Surface Areas and Volumes Test - 26

$$\displaystyle V=\dfrac{1}{3}\times 30\left ( 16+36+\sqrt{16\times 36} \right )cm^{3}=760cm^{3}$$

Let the plane XY divide the height AD of cone ABC such that $$AE:ED=1:2$$,
 where AED is the axis of the cone. Let $$r_2$$ and $$r_1$$ be the radii of the circular section XY and the base BC of the cone respectively and let $$h_ 1h$$ and $$h_1$$ be their heights.
