600 cm²

Given, base of a triangle = 3 cm
and height of a triangle = 2 cm 
\(\therefore\)  Area of a triangle = \(\frac{1}{2}\times\) Base \(\times\) Height
                              = \(\frac{1}{2}\times\) 3 \(\times\) 2 = \(\frac{1}{2}\times\) 6 = 3 cm².

Given, base of a parallelogram = 5 cm
and height of a parallelogram = 3 cm
\(\therefore\) Area of a parallelogram = Base x Height
                                     = 5 x 3 = 15 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, length of rectangle = 6 cm
and breadth of rectangle = 4 cm
\(\therefore\)  Area of a rectangle = Length X Breadth = 6 x 4 = 24 cm.
From the given figure, it is clear that rectangle is divided into 4 congruent polygons.
\(\therefore\)  Area of each polygon = \(\frac{1}{4}\times\) Area of a rectangle
                                    = \(\frac{1}{4}\times\) 24 = 6 cm².

2500 m.

420 cm²

We need to find the area of rectangular swimming pool to cover it with plastic sheer.

We need to find the perimeter of a rectangular flower bed, which gives the required length of the wire required to fence a rectangular flower bed.

Given, length = 15 cm and breadth = 8 cm
\(\therefore\) Area of a rectangle = Length x Breadth = 15 x 8 = 120 cm²
Hence, area of rectangle is 120 cm².

Given, circumference of a circular sheet = 154 m
\(\Rightarrow 2\pi r=154\Rightarrow 2\times \frac{22}{7}\times r=154\Rightarrow r=\frac{154\times 7}{2\times 22}=\frac{49}{2}\)m
Now, area of the circular sheet = \(\pi r^2\)
    = \(\frac{22}{7}\times \frac{49}{2}\times \frac{49}{5}=\frac{154\times 49}{4}=\frac{7546}{4}\)= 1886.5 m².
Hence, the radius and area of the circular sheet are 49/2 m and 1886.5 m², respectively.

If base (PS) = 8 cm and height = QN, then
Area of the parallelogram = 91.2 cm²
                      [since, parallelogram is same in position]
\(\Rightarrow\)  PS x QN = 91.2  \(\Rightarrow\)  8 x QN = 91.2
\(\therefore\)  QN = \(\frac{91.2}{8}\) = 11.4 cm.

(a) Given, base of the parallelogram = 20 cm and area of the parallelogram = 246 cm²
We know that area of the parallelogram = Base x Height
\(\therefore\)  Base x Height = 246  \(\Rightarrow\) 20 x Height = 246
\(\Rightarrow\) Height = \(\frac{246}{20}\) = 12.3 cm
Hence, the height of the parallelogram is 12.3 cm.
(b) Given, height of the parallelogram = 15 cm and area of the parallelogram = 154.5 cm²
We know that, area of the parallelogram = Base x Height
\(\therefore\) Base x Height = 154.5  \(\Rightarrow\) Base x 15 = 154.5
\(\Rightarrow\)  Base = \(\frac{154.5}{15}\) = 10.3 cm
Hence, the base of the parallelogram is 10.3 cm.
(c) Given, height of the parallelogram = 8.4 cm and area of the parallelogram = 48.72 cm²
We know that, area of the parallelogram = Base x Height
\(\therefore\) Base x Height = 48.72  \(\Rightarrow\) Base x 8.4 = 48.72
\(\Rightarrow\)  Base = \(\frac{48.72}{8.4}\) = 5.8 cm
Hence, the base of the parallelogram is 5.8 cm.
(d) Given, base of the parallelogram = 15.6 cm and area of the parallelogram = 16.38 cm²
We know that, area of the parallelogram = Base x Height
\(\therefore\) Base x Height = 16.38  \(\Rightarrow\) 15.6 x Height = 16.38
\(\Rightarrow\)  Height = \(\frac{16.38}{15.6}\) = 1.05 cm
Hence, the height of the parallelogram is 1.05 cm.
On putting the missing values, the complete table is

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 (a), in \(\Delta\)AMD, by using Pythagoras theorem,
                  AD = \(\sqrt{MG^2+AM^2}\)
                        = \(\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\) units
\(\therefore\)  Perimeter = 2(AD + AB) = 2(\(\sqrt{10}\) + 5) units.
Hence, the required complete table is as follows:

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.

\(\therefore\) Side of square pizza = 45 cm
\(\therefore\) Area of a square pizza = (Side)²
                                     = (45)² = 2025 cm²
Diameter of circular pizza = 50 cm
Radius = \(\frac{50}{2}\) = 25 cm.
\(\therefore\) Area of a circle = \(\pi r^2\)
\(\therefore\) Area of the circular pizza = \(\pi r^2\)
                                         = \(\frac{22}{7}\times 25\times 25=\frac{22}{7}\times 625\)
                                         = \(\frac{13750}{7}\) = 1964.28 cm²
Price of 1 cm square pizaa  = \(\frac{2025}{150}\) = Rs 135
Price of 1 cm circular pizaa  = \(\frac{1964.29}{160}\) = Rs 12.27
Hence, the circular pizza is a better deal.

A wall of a room is of dimensions 5 m x 4 m.
Length of the room = 5 m
Breadth of the room = 4 m
\(\therefore\) Area of the room = 5 x 4= 20 m²
Length of the window = 1.5 m
Breadth of the window = 1 m
\(\therefore\) Area of the window = 1.5 x 1 = 1.5 m²
Length of the door = 2.25 m
Breadth of the door = 1 m
\(\therefore\) Area of the door = 2.25 x 1 = 2.25 m²
The area of the wall to be painted = Area of the room - Area of the window - Area of the door
                                                 = 20 - 1.5 - 2.25
                                                 = 20 - 3.75 = 16.25 m².