\(\therefore\)   Area of a rectangle = Length x Width
\(\Rightarrow\)   600 = 30 x Width   \(\Rightarrow\) Width = \(\frac{600}{30}=\frac{60}{3}\) = 20 m.

Circumference of the circle = 44 cm [given]
\(\therefore\) Circumference of a circle = 2\(\pi r\)
\(\therefore 2\pi r=44\Rightarrow \pi r =\frac{44}{2}=22\Rightarrow \pi r =22\Rightarrow r=22\times \frac{7}{22}\) = 7 cm.
Then, area of the Circle = \(\pi r^2=\frac{22}{7}\times 7\times 7\) = 154 cm².

Given, base of a triangle = 5 cm
and height of a triangle = 3.2 cm 
\(\therefore\)  Area of a triangle = \(\frac{1}{2}\times\) Base \(\times\) Height
                              = \(\frac{1}{2}\times\) 5 \(\times\) 3.2 = \(\frac{1}{2}\times\) 16 = 8 cm².

Given, base of the parallelogram ABCD, AB = 7.2 em
and length of perpendieular from C on AB = 4.5 em
\(\therefore\) Area of the parallelogram = Base x Height
                                         = 7.2 x 4.5 = 32.4 cm².

Given, area of a triangle is 16 cm 2 in which AB = 2x, BC = x,  \(\Rightarrow\) Area of a triangle = \(\frac{1}{2}\) x BC x AC
\(\Rightarrow\)  16 = \(\frac{1}{2}\times x\times2x\Rightarrow 16=x^2\Rightarrow x=4\)
So, BC = 4 cm, AB = 2 x 4 = 8 cm.

Area covered by swimming pool
=30 m x 20 m
= 600m².
Length of outer rectangle = (30 + 8 + 8) m = 46 m
and its breadth = (20 + 5 + 5) m = 30 m
So, the area of outer rectangle
= 46 m x 30 m = 1380 m²
Area of cemented path = Area of outer rectangle - Area of swimming pool
= (1380 - 600) m² = 780 m²
Cost of cementing 1 m² path = Rs. 200
So, total cost of cementing the path
= 780 x Rs.200
= Rs.1,56,000

Given, ABCD is a quadrilateral in which AC = 22 cm, BM = 3 cm, DN = 3 cm and BM \(\bot\) AC, DN \(\bot\) AC.
\(\therefore\)  Area of  \(\Delta\)ABC = \(\frac{1}{2}\times\) AC x BM          [\(\therefore\) Area of triangle = \(\frac{1}{2}\times\)Base x Height]
                            = \(\frac{1}{2}\times\)22 x 3 = \(\frac{66}{2}\) cm² = 33 cm²
and area of \(\Delta\)ABD = \(\frac{1}{2}\times\)AB x DN = \(\frac{1}{2}\times\)22 x 3
                            = \(\frac{66}{2}\) cm² = 33 cm²
Area of quadrilateral ABCD = Area of \(\Delta\)ABC + Area of \(\Delta\)ACD
                                       = 33 + 33 = 66 cm²
Hence, the area of quadrilateral ABCD is 66 cm².

Let the radii of the given circles be r₁ and r₂. Let their circumference be 'C₁' and 'C₂' respectively.
\(\therefore\)C₁ = 2\(\pi\)r₁ and C₂ = 2\(\pi\)r₂
Since, C₁ : C₂ = 5 : 6
\(\therefore\)(2\(\pi\)r₁):(2\(\pi\)r₂)=5:6
\(\Rightarrow\)\(\frac { 2\pi r_{ 1 } }{ 2\pi r_{ 2 } } =\frac { 5 }{ 6 } \Rightarrow \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 5 }{ 6 } \)
Now, let the area of the given circles be A₁ and A₂
\(\therefore\)A₁ = \(\pi\)r₁² and A₂ = \(\pi\)r₂²
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\frac { { { \pi r }_{ 1 } }^{ 2 } }{ { { \pi r }_{ 2 } }^{ 2 } } =\frac { { { r }_{ 1 } }^{ 2 } }{ { { r }_{ 2 } }^{ 2 } } \Rightarrow \frac { A_{ 1 } }{ A_{ 2 } } =\left( \frac { { { r }_{ 1 } } }{ { { r }_{ 2 } } } \right) ^{ 2 }\)
\(\Rightarrow\)\(\frac { { A }_{ 1 } }{ A_{ 2 } } =\left( \frac { 5 }{ 6 } \right) ^{ 2 }=\frac { 25 }{ 36 } \Rightarrow \)A₁:A₂
= 25:36
Thus, the areas of the two given circles are in the ratio 25 : 36.

ABCD is the rhombus such that its diagonals AC and BD intersect at O. Here BD = 12 cm.

Suppose AC = x cm.
Since, the area of a rhombus =\(\frac{1}{2}\)\(\times\)[Product of 2 its diagonals]
\(\therefore\)Area of the rhombus ABCD =\(\frac{1}{2}\)\(\times\) 12 \(\times\) x cm²
2 But the area of the rhombus ABCD = 96 cm²
\(\Rightarrow\)\(\frac{1}{2}\)\(\times\)12 \(\times\) x = 96
\(\Rightarrow\)x=\(\frac{96\times2}{12}\)cm \(\Rightarrow\)x=8\(\times\)2cm
\(\Rightarrow\) x=16 cm
Since, the diagonals of a rhombus, bisect each other at right angles.
\(\therefore\)\(\angle \)COD = 90°, OC =\(\frac{1}{2}\)\(\times\) 16 cm=8 cm
and Od=\(\frac{1}{2}\)\(\times\) 12 cm =6 cm
Now, in right \(\Delta\)COD, we have
\(\Rightarrow\)OD² + OC² = CD² \(\Rightarrow\) 6² + 8² = CD²
\(\Rightarrow\)36 + 64 = CD² \(\Rightarrow\)100 = CD²
\(\Rightarrow\) 10² = CD² \(\Rightarrow\) CD = 10 cm.
Since, all the sides of rhombus are equal and perimeter of a rhombus = 4 x Side.
\(\therefore\) Perimeter ofthe rhombus = 4 x 10 em = 40 cm

Here, Length of the rectangular park = 70 m
Breadth of the rectangular park = 45 m
The cross paths are shown by EFGH and PQRS in the figure.
Now, PQ =3 m and PS = 45 m
EH = 3 m and EF = 70 m
KL = 3 m and KN = 3 m
Now, area of the path = [Area of rectangle PQRS] + [Area of rectangle EFGH]- [Area of square KLMN]
= [PS x PQ] + [EF x EH] - [KL x KN]
= [45 x 3] m² + [70 x 3] m² - [3 x 3] m²
= 135 m² + 210 m² - 9 m² = 336 m²
Now, cost of constructing the path = Rs. 110 x 336 = Rs. 36960.
Note: While finding the area of cross roads, the area of the middle square (here, KLMN) is taken twice
which is to be subtracted once.