Solution:
Each edges of cube = a cm
∴ Total surface area of cube = 6a²
⇒ 6a² = 864
⇒ a² = $$\frac{864}{6}$$ = 144
⇒ a = $$\sqrt {144} $$ = 12 cm
∴ Volume of cube :
= a³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³

Solution:
$$\eqalign{ \Leftrightarrow \frac{1}{2}{d_1} \times 2{d_1} = 25 \cr \Leftrightarrow d_1^2 = 25 \cr \Leftrightarrow {d_1} = 5 \cr} $$
∴ Sum of lengths of diagonals = (5 + 10) cm = 15 cm

Solution:
$$\eqalign{ {{\text{A}}_{\text{1}}} = \frac{{\sqrt 3 }}{2}{a^2}{\text{ and }} \cr {{\text{A}}_{\text{2}}} = \frac{{\sqrt 3 }}{2}{\left( {2a} \right)^2} \cr \,\,\,\,\,\,\,\,\,\, = 4 \times \frac{{\sqrt 3 }}{2}{a^2} \cr \,\,\,\,\,\,\,\,\,\, = 4{{\text{A}}_{\text{1}}} \cr \therefore {\text{K}} = 4 \cr} $$

Solution:
Let the area of the whole painting be x cm²
Then,
$$\eqalign{ \Rightarrow \frac{1}{4}x + 100 = \frac{3}{4}x \cr \Rightarrow \frac{1}{2}x = 100 \cr \Rightarrow x = 200 \cr} $$
∴ Area of painting = 200 cm²
And, Height = 10 cm
Length of painting :
$$\eqalign{ = \left( {\frac{{200}}{{10}}} \right)cm \cr = 20\,\,cm \cr} $$

Solution:
Distance mode by the skater in 44 sec
= Circumference of the circle
= $$\left( {2 \times \frac{{22}}{7} \times 28} \right)$$     m
= 176 m
Speed of skater :
= $$\left( {\frac{{176}}{{44}}} \right)$$   m/sec
= 4 m/sec
Perimeter of hexagon :
= (6 × 48) m
= 288 m
∴ Required difference :
= $$\left( {\frac{{288}}{{4}}} \right)$$   sec
= 72 sec

Solution:
Required ratio :
$$\eqalign{ = \left( {6 \times 6} \right):\left( {\frac{{\sqrt 3 }}{4} \times 6 \times 6} \right) \cr = 4:\sqrt 3 \cr} $$

Solution:
Since (20)² + (21)² = (29)²
So, it is a right-angled triangle with base = 20 m and height = 21 m
∴ Area = $$\left( {\frac{1}{2} \times 20 \times 21} \right)$$   m² = 210 m²