Volume And Surface Area

Details

$$\frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = 27$$
$$ ⇒ r = 81\,cm$$

$$4\pi {R^2} = 6{a^2}$$
$$ ⇒ \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }}$$
$$ ⇒ \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }}$$
$$\therefore \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} $$
$$ = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}$$
$$= \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3}$$
$$= \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }}$$
$$= \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }}$$
$$= \frac{{\sqrt {12} }}{{\sqrt {2\pi } }}$$
$$= \frac{{\sqrt 6 }}{{\sqrt \pi }}$$
$$\text{or, }\sqrt 6 :\sqrt \pi $$

Since level of water increased in cylinder by height 4. 

This is because of the rectangular block .

Therefore , area of rectangular block immersed in water is 

= $$\frac{22}{7}\×{6}\×{6}\×{4}$$  

Thats why portion of block immersed in water is 

= (2$$\frac{2}{7}$$ × 36 × 4) / total vol. of rectangle 

= (2$$\frac{2}{7}$$ × 36 × 4)/ (7×5×6) 

= (2$$\frac{2}{7}$$ × 24)/ 35.

Given length of width of swimming pool is 9 m and 12 m respectively
Volume of swimming pool :
$$ = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right)$$
$$ = 9 \times 12 \times \frac{5}{2}$$
$$ = 270\,\text{cu. metre}$$

Given height of the conical tent = 3 m

Diameter = 10 ⇒ Radius = D/2 = 10/2 = 5 m 

Now, as the tent is in the conical form

Volume of the conical tent = $$π}r}2}\frach}3}$$ 

⇒ (22×5×5×3) / (7×3) 

= 22×25/7 

= 78.54 cu.m 

Given each person requires 3.92 cu.m of air 

⇒ Number of persons can be seated in the tent = 78.54/3.92 = 20.03 =~ 20

4/3 π×10×10×10 = 8×4/3 π×r×r×r
r = 5
4π×5×5 = 100π

Volume of bucket :
= 28.490 litres
= (28.490 ×1000) cm³
= 28490 cm³
Let the height of the bucket be h cm
We have : r = 21 cm. and R = 28 cm
$$\therefore \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$     $$ = 28490$$
$$ ⇒ h\left( {784 + 441 + 588} \right) = $$     $$\frac{{28490 \times 21}}{{22}}$$
$$ ⇒ 1813h = 27195$$
$$ ⇒ h = \frac{{27195}}{{1813}}$$
$$ ⇒ h = 15\,cm$$

Let the length of each side of the square base be×metres
Then,
$${x^2} + {x^2} = {\left( 3 \right)^2}$$
$$ ⇒ 2{x^2} = 9$$
$$ ⇒ {x^2} = \frac{9}{2}$$
$$ ⇒ x = \frac{3}{{\sqrt 2 }}$$
∴ Volume of parallelopiped :
$$ = \left( {\frac{3}{{\sqrt 2 }} \times \frac{3}{{\sqrt 2 }} \times 10} \right){{\text{m}}^{\text{3}}}$$
$$ = \frac{{90}}{2}{{\text{m}}^{\text{3}}}$$
$$ = 45\,{{\text{m}}^{\text{3}}}$$

Area of base :
$$ = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2}$$
$$ = \frac{{\sqrt 3 }}{4}{m^2}$$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$ = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2}$$
$$\left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right]$$
$$ = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2}$$
$$ = \frac{{\sqrt {35} }}{4}{m^2}$$
∴ Whole surface area of the pyramid :
$$ = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2}$$
$$ = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2}$$