Solution:
We have :
$$lb$$ = 60 and $$\sqrt {{l^2} + {b^2}} + l = 5b$$
Now,
$$\eqalign{ {l^2} + {b^2} = {\left( {5b - l} \right)^2} \cr \Rightarrow 24{b^2} - 10lb = 0 \cr \Rightarrow 24{b^2} - 600 = 0 \cr \Rightarrow {b^2} = 25 \cr \Rightarrow b = 5 \cr \therefore l = \left( {\frac{{60}}{b}} \right) \cr \,\,\,\,\,\,\,\, = \left( {\frac{{60}}{5}} \right)m \cr \,\,\,\,\,\,\,\, = 12\,m \cr} $$
So, length of the carpet = 12 m

Solution:
Side of the square :
$$\eqalign{ = \frac{{120}}{4}cm \cr = 30\,cm \cr} $$
Radius of the required circle :
$$\eqalign{ = \left( {\frac{1}{2} \times 30} \right)cm \cr = 15\,cm \cr = \pi \times {r^2} \cr = \left[ {\frac{{22}}{7} \times {{\left( {15} \right)}^2}} \right]c{m^2} \cr} $$

Solution:
Let the radius of the ground be R metres.
Then,
$$\eqalign{ \Rightarrow 2\pi R = 88 \cr \Rightarrow R = \left( {\frac{{88 \times 7}}{{2 \times 22}}} \right) \cr \Rightarrow R = 14\,m \cr} $$
Area of land strip :
$$\eqalign{ = \pi \left[ {{{\left( {14} \right)}^2} - {{\left( {11} \right)}^2}} \right]{m^2} \cr = \left( {\frac{{22}}{7} \times 25 \times 3} \right){m^2} \cr = \left( {\frac{{1650}}{7}} \right){m^2} \cr} $$
∴ Cost of levelling :
$$\eqalign{ = {\text{Rs}}{\text{.}}\left( {\frac{{1650}}{7} \times 7} \right) \cr = {\text{Rs}}{\text{. 1650}} \cr} $$

Solution:
$$\eqalign{ {\text{Area}}\,{\text{to}}\,{\text{be}}\,{\text{plastered}} \cr = \left[ {2\left( {l + b} \right) \times h} \right] + \left( {l \times b} \right) \cr = \left\{ {\left[ {2\left( {25 + 12} \right) \times 6} \right] + \left( {25 \times 12} \right)} \right\}{m^2} \cr = \left( {444 + 300} \right){m^2} \cr = 744{m^2} \cr \therefore {\text{Cost}}\,{\text{of}}\,{\text{Plastering}} \cr = Rs.\,\left( {744 \times \frac{{75}}{{100}}} \right) \cr = Rs.\,558 \cr} $$

Solution:
Area of the field :
= (25 × 15) m²
= 375 m²
Area of the passages :
= (25 × 2 + 15 × 2 - 2 × 2) m²
= 76 m²
Area under grass :
= (375 - 76) m²
= 299 m²

Solution:
Note : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Since, XY || BC, we have :
$$\eqalign{ \angle AXY = \angle ABC{\text{ and }} \cr \angle AYX = \angle ACB \cr {\text{Aslo, }}\angle A = \angle A\,\,{\text{(common)}} \cr {\text{So, }}\vartriangle AXY \sim \,\vartriangle ABC \cr {\text{Let area (}}\vartriangle ABC{\text{)}} = x\,\text{sq. units} \cr {\text{Then,}} \cr {\text{Area (}}\vartriangle AXY{\text{) = }}\frac{x}{2}\text{sq. units} \cr \frac{{{{\left( {AB} \right)}^2}}}{{{{\left( {AX} \right)}^2}}} = \frac{x}{{\left( {\frac{x}{2}} \right)}} \cr \Rightarrow \frac{{AB}}{{AX}} = \sqrt 2 \cr \Rightarrow \frac{{AX + BX}}{{AX}} = \sqrt 2 \cr \Rightarrow 1 + \frac{{BX}}{{AX}} = \sqrt 2 \cr \Rightarrow \frac{{BX}}{{AX}} = \left( {\sqrt 2 - 1} \right) \cr \Rightarrow \frac{{AX}}{{BX}} = \frac{1}{{\left( {\sqrt 2 - 1} \right)}}{\text{ Or 1}}:\left( {\sqrt 2 - 1} \right) \cr} $$

Solution:
Radius of incircle $$ = \frac{a}{{2\sqrt 3 }}$$
Area of incircle $$ = \left( {\frac{{\pi \times {a^2}}}{{12}}} \right)c{m^2}$$
$$\eqalign{ \therefore \frac{{\pi {a^2}}}{{12}} = 154 \cr \Rightarrow {a^2} = \frac{{154 \times 12 \times 7}}{{22}} \cr \Rightarrow a = 14\sqrt 3 \cr} $$
∴ Perimeter of the triangle :
$$\eqalign{ = \left( {3 \times 14\sqrt 3 } \right)cm \cr = \left( {42 \times 1.732} \right)cm \cr = 72.7\,cm\,(\text{approx}) \cr} $$