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.

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.

Condition is given, here a square and a rectangular field have the same perimeter.
Given, side of a square field = 60 m
∴ Perimeter of square field = 4 Side = 4\(\times\) 60m = 240 m
and area of square field = (side)²= (60)² = 60m\(\times\) 60m
 = 3600 m²
Also, given lengrh of rectangular field = 80 m
Let breadth of rectangular field be x m.
According to the question,
Perimeter of square = Perimeter of rectangle
Perimeter of rectangular field = 240 m
Also, perimeter of a rectangular field = 2 (Length + Breadth)
∴  2(80 + x) = 240 ⇒ (80+x) = \(\frac { 240 }{ 2 } \)
⇒  80 + x = 120
⇒ x = 120- 80 ⇒ x = 40
Now, area of rectangular field = Length \(\times\) Breadth
= 80 m x 40m = 3200 m²
Here, it is clear that, 3600 > 3200
∴  Area of square field> Area of rectangular field
Hence, square field has a larger area.

Given, length = 18 m, breadth = 14 m and height = 7 m
∴ Volume of the cuboid = Length\(\times\)Breadth\(\times\)Height
 = 18\(\times\)14\(\times\)7=1764m³
Surface area of the cuboid = 2 (lb + bh + lh)
= 2(18\(\times\)14 + 14\(\times\)7 + 18\(\times\)7)
= 2(252 + 98 + 126)=2\(\times\)476 = 952m²

Given, side of the cube = 4 cm
∴ Volume of the cube = (Side)³ = (4)³
 = 4\(\times\)4\(\times\)4 =64 cm³

10 m

Given, one of the parallel sides of the trapezium, a = 15 m
Its height (h)= 10 m
Let another side be b m and then, area of trapezium = 450 m²
Area of trapezium = h(a+b)
⇒ \(450=\frac { 1 }{ 2 } \times 10\times (15=b)\)  ⇒\(\frac { 450\times 2 }{ 10 } \)=15+b    ⇒ 90=15+b
∴ b =90-15 =75
Hence, the other parallel side of trapezium is 75 m.

180 cm²

Given,diameter of base of cylindrical container = 14 cm
∴ Radius of base of cylindrical container, r =\(\frac { 14 }{ 7 } \) =7cm and height of cylindrical container = 20 cm
Now, label is placed 2 ern from top and bottom.
∴  Height of label, h = (20 - 2 - 2) = 16 cm
Now, area of label = Cured surface area of cylindrical portion of which label is placed
 =   \(2\pi rh=2\times \frac { 22 }{ 7 } \times 7\times 16=704\quad cm^{ 2 }\)  [∵ radius = 7 cm]
Hence, area of the label is 704 cm²

Let the side of the first cube be a and other be b.
∴  \(\frac { { a }^{ 3 } }{ b^{ 3 } } =\frac { 1 }{ 64 } \Rightarrow \frac { { a }^{ 3 } }{ b^{ 3 } } =\left( \frac { 1 }{ 4 } \right) ^{ 3 }\)
⇒ \(\frac { a }{ b } =\frac { 1 }{ 4 } \Rightarrow a=\frac { b }{ 4 } \)
∴ Area of face of first cube = a² = \(\left( \frac { b }{ 4 } \right) ^{ 2 }=\frac { { b }^{ 2 } }{ 16 } \)
and area of face of other cube = b²
∴ Required ratio =\(​​\frac { { b }^{ 2 } }{ 16 } \)= b² =1:16

Opposite faces of a cuboid are equal in area

Volume of a cylinder with radius r and height h is \(\pi\)r²h