Here, the circumference = 88 em.
Let radius of the circle be 'r',
\(\therefore\)         2\(\pi\)r=88.
or  2\(\times\)\(\frac{22}{7}\)\(\times\)r=88
or  r=\(\frac{88\times7}{2\times22}\)=14 cm
Now,   area of the circle =\(\pi\)r²=\(\frac{22}{7}\)\(\times\)14\(\times\)14 cm²=22\(\times\)2\(\times\)14 cm²
=616 cm²​​
Thus, the area of the circle is 616 cm².

Here, radius (r) = 14 cm
\(\because\)Circumference of circle =2\(\pi\)r
\(\therefore\) Circumference of the semi-circular disc =\(\frac{1}{2}\)(2\(\pi\)r)=\(\pi\)r
=\(\frac{22}{7}\)\(\times\)14== 44 cm
Since,   diameter = 2 x r
= 2 x 14 em = 28 em
\(\therefore\)Perimeter of the semi-circular disc = 44 cm + 28 cm = 72 cm

DE = EA + AD = (8 + 5) cm = 13 cm
DE is the radius of the circle.
Also, DB is the radius of the circle.
Next, AC = DB [Since diagonals of a rectangle are equal in length]
Therefore AC = 13cm.
From \(\Delta\)ADC, DC² = AC²-AD² = 13^{2 -} 5²
= 169 - 25 = 144 = 122
So, DC = 12
Thus, length of DC is 12 cm.
Hence, perimeter of the rectangle ABCD
= 2(12 + 5) Cm = 34 cm.

Let the radius of the circle be r.
Then, 2\(\pi\)r = 33
i.e., r = \(\frac { 33 }{ 2\pi } \)=\(\frac { 33 }{ 2 } \) x \(\frac { 7 }{ 21 } \)=\(\frac { 21 }{ 4 } \)
Thus, radius is \(\frac { 21 }{ 4 } \)cm
So, area of the circle = \(\pi\)r² = \(\frac { 22 }{ 7 } \). \(\frac { 21 }{ 4 } \).\(\frac { 21 }{ 4 } \) = \(\frac { 693 }{ 8 } \)
Thus, area of the circle is \(\frac { 693 }{ 8 } \) cm² .

Height = 3 cm
Area of traingle = \(\frac { 1 }{ 2 } \)bh
36 = \(\frac { 1 }{ 2 } \)bh
\(\Rightarrow\) 36 = \(\frac { 1 }{ 2 } \)xbx3
\(\Rightarrow\) 72 = b x 3
\(\Rightarrow\) \(\frac { 75 }{ 3 } \) = b
\(\Rightarrow\) b = 24 cm
\(\therefore\) Base is 24 cm.

Let ABCD be the rectangular park and the shaded region represents the 2.5 wide path to be constructed outside the park.

We have, 
PQ = (45 + 2.5 + 2.5) m = 50 m,
PS = (30 + 2.5 + 2.5) m = 35 m
Clearly,
Area of the path = Area of rectangle PQRS - Area of rectangle ABCD
= (50 x 35) m² - (45 x 30) m²
= 1750 m^{2 -} 1350 m² = 400 m²

\(\therefore\) Breadth of outer figure = 12 cm
Breadth of inner figure = 12 - 4 - 4 = 4 cm
Length of outer figure = 18 cm
Length of inner figure = 18 - 4 = 14 cm
So, area of inner rectangular figure
            = 14 x 4 = 56 cm².

Given, circumference of the circular sheet = 154 m
\(\therefore\) Circumference of a circular sheet = 2\(\pi\)r
\(\Rightarrow \)         154 = 2\(\pi\)r
\(\Rightarrow \)             r = \(\frac{154}{2\pi}=\frac{154\times 7}{2\times 22}=\frac{7\times 7}{2}=\frac{49}{2}\)m
\(\therefore\)  Diameter = 2 x Radius
                   = 2 x \(\frac{49}{2}\)
                   = 49 cm.

