Since both breadth and height are given in cm, it is necessary to convert the length also in cm.
So we get, l = 120 mm = \(\frac{120}{10}\) = 12cm and take b = 10 cm, h = 8 cm as such.
Volume of a cuboid = l × b × h

= 12 ×10 × 8
= 960 cm³
Volume of 10 such cuboids= 10 × 960
= 9600 cm³

Side of a cube = 7 cm
Now length of the resulting cuboid (l) = 7+7 =14 cm
Breadth (b) = 7 cm, Height (h) = 7 cm
So, Total Surface Area = 2(lb + bh + lh)
= 2[(14 \(\times\) 7)+(7 \(\times\) 7)+(14 \(\times\) 7)]

= 2(98 + 49 + 98)
= 2 × 245
= 490 cm²
Lateral Surface Area = 2(l + b) × h
= 2(14 + 7) × 7 = 2 × 21× 7
= 294 cm²

Here, Total Surface Area of the cube = 486 cm²
6a² = 486 \(\Rightarrow\) a² = \(\frac{486}{6}\) and so, a² = 81 . This gives a = 9.
The side of the cube = 9 cm
Lateral Surface Area = 4a² = 4 × 9² = 4 × 81 = 324 cm²

The side of the cube (a) = 5 cm
Total Surface Area = 6a² = 6(5²) = 150 sq. cm
Lateral Surface Area = 4a² = 4(5²) = 100 sq. cm

Here, length (l) = 25 m, breadth (b) =15 m, height (h) = 5 m.
Area of four walls = LSA of cuboid
= 2(l + b) × h
= 2(25 +15) × 5
= 80 × 5 = 400 m²
Area of the floor = l × b

= 25 ×15
= 375 m²
Total renovating area of the hall = (Area of four walls + Area of the floor) = (400 + 375) m² = 775 m²
Therefore, cost of renovating at the rate of Rs.80 per m² = 80 × 775
= Rs. 62,000

Here, length (l) = 6 m, breadth (b) = 1.5 m, height (h) =\(\frac{300}{100}\)m = 3m
The wooden box is in the shape of cuboid.
The painting area of the wooden box = Total Surface Area of cuboid = 2(lb + bh + lh)

= 2(6 ×1.5 +1.5 × 3 + 6 × 3)
= 2(9 + 4.5 +18) = 2 × 31.5
= 63m²
Given that cost of painting of 1 m² is Rs. 50
The cost of painting area for 63 m² = 50 × 63 = Rs. 3150.

Given the dimensions of the cuboid;
that is length (l) = 7.5 m, breadth (b) = 3 m and height (h) = 5 m.
TSA = 2(lb + bh + lh)

= 2[(7.5 × 3) + (3 × 5) + (7.5 × 5)]
= 2(22.5 +15 + 37.5)
= 2 × 75
= 150 m²
LSA = 2(l + b) × h
= 2(7.5 + 3) × 5
= 2 ×10.5 × 5
= 105 m²

Given the sides are in the ratio 9:10:11, let the sides be 9k, 10k, 11k
The perimeter of the triangular park = 300 m
9k +10k +11k = 300m
30k = 300
k = 10m
Therefore, the sides are a = 90 m, b = 100 m, c = 110 m
s=\(\frac{1+b+c}{2}=\frac{90+100+110}{2}=\frac{300}{2}\)= 150 m
Hence, Area of triangular park = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 150\times (150-90)(150​​-100)(150-110) } \)
= \(\sqrt { 150\times 60​​\times 50\times 40 } \)
= \(\sqrt { 3\times 50\times 20​​\times 3\times 50\times 2\times 20 } \)
= 50 \(\times\) 20 \(\times\) 3\(\sqrt{2}\)
= 3000 \(\times\)1.414 = 4242 m²

For clarity, let us draw a rough figure indicating the measurements:

(i) The semi-perimeter of Fig.1, s \(\frac{30+40+50}{2}\)= 60 m
Fig. 2, s = \(\frac{35+40+45}{2}\) = 60 m
Fig. 3, s = \(\frac{40+40+40}{2}\) = 60 m
Note that all the semi-perimeters are equal.
(ii) Area of triangle using Heron’s formula:
In fig. 1, Area of triangle = \(\sqrt { 60(60-30)(60-40)(60-50) } \)
= \(\sqrt { 60\times 30​​\times 20\times 10 } \)
= \(\sqrt { 30\times 2​​\times 30\times 2\times 10\times 10 } \)
= 600 m²
In fig. 2, Area of triangle = \(\sqrt { 60(60-35)(60-40)(60-45) } \)
= \(\sqrt { 60\times 25\times 20\times 15 } \)
= \(\sqrt { 20\times 3​​\times 5\times 5\times 20\times 3 \times 5 } \)
= 300\(\sqrt{5}\) (since \(\sqrt{5}\) = 2.236)
= 670.8 m²
In fig. 3, Area of triangle = \(\sqrt { 60(60-40)(60-40)(60-40) } \)
= \(\sqrt { 60\times 22\times 20\times 20 } \)
= \(\sqrt { 3​​\times 20\times 20\times 20\times 20 } \)
= 400\(\sqrt{3}\) (since \(\sqrt{3}\) = 1.732)
= 692 .8 m²
We find that though the perimeters are same, the areas of the three triangular plots are different. The area of triangle in fig. 3 is the greatest among these; the buyer can be suggested to choose this since it is more spacious.

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14​​\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.