\left( {l + b} \right) = 10 \cr
& \Rightarrow l + b = 5 \cr
& \Rightarrow b = \left( {5 - l} \right)cm \cr} $$
Area of the rectangle :
$$\eqalign{
& = l\left( {5 - l} \right)c{m^2} \cr
& = \left( {5l - {l^2}} \right)c{m^2} \cr} $$
Area of the square :
$$\eqalign{
& = 2\left( {5l - {l^2}} \right)c{m^2} \cr
& = \left( {10l - 2{l^2}} \right)c{m^2} \cr} $$
$$\eqalign{
& \therefore \left( {l + 1} \right)\left( {6 - l} \right) = \left( {10l - 2{l^2}} \right) \cr
& \Rightarrow {l^2} - 5l + 6 = 0 \cr
& \Rightarrow \left( {l - 3} \right)\left( {l - 2} \right) = 0 \cr
& \Rightarrow l = 3 \cr} $$
Area of the square :
$$\eqalign{
& = \left( {10 \times 3 - 2 \times 9} \right)c{m^2} \cr
& = 12\,c{m^2} \cr} $$
∴ Side of the square $$ = \sqrt {12} \,cm = 2\sqrt 3 $$

Answer: Option C. -> Rs. 6000
$$\eqalign{
& {\text{Area of the field}} = \left( {{\text{Base}} \times {\text{Height}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {150 \times 80} \right){m^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12000\,{m^2} \cr} $$
∴ Cost of watching :
$$\eqalign{
& = {\text{Rs}}{\text{. }}\left( {12000 \times 0.50} \right) \cr
& = {\text{Rs}}{\text{. 6000}} \cr} $$

Answer: Option A. -> 4 cm
Let the length of each of the sides containing the right angle be x cm
Then,
Hypotenuse :
$$\eqalign{
& = \sqrt {{x^2} + {x^2}} \,cm \cr
& = \sqrt {2{x^2}} \,cm \cr
& = \sqrt 2 x\,cm \cr} $$
Perimeter of the triangle :
$$\eqalign{
& = \left( {x + x + \sqrt 2 x} \right)cm \cr
& = \left( {2x + \sqrt 2 x} \right)cm \cr
& = \sqrt 2 x\left( {\sqrt 2 + 1} \right)cm \cr
& \therefore \sqrt 2 x\left( {\sqrt 2 + 1} \right) = \left( {4\sqrt 2 + 4} \right) \cr
& \Rightarrow \sqrt 2 x\left( {\sqrt 2 + 1} \right) = 4\left( {\sqrt 2 + 1} \right) \cr
& \Rightarrow \sqrt 2 x = 4 \cr
& \Rightarrow x = 2\sqrt 2 \cr} $$
Hence, hypotenuse :
$$\eqalign{
& = \left( {\sqrt 2 \times 2\sqrt 2 } \right)cm \cr
& = 4\,cm \cr} $$