Let ‘a’ be the side of the cube.
Given that, total surface area = 864 cm²
6a²= 864
a² = \(\frac{864}{6}\)
a² = 144
Therefore, side (a) = 12 cm
Now, volume of the cube = a³
= 123 = 12 ×12 ×12 = 1728 cm³

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

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

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

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.

TSA of a cube 6a² = 726

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

(i) l = 12 cm
b = 8 cm
h= 6 cm
Volume of the cuboid = lbh
= 12 \(\times\) 8 \(\times\) 6 cm³ = 576 cm³
(ii) l = 60 m 
b = 25 m 
h = 1.5 m
Volume of the cuboid = l b h = 60 \(\times\) 25\(\times\)1.5 m³ = 2250 m³

l \(\times\) b \(\times\) h = 6 m \(\times\) 400 cm \(\times\) 1.5 m
l = 6m, b = 4 m, h = 1.5 m
Total surface area of the cuboid = Outer surface area
= 2 (lb + bh + hI)
= 2 ((6 \(\times\) 4) + (4 \(\times\) 1.5) + (1.5 \(\times\) 6))
= 2 (24 + 6 + 9) = 2 (39) m²
Cost of painting 1 m² = Rs. 22
Cost of painting 78 m²= 78 \(\times\) 22 = Rs. 1716

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 ABCD
= Area of the Δ ABD + Area of the Δ BCD
Sides of the triangle ABD are 13 cm, 14 cm, 15 cm.
s = \(\frac { 13+14+15 }{ 2 } \)cm
= \(\frac { 42 }{ 2 } \) = 21 cm
Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 21(21-13)(21-14)(21-15) } \)
=\(\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 7056 } \)=84 cm²
sides of the triangle
BCD are 12 cm, 9 cm, 15 cm
∴ s = \(\frac { 12+9+15 }{ 2 } =\frac { 36 }{ 2 } \) = 18 cm
Area =\(\sqrt { 18(18-2)(18-9)(18-15) } \)
=\(\sqrt { 18\times 6\times 9\times 3 } =\sqrt { 2916 } \) = 54 cm²
∴ Area of the quadrilateral
= 84 cm² + 54 cm² =138 cm²

sides of a triangle = 48 cm, 20 cm, 52 cm,
s = \(\frac { 48+20+52 }{ 2 } =\frac { 120 }{ 2 } \) = 60 cm
(i) Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 60(60-48)(60-20)(60-52) } \)
=\(\sqrt { 60\times 12\times 40\times 8 } =\sqrt { 230400 } =\sqrt { 480\times 480 } \) = 480 sq.m
(ii) Area of the parallelogram = Area of the triangle (given)
bh = 480 cm²

h = 24 cm

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