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

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

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

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

Volume of wood = 12 \(\times\) 10\(\times\) 9 - 10 \(\times\) 8 \(\times\) 7
= 1080 - 560 = 520 cm³ \(\times\)2.00 = Rs. 1040

6a² = 96 cm²
a² =\(\frac { 96 }{ 6 } \)
a = 6
Volume = a³ = 6³ = 36 \(\times\) 6 = 216 cm³.

(i) TSA = 6a² = 6 \(\times\) 27² = 6 \(\times\) 729 = 4374 cm²
(ii) TSA = 6a² = 6 \(\times\)3² = 54 cm²
(iii) TSA = 6a² = 6 \(\times\) 6² = 6 \(\times\) 36 = 216 cm²
(iv) TSA = 6a² = 6 \(\times\) 2.1² = 6 \(\times\) 4.41 = 26.46 cm²

Area of its base = \(\frac { TSA-LSA }{ 2 } \)
= \(\frac { 40-26 }{ 2 } \) m²
= 7 m²

PR =\(\sqrt { 225+64 } =\sqrt { 289 } \) = 17 cm
s = \(\frac { 15+17+18 }{ 2 } \) = 20 cm
Area of Δ PQR = \(\\ \sqrt { 20\times 5\times 12\times 3 } =\sqrt { 5\times 4\times 5\times 4\times 3\times 3 } \)=60
ΔPRS
Area = \(\sqrt { 27(27-17)(27-25)(27-12) } =\sqrt { 27\times 10\times 2\times 15 } \)
=\(\sqrt { 3\times 3\times 3\times 5\times 2\times 2\times 5\times 3 } \)=9 x 5 x 2 =90
Total area = 90 + 60 =150 cm².

3a = 150
a = \(\frac { 150 }{ 3 } \) = 50 m