Volume of a cuboid = Length \(\times\) Breadth \(\times\) Height
Here, Length of the cuboid = 50 cm
Breadth of the cuboid = 45 cm
Height of the cuboid = 30 cm
\(\therefore\) Volume of the cuboid = 50 x 45 x 30 cm³
Volume of the small cube = Side \(\times\) Side \(\times\) Side = 5 \(\times\) 5 \(\times\) 5 cm³
Let the number of the required small cubes be x
\(\therefore\)x\(\times\) (5 \(\times\) 5 \(\times\) 5) = 50 \(\times\) 45 \(\times\) 30
\(\Rightarrow x=\frac{50\times45\times30}{5\times5\times5}=10\times9\times6=540\)
Thus, 540 small cubes can be placed in the given cuboid.

A road roller is a cylinder such that Radius \(\frac{84}{2}\ cm=42\ cm\)
Length (h) = 1m = 100 cm
\(\because\) Lateral surface area of a cylinder = 2\(\pi\)rh
\(\therefore\) Lateral surface area of the road roller \(=2\times\frac{22}{7}\times 42\times 100\ cm^2\)
\(=2\times 22\times 6\times 100\ cm^2\)
\(\therefore\) Area of road levelled in 1 revolution = 26400 cm²
\(\Rightarrow\) Area of road levelled in 250 revolutions
\(=250\times 26400\ cm^2=\frac{250\times26400}{100\times100}m^2\)
\(=\frac{25\times264}{10}m^2=\frac{6600}{10}m^2=660 m^2\)

Here, Radius (r) = 7 cm
Height (h) = 5 cm
Since, total surface area of the closed cylinder = 2\(\pi\)r (r + h)
\(\therefore\) Total surface area of the cylindrical tank
\(=2\times\frac{22}{7}\times(7)\times(7+5)m^2\)
\(2\times22\times12 m^2=528\ m^2\)
Let length of the sheet required be'l' m
\(\therefore\) Area of the rectangular sheet = I x 10 m² 
\(\Rightarrow l\times 10=528\Rightarrow l=\frac{528}{10}m=52.8 m\)
Thus, the required length of the sheet = 52.8 m.​​​​​​​

The edge (side) of the given cube = 2 cm
Since, three such cubes are joined, then
Total length (/)= (2 + 2 + 2) cm = 6 cm
Breadth (b) = 2 cm
Height (h) = 2 cm
\(\therefore\) Total surface area of the resultant cuboid

= 2[lb + bh + hI]
= 2[6 x 2 + 2 x 2 + 2 x 6] cm²
= 2[12 + 4 + 12] cm²
= 2[28] cm² = 56 cm²
Thus, the required total surface area of the cuboid = 56 cm².

Here, side of the square = 6 cm
\(\therefore\) Area of the square = Side x Side
= 6 cm x 6 cm = 36 cm²
Since, [Area of the rhombus]
= [Area of the square]
\(\therefore\) Area of the rhombus = 36 cm²
One of the diagonal of the rhombus = 4 cm
Let the other diagonal of the rhombus be
\(\therefore\) Area of rhombus
\(\therefore\frac{1}{2}\times Product \ of\ the\ diagonals\)
\(\therefore\) Area of the rhombus \(=\frac{1}{2}\times 4\times d\)
Now, \(\frac{1}{2}\times 4\times d=36\Rightarrow d=\frac{36\times 2}{4}=18\)
Thus, the required length of the rhombus = 18 cm.

Let the length of the diagonal BD be x cm
\(\because\) Area of a quadrilateral  \(=\frac{1}{2}\times(diagonal)\times\) (Sum of the length of perpendiculars on the diagonal from the opposite vertices)
\(\therefore\) Area of the quadrilateral ABCD
\(=\frac{1}{2}\times BD\times(AP+CQ)\)
\(=\frac{1}{2}\times x\ cm\times(4.5\ cm+3.5\ cm)\)
\(=\frac{1}{2}\times x\times 8\ cm^2\)
Since area of the quadrilateral ABCD = 32 cm²
\(\therefore \frac{1}{2}x\times 8=32\Rightarrow x=\frac{32\times2}{8}\ cm=8\ cm\)
Thus, the required length of the diagonal BD = 8 cm.

Here, parallel sides are 25 cm and 15 cm.
Height (i.e. distance between the parallel sides) = 12 cm
\(\because\) Area of a trapezium ABCD
\(=\frac{1}{2}\times(Sum\ of\ parallel\ sides)\times \ Height\)
\(\therefore\) Area of the trapezium ABCD
\(=\frac{1}{2}\times(25\ cm+15\ cm)\times12\ cm\)
\(=\frac{1}{2}\times40\times12\ cm^2=240\ cm^2\)
Thus, the required area of the trapezium ABCD = 240 cm².

