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

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

Area of the parallelogram = 2 \(\times\) Area of the Δ ABC
s = \(\frac { 34+20+42 }{ 2 } =\frac { 96 }{ 2 } \)  = 48 m
Area of Δ ABC =\(\sqrt { 48(48-34)(48-20)(48-42) } \)
=\(\sqrt { 48\times 14\times 28\times 6 } =\sqrt { 112896 } \)=  336 m²
∴ Area of the parallelogram = 2 \(\times\) 336 m² = 672 m².

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

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

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