I \(\times\) b \(\times\) h = 440 cm³
h = \(\frac { 440 }{ 88 } \) = 5 cm

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.

The dimensions of a water tank
l = 12 m b = 10 m h= 8 m
V = l \(\times\) b \(\times\) h = 12 \(\times\) 10 \(\times\) 8 = 960 m³
The volume of water to be filled to the height of 3m.
V = 12 \(\times\) 10 \(\times\) 3
= 360 m³ = 360 \(\times\) 1000 = 360000 litres.

TSA of a cube 6a² = 726

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

(i) side of a cube (a) = 5 cm
Volume of a cube V = a³ = 5 \(\times\) 5 \(\times\) 5 = 125 cm³
a = 3.5 m
V = a³ = 3.5 \(\times\) 3.5 \(\times\) 3.5 m³ = 42.875 m³
a = 21 cm
V = a³ = 21 \(\times\) 21 \(\times\) 21 cm³ = 9261 cm³.

Dimensions of a match box = 6 cm \(\times\) 3.5 cm \(\times\) 2.5 cm
V = (6 \(\times\) 3.5 \(\times\) 2.5) cm³ = 52.5 cm³
Volume of 12 such boxes = 12 \(\times\) 52.5 cm³ = 630 cm³.

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.

Perimeter of an equilateral triangle = 180 cm
∴ one side (a) = \(\\ \frac { 180 }{ 3 } \) = 60 m.
Area of an equilateral triangle =\(\frac { \sqrt { 3 } }{ 4 } \) a² sq.units

= 900\(\sqrt { 3 } \) m²
= 900 \(\times\) 1.732 = 1558.8 m²

side: 22 m, 120 m, 122 m
Using Heron's formula
s = \(\frac { 22+120+122 }{ 2 } =\frac { 264 }{ 2 } \) = 12 m
Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 132(132-22)(132-120)(132-122) } \)
= \(\sqrt { 132\times 110\times 12\times 10 } \)
= \(\sqrt { 1742400 } =\sqrt { 11\times 11\times 3\times 3\times 2\times 2\times 2\times 2\times 10\times 10 } \)
= 11 \(\times\) 3 \(\times\) 2 \(\times\) 10 = 1320 m²
cost of levelling 1 m² = Rs. 20
∴ cost oflevelling 1320 m²= 1320 \(\times\) 20 = Rs. 26400