For box A,
Given, length of box = 20 cm
Breadth of box = 8 cm
and height of box = 3 cm
∴ Volume of the box = I \(\times\)b \(\times\)h = 20\(\times\)8\(\times\)3 cm³ =480 cm³
Total surface area of the box = 2(lb + bh + hl)
= 2(20\(\times\) 8 + 8\(\times\)3 + 3\(\times\)20) = 2\(\times\)(160+24+60) =2\(\times\)244 = 488 cm³
For box B,
Given, length of box = 12 cm
Breadth of box = 10 cm and height of box = 4 cm
∴ Volume of the box = I \(\times\) b \(\times\) h = 12 \(\times\) 10 \(\times\) 4 = 480 cm³
Total surface are of the box = 2 (lb + bh + hl)
 = 2(120 + 40 + 48) = 2\(\times\)208 = 416cm²
It is clear from above that both boxes have same volume, but the surface area of type B box is less than that of type A box.
Thus, material required for box B is less.
∴ Box B is more economical than box A.
Now, let another box of size be 12 cm\(\times\)8 crn\(\times\)5cm
Volume of this box = 12\(\times\)8\(\times\)5 = 480 cm³
and its surface area =2(lb + bh + hl) = 2(12\(\times\)8 + 8\(\times\)5 + 5\(\times\)12)
= 2(96 + 40 + 60) = 2\(\times\)196 = 392
Clearly, surface area of this box is less than that of box B
Hence, we can suggest any other box of size 12 cm\(\times\)8 cm \(\times\)5 crn, which has the same volume buts is more economical than the given boxes.

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.

Volume of godowns, reservoirs, etc., can be found by this method

A cylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is 20 cm.
Height of the cylinder = h = 14 cm
Radius = r = 20 cm
Volume of the cylinder = V = \(\pi\) r² h
\(=\frac{22}{7} \times 20 \times 20 \times 14=17600 \mathrm{~cm}^{3}\) 
Hence, the volume of the cylinder is 17600 cm³.

Given, length of cabinet (I) = 1m
Breadth of cabinet (b) = 2 m and height of cabinet (h) =1.5 m
∴ Total painted (surface) area of cabinet = Total surface area - Area of bottom
= 2(lb + bh + hI) - Ib = Ib + 2bh + 2hl
= (1\(\times\)2+2\(\times\)2\(\times\)1.5+2\(\times\)1.5\(\times\)1)=(2 + 6 + 3)m² = (2 + 6 + 3)m² = 11 m²
Hence, she covers 11 m² area of cabinet.

In a parallelogram, opposite sides are parallel and equal, so by drawing a diagonal, it can be divided into two congruent triangle but in a trapezium, only one pair of opposite sides is parallel, so it cannot be divided into two congruent triangles

180 cm²

1 cubic cm = 1 cm x 1 cm x 1 cm
= 1cm³
= 10 mm x 10 mm x 10 mm
= 1000 mm³
1 cubic m = 1 m x 1 m x 1 m = 1 m³
= 100cm x l00cm x l00cm
= 1000000 cm³
1 cubic mm = 1 mm x 1 mm x 1 mm
= 1 mm³
= 0.1 cm x 0.1 cm x 0.1 cm
=  0.001 cm³

40 m²

Diagonal LN = 8 cm
Perpendicular MX = 4.5 cm
Perpendicular OY = 3.5 cm
\(\therefore\) Sum of the perpendiculars = MX + OY
= 4.5 cm + 3.5 cm = 8 cm
\(\therefore\) Area of the quadrilateral \(=\frac{1}{2}\times(diagonal)\times\)(Sum of the lengths of perpendiculars on it from opposite vertices)
\(=\frac{1}{2}\times8\times8\ cm^2=32\ cm^2\)

Given,volume of the cylinder
1.54 m³= \(\frac { 154 }{ 100 } \times (100)^{ 3 }\) cm³  [∵ 1m³= (100)³ cm³]
and diameter of the base = 140 cm
∴ Radius of the base =\(\frac { 140 }{ 2 } \) =70 cm
Let height of the cylinder be h cm.
We know that, volume of the cylinder = \(\pi\)r²h
∴ 154\(\times\)(100)² =\(\frac { 22 }{ 7 } \) \(\times\)(70)²\(\times\)h
⇒  154\(\times\)(100)² =\(\frac { 22 }{ 7 } \) \(\times\)70\(\times\)70\(\times\)h
⇒   154\(\times\)(100)² = 22\(\times\)700\(\times\)h
⇒ h = \(\frac { 154\times (100)^{ 2 } }{ 22\times 700 } =\frac { 154\times 100\times 10 }{ 22\times 700 } \) = 100 cm
Hence,the height of the cylinder is 100 cm.

Here, the given trapezium is divided into three parts out of which two are triangles WLZ and XMY and one is a rectangle LMYZ.
∴ Area of trapezium WXYZ = Area of ΔWLZ + Area of rectangle LMYZ + Area of ΔXMY
= \(\left( \frac { 1 }{ 2 } \times c\times b \right) =(b\times h)+\left( \frac { 1 }{ 2 } \times d\times h \right) \)
 \(\left[ \because area\ of\ triangle\ =\frac { 1 }{ 2 } \times base\times height\ and\ area\ of\ rectangle\ =\ length\ \times breadth \right] \)
=\(\frac { 1 }{ 2 } \) (ch + db) + bh
=\(\frac { h }{ 2 } \) (c + d + 2b) = \(\frac { 1 }{ 2 } \)b{(c + d + b) + b}
=\(\frac { h }{ 2 } \) (a + b) [from the figure ,c + b + d = a]

Height = \(\frac { 80 }{ 20 } \) = 4 m

None: on cutting a cube into number of cubes, we have no change in volume of the cube. So, there will be no percentage increase in the volume.