Let ‘a’ be the side of cubical tank.
Here, volume of the tank = 64,000 litres
i.e., a³ = 64,000 = \(\frac{64000}{1000}\) [since,1000 litres = 1m³ ]
a³ = 64 m³
a =\(\sqrt [ 3 ]{ 64 } \) a = 4 m
Therefore, length of the side of the tank is 4 metres.

Given that side (a) = 10 cm
volume of the cube = a³
= 10 × 10 × 10
= 1000 cm³

Here, the dimensions of a sweet box are Length (l) = 22cm, breadth (b) = 18cm,
height (h) = 10 cm.
Volume of a sweet box = l × b × h
= 22 × 18 × 10 cm³
The dimensions of a carton are
Length (l) = 1m= 100 cm, breadth (b) = 88 cm,
height (h) = 63 cm.

Volume of the carton = l × b × h
= 100 × 88 × 63 cm³
The number of sweet boxes packed =\(\frac{volume \ of \ the \ carton}{volume\ of \ a \ sweet \ box}\)
= \(\frac{100 \times 88\times63}{22\times18\times10}\)
= 140 boxes

Length of the fish tank l =3.8 m
Breadth of the fish tank b =2.5 m ,
Height of the fish tank h =1.6 m
Volume of the fish tank = l × b × h
= 3.8 × 2.5 ×1.6
= 15.2 m³
= 15.2 ×1000 litres
= 15200 litres

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, 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

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.

V = 30 \(\times\) 25 \(\times\) 20 cm³
External box volume = 15000 cm³
Thickness of the wood = 2 cm all around
∴ Volume of the inner cuboid = 26 \(\times\) 21 \(\times\) 16 = 8736 cm³
The volume of the wooden part
Required Volume = Volume of the external box - Volume of the inner box
= 15000 - 8736
= 6264 cm³.

V = a³ = 125000 litres
= 125 m³
a = \(\sqrt [ 3 ]{ 125{ m }^{ 3 } } =\sqrt [ 3 ]{ 5\times 5\times 5 }\)
∴ length of its side = 5 m.

TSA of a cube 6a² = 726

a² = 121
a = 11 cm
Its volume = a³ = 11 \(\times\) 11 \(\times\) 11 cm³ = 1331 cm³.

V = l \(\times\) b \(\times\) h = 1440 m³

height = 12 m

I = 24 cm b = 12 cm h = 8 cm
Volume of the brick = I x b x h = (24 \(\times\) 12 \(\times\) 8) cm³= 2304 cm³
Wall dimensions are:
L = 20 m = 2000 cm
B = 48 cm
H = 6 m = 600 cm
No. of brieks required to build the wall = \(\frac { Volume\ of\ the\ wall }{ Volume\ of\ a\ bricks } \)

No. of brieks required = 25000

Let 1 ratio = x then 5:4:3
⇒  5x: 4x: 3x
5x \(\times\) 4x \(\times\) 3x = 60x³ = 7500 cm³

x = 5 cm
∴ 5x = 5 \(\times\) 5 = 25 cm
4x = 4 \(\times\) 5 = 20 cm
3x = 3 \(\times\) 5 = 15 cm
∴ Dimensions of a chocolate box are 25 cm \(\times\) 20 cm \(\times\) 15 cm.

a = 6.5 m
6a² = 6 \(\times\) 6.5 \(\times\) 6.5 = 253.5 m²
Area to be painted = 253.5 m²
Cost of painting 1 m² = Rs. 24
∴ Cost of painting 253.5 m²= 253.5 \(\times\)24 = Rs. 6084

Dimensions of a hall 10 m \(\times\) 9 m \(\times\) 8 m
l = 10 m
b = 9 m
h = 8 m
White washing to be done for the area of the surface
= 2 (lh + bh) + lb
= 2 (10 \(\times\) 8 + 9 \(\times\) 8) + 10 \(\times\) 9
= 2 (80 + 72) + 90 =2 \(\times\) 152 + 90
= 304 + 90 = 394 m²
Cost of white washing per m²= Rs. 8.50
Cost of white washing 394 m² = 394 \(\times\) 8.50
Total cost = Rs. 3349

In the figure AP 丄 CD and BQ 丄 CD
⇒ PQ = AB = 10, AP = BQ = h
Assume DP = x  ⇒ QC = 15 - (10 + x)
= 15 - 10 - x = 5 - x
In Δ ADP,  AP² = 7²- x²= 49 - x²
In Δ BQC, BQ² = 8²- (5 - x)² = 64 - (25 - 10x + x²) .
= 64 - 25 + 10x - x²
=39 - x²+  10x
Equating AP² = BQ²
49 - x² = 39 - x² + 10x

⇒ AP=\(\sqrt { { 7 }^{ 2 }-{ x }^{ 2 } } =\sqrt { 49-1^{ 2 } } =\sqrt { 48 } \)
∴ Area of the trapezium = \(\frac{1}{2}\) h(a+b) sq.units
=\(\frac { 1 }{ 2 } \times \sqrt { 48 } \times (10+15)=\frac { 1 }{ 2 } \times \sqrt { 3\times 16 } \times 25\)

