Given, circumference of a circle = 31.4 ern
Let r be the radius of the circle.
\(\Rightarrow\)   2\(\pi\)r = 31.4  \(\Rightarrow\) 2 x 3.14r = 31.4
\(\Rightarrow r=\frac{31.4\times 100}{2\times 31.4\times 10}=\frac{10}{2}\)= 5 cm.
\(\therefore\) Area of the circle = \(\pi\)r² = 3.14 x (5)²
                             = 3.14 x 25 = 78.5 cm²
Hence, the radius and the area of the circle are 5 cm and 78.5 cm², respectively.

Given, diameter of the circular table top = 1.6 m
Then, radius of the circular table top = \(\frac{1.6}{2}\) = 0.8 m
\(\therefore\)  Area of the circular table top = \(\pi\)r² = 3.14 x (0.8)2
                                              = 3.14 x 0.8 x 0.8 = 2.0096 m²
Now, cost of polishing 1 sq m = Rs 15
\(\therefore\) Cost of polishing 2.0096 m² = Rs (15 x 2.0096) = Rs 30.144
                                             = Rs 30.14 (approx.)
Hence, the cost of polishing the circular tabletop at the rate of Rs 15 per sq m² is Rs 30.14 (approx.)

Draw two circles of radii 1 em and 2 cm on the graph paper.
By using the graph paper,
For smaller circle,
Number of more than half-filled squares = 4
\(\therefore\) Area = 4 squares (approx.)
                     [for more than half squares, estimate area = 1]
For larger circle,
Number of full-filled squares = 4
Number of more than half-filled squares = 8
Number of less than half filled squares = 4
\(\therefore\) Required area = 4 + 8 + 0 = 12 sq units (approx.)
                   [for less than filled squares, estimate area = 0)
Hence, by using formula, we get more accurate area.
By using the formula,
Area of smaller Circle = \(\pi r^2 = \frac{22}{7}\times (1)^2\)
                               = \(\frac{22}{7}\) = 3.14 cm² (approx..)
an area of larger Circle = \(\pi R^2 = \frac{22}{7}\times (2)^2\)
                                 = \(\frac{22}{7}\times 4=\frac{88}{7}\)
                                 = 12.57 cm² (approx....)

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.

Here, base = 8 em and height = 2.5 cm
\(\therefore\) Area of the parallelogram = Base x Height
                                         = 8 x 2.5 = 20 cm².

Here, base = 8 cm and height = 3.5 cm
\(\therefore\) Area of the parallelogram = Base X Height
                                        = 8 x 3.5 = 28 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.

44%

Perimeter of the rectangle
= 2\(\times\)(l+b)
⇒ 130 = 2\(\times\)(l + 30)
⇒ l+30 = \(\frac { 130 }{ 2 } \)
⇒ l+30= 65
⇒ l = 65 - 30 = 35 cm
Hence, the length of the rectangle is 35 cm.
Area of the rectangle
= l\(\times\)b
= 35\(\times\)30 = 1050 cm²
Hence, the area of the rectangle is 1050 cm²

84 cm

\(\therefore\) Area of rectangle = I x b and area of circle = \(\pi\)r²
\(\therefore\) Area of total rectangular land = 15 m x 10 m
                                              =150 m²
Area of land covered by plants = 9 m x 1 m = 9 m²
Area of land covered by camel = 5 m x 3 m = 15 m²
Region of land covered by ox is circular area.
So, the diameter is given, d = 2.8 m
\(\therefore\)  Radius = \(\frac{d}{2}=\frac{2.8}{2}\) = 1.4 m.
\(\therefore\) Area of land covered by ox = \(\pi\)r²
                                           = \(\frac{22}{7}\) x 14 x 14 = 616 m²
Total area covered by plants, camel and ox
                   = 9 + 15 + 6.16 = 30.16 m²
Remaining land for living = Total area - Area covered by plants and animals
                                    = (150 - 30.16 ) m² = 119.84 m²
\(\therefore\) Ratio of areas kept for animals and plants to the living area
                      = 30.16: 119.84 = 3016: 11984 = 377: 1498
The value depicted here is that we should save our environment and balance environment.

Perimeter of an equilateral triangle = 9 cm
Side of an equilateral triangle = \(\frac{9}{3}\) = 3 cm.
\(\therefore\)  Area of an equilateral triangle = \(\frac{\sqrt{3}}{a}a^2\)
So area = \(\frac{\sqrt{3}}{a}\times (3)^2 =\frac{\sqrt{3}}{4}\times 9=\frac{9\times 1.73}{4}=\frac{15.57}{4}\)
            = 3.89 cm²
So, the area of triangle is 3.89 cm².