Let ABCDE be a given pentagonal shaped park.
Given, length of each side of regular peatagon = 15 m
Now, Jyoti divide it into two congruent trapeziums by joining DP, where P is the mid-point of AB. Here, AE and DP are parallel sides and AP is perpendicular distance.
Given, length of DP = 30 m
and length of AP = \(\frac { AB }{ 2 } =\frac { 15 }{ 2 } m\)
∴ Area of pentagonal park ABCDE
 = 2 \(\times\)Area of trapezium APDE
[∵ both trapeziums are congruent, so their area will be same]
= 2\(\times\) [\(\frac { 1 }{ 2 } \)\(\times\)sum of parallel sides \(\times\) perpendicular distance between parallel sides]
\(2\times \left[ \frac { 1 }{ 2 } \times (30+15)\times \frac { 15 }{ 2 } \right] =2\times \frac { 1 }{ 2 } \times 45\times \frac { 15 }{ 2 } \)
\(\frac { 45\times 15 }{ 2 } =\frac { 675 }{ 2 } \) = 337.5 m²
Now, Kavita divide it into two parts by joining EC our of which one is ΔDEC and other is a square ABCE

Then, altitude of ΔDEC = DQ = 30 m-15 m = 15 m
∴ Area of pentagon ABCDE
= Area of ΔDEC + Area of square ABCE
= \(\frac { 1 }{ 2 } \)\(\times\)EC \(\times\)DQ+AB \(\times\)BC
=\(\frac { 1 }{ 2 } \) \(\times\)15 \(\times\)15 +15\(\times\) 15
=  \(\frac { 225 }{ 2 } +225=\frac { 225+450 }{ 2 } =\frac { 675 }{ 2 } \) =337.5 m²
Hence,area of park using both ways is serve, i.e. 337.5 m²
Yes,we can find the area of pentagonal park to two another ways by drawing the figure given below and then find its area:
 

(a)Here, the food-piece is in the shape of semi-circle and diameter of semi-circle = 2.8 cm
∴ Radius of semi circle = \(\frac { Diameter }{ 2 } =\frac { 2.8 }{ 2 } \) = 1.4cm
Perimeter of a semi- circle = \(\frac { 2\pi r }{ 2 } =\pi r\)
∴ Perimeter of given food-piece =\(\pi\) r+ Diameter of semi circle
=  \(\left( \frac { 22 }{ 7 } \times 1.4+2.8 \right) \)cm=(4.4+2.8)cm =7.2 cm
(b)Here, given piece of food has semi-circular shape in one side.
So perimeter od semi-circular part 
\(\pi r=\frac { 22 }{ 7 } \times 1.4\) =4.4 cm         \(\left[ \because radius\quad =\frac { 2.8 }{ 2 } =1.4cm \right] \)
∴  Perimeter of given food piece 
= 4.4 cm + 1.5 cm + 2.8 cm + 1.5 cm = 10.2 cm
(c)Here, given piece of food has semi-circular shape on the upper side.
So, perimeter of semi-circular part = \(\pi\)r
= \(\frac { 22 }{ 7 } \times 1.4\) =4.4 cm
∴  Perimeter of given food-piece
= 4.4 cm + 2 cm + 2 cm = 8.4 cm
Now 10.2> 8.4> 7.2
So, it is clear that, the ant would has to take a longer round for food-piece in figure (b), since it has larger perimeter.

Given, the shape of a garden is rectangular in middle and semi-circular at both ends.
Also, diameter of circular part = 7 m
∴ Radius of circular part = \(\frac { 7 }{ 2 } \)= 3.5 m
Then, length of rectangular part = 20m -(3.5+3.5)m and breadth of rectangular part = 7 m
Now,area of one semi-circular part = \(\frac { 1 }{ 2 } { \pi r }^{ 2 }\)
where, r is the radius of semi-circle.
∴ Area of both semi-circular parts = \(2\times \frac { 1 }{ 2 } { \pi r }^{ 2 }={ \pi r }^{ 2 }\)
 =  \(\frac { 22 }{ 7 } \times \left( \frac { 7 }{ 2 } \right) ^{ 2 }\)    \(\left[ \because \pi =\frac { 22 }{ 7 } \right] \)
= \(\frac { 22 }{ 7 } \times \frac { 49 }{ 7 } =\frac { 77 }{ 2 } \) m²
Area of rectangular part = Length \(\times\) Breadth
= 13m\(\times\) 7m = 91m²
∴ Area of garden = Area of both circular parts + Area of rectangular part
=\(\frac { 77 }{ 2 } \) + 91 m²
=\(\left( \frac { 77+182 }{ 2 } \right) { m }^{ 2 }\)= 129.5 m²
Now, Perimeter of one semi-circle part = \(\frac { 2\pi r }{ 2 } =\pi r\)
∴ Perimeter of both semi-circle parts = \(2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \) =22m
Here, rectangular part is in between two semi-circular parts, so for perimeter of garden, we will take only length of rectangular part.
∴  Perimeter of garden = Perimeter of both circular parts + 22 \(\times\) Length of rectangular part
= 22 + 2 \(\times\)13 = (22+26) m = 48m