= 2 \(\times\) 25 \(\times\) 1.732 = 86.6 m²

Perimeter of the rhombus land = 160 m
4a = 160 m
a = 40 m
One of the diagonal = 48 m
∴ Area of the land = 2 \(\times\) Area of the Δ ABC
s = \(\frac { 40+40+48 }{ 2 } =\frac { 128 }{ 2 } \)= 64 m
Area of Δ ABC =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 64(64-40)(64-40)(64-48) } \)
=\(\sqrt { 64\times 24\times 24\times 16 } =\sqrt { 589824 } \)
= 768 m²
∴ Area of the land = 2 \(\times\) 768 m² = 1536 m²

Area of the quadrilateral
= Area of Δ ABD + Area of Δ BCD
Δ ABD is right angled triangle
∴ Area = \(\frac{1}{2}\)b h
=
In Δ ABD, BD²= AD² + AB²
= 15²+ 20²= 225 + 400 = 625 m²
BD =\(\sqrt { 625 } \) = 25 m
∴ In ΔBCD, s = \(\frac { 25+25+17 }{ 2 } \)
= \(\frac { 68 }{ 2 } \) = 4
Area of Δ BCD = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 34(34-25)(34-26)(34-17) } \)
= \(\\ \sqrt { 34\times 9\times 8\times 17 } =\sqrt { 41616 } \)
∴ Area of the quadrilateral
= (150 + 204) m² = 354 m².

By the Pythagoras theorem
AB² = AD² + DB²
= 12²+16² = 144 + 256 = 400
AB = 20 cm
s = \(\frac { 34+20+42 }{ 2 } =\frac { 96 }{ 2 } \) = 48
∴ Area of the Δ ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 48(48-34)(48-20)(48-42) } \)
= \(\sqrt { 48\times 14\times 28\times 6 } =\sqrt { 112896 } \)
= \(\\ \sqrt { 336\times 336 } \) = 336 sq.cm
Area of the triangle ABD

∴ Area of the unshaded region
= Area of Δ ABC - Area of Δ ABD
= 336 - 96 = 240 cm².

Area of an isoeeles triangle
h = \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } =\sqrt { 169-25 } =\sqrt { 144 } \)=12 m
∴ Area of the triangular board
= \(\frac { 1 }{ 2 } \times bh=\frac { 1 }{ 2 } \) \(\times\) 10 \(\times\) 12 = 60 m²
cost of painting 1m²= Rs. 17.50
cost of painting 60m² = 60 \(\times\) 17.50 = Rs. 1050

s = 600 m
side s are in the ratio 5 : 12 : 13
5x + 12x + 13x = 30x
s = 600 ⇒ \(\frac { 30x }{ 2 } \) = 600
30x = 1200
x = 40
∴ sides are 200 m, 480 m, 520 m.
∴ Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 600(600-200)(600-480)(600-520) } \)
=\(\sqrt { 600\times 400\times 120\times 80 } =\sqrt { 2304000000 } \)
=\(\sqrt { 48\times 48\times 1000\times 1000 } \) = 48 \(\times\) 1000 =48000 sq.m

(i) sides: 10 cm, 24 cm, 26 m
Using Heron's formula
Area of the triangle = \(\sqrt { s(s-a)(s-b)(s-c) } \) sq. units
s = \(\frac { a+b+c }{ 2 } =\left( \frac { 10+24+26 }{ 2 } \right) cm=\frac { 60 }{ 2 } \) = 30 cm
∴ Area =\(\sqrt { 30(30-10)(0-24)(0-26) } \)
= \(\sqrt { 30\times 20\times 6\times 4 } =\sqrt { 600\times 24 } =\sqrt { 14400 } \) = 120 cm²
(ii) Sides: 1.8 m, 8m, 8.2 m
s = \(\\ \frac { 1.8+8+8.2 }{ 2 } =\frac { 18 }{ 2 } \) = 9
∴ Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 9(9-1.8)(9-8)(9-8.2) } =\sqrt { 9\times 7.2\times 0.8 } \)
= \(\sqrt { 51.84 } \) = 7.2 m²

Let ABCD be the rhombus.
Its perimeter = 4 × side = 400 m
Therefore, each side of the rhombus = 100 m
Given the length of the diagonal AC = 120 m
In \(\triangle\)ABC, let a =100 m, b =100 m, c = 120 m

s = \(\frac{a+b+c}{2}=\frac{100+100+120}{2}\) = 160 m
Area of \(\triangle\)ABC =\(\sqrt{160(160-100)(160-100)(160-120)}\)
= \(\sqrt{160 \times 60\times 60 \times40}\)
= \(\sqrt{40 \times 2 \times \times2\times60\times60\times40}\)
= 40 × 2 × 60 = 4800 m²
Therefore, Area of the field ABCD = 2 × Area of \(\triangle\)ABC = 2 × 4800 = 9600 m²

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²