Given, MBC is isosceles with AB = AC = 75 cm and BC = 9 cm.
Let BC be base of triangle, then height (AD) = 6 cm
\(\therefore\) Area of MBC = \(\frac{1}{2}\times\) Base x Height
                       = \(\frac{1}{2}\times\) 9 x 6
                       = \(\frac{54}{2}\) = 27 cm² 
Now, let side AB = 75 cm be base.
Then, CE (the height from C to corresponding side AB) be the height.
\(\therefore\)  Area of  \(\Delta\)ABC = \(\frac{1}{2}\times\) x AB x CE
\(\therefore\)     \(\frac{1}{2}\times\) AB x CE = 27
\(\Rightarrow\)  \(\frac{1}{2}\times\) 7.5 x CE = 27
\(\therefore\)                   CE = \(\frac{27\times 2}{7.5}\) = 7.2 cm
Hence, the height from C to AB i.e. CE is 7.2 cm

Firstly, on counting the squares enclosed within the figures of the parallelogram, we find in each case these are 15 in numbers. So, the area of each parallelogram = 15 sq units, base = 5 units and height = 3 units in each case. Now, draw a perpendicular DM to the base AB or produced AB (if necessary) in each figure as shown below:

For figure (b), AD = \(\sqrt{MD^2+AM^2}\)
                         = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\) units
\(\therefore\) Perimeter = \(2(\sqrt{13}+5)\) units.
Hence, the required complete table is as follows:

Given, length of the door = 2 m and breadth of a door = 1m
Since door is in the shape of a rectangle
\(\therefore\)  Area of the door = Length x Breadth = 2 x 1= 2 m
Also, length of the wall = 4.5 m and breadth of the wall = 3.6 m
\(\therefore\)  Area of the wall including door = Length x Breadth = 4.5 x 3.6 = 16.2 m²
Now, area of the wall for whitewash = Area of the wall including door - Area of the door = 16.2 - 2 = 14.2 m²
\(\therefore\) Cost of whitewashing the wall at the rate of Rs 20 per m²
                                 = Rs (14.2 x 20) = Rs 284
Hence, the cost of whitewashing the wall is Rs 284.

Let b be the breadth of the rectangular park.
Given, side of a square park, a = 60 m and length of the rectangular park, I = 90 m
Now, area of a square park = (Side)² = (60)² = 3600 m²
According to the question,
Area of a rectangular park = Area of a square park
\(\Rightarrow\)  I x b = 3600 [\(\therefore\) area of a rectangular park = I x b]
\(\Rightarrow\) 90 x b = 3600  \(\Rightarrow\) b = \(\frac{3600}{90}\Rightarrow\) b = 40 m
Hence, the breadth of a rectangular park is 40 m.

Given, perimeter of a rectangular sheet = 100 cm and length of a rectangular sheet, I = 35 cm
Let b be the breadth of a rectangular sheet .
We know that, perimeter of a rectangular sheet = 2 (/ + b)
\(\Rightarrow\)  2(/ + b) = 100  \(\Rightarrow\) 2(35 + b) = 100
\(\Rightarrow\) \(35+b=\frac{100}{2}\)    [dividing by 2 on both sides]
\(\Rightarrow\) 35 + b = 50  \(\Rightarrow\) b = 50 - 35  \(\Rightarrow\) b = 15 cm
\(\therefore\) Area of a rectangular sheet = I X b = 35 x 15 = 525 cm²
Hence, the breadth and area of a rectangular sheet are 15 cm and 525 cm², respectively.

We know that, perimeter of a square park = 4 x Side
Given,  perimeter of a square park = 320 m
Now,  4 x Side = 320
\(\Rightarrow\)        \(side=\frac{320}{4}=80m\)
Area of a square park = Side x Side = 80 x 80
                                                     = 6400 m²
Hence, the area of a square park is 6400 m².

Given, length of a rectangular piece of land, I = 500 m and breadth of a rectangular piece of land, b = 300 m
Cost of 1m² land = Rs 10000
\(\therefore\) Cost of 150000 m² land = Rs 10000 x 150000
                                       = Rs 1500000000
Hence, the cost of 150000 m² land at the rate of 10000 per m² is Rs 1500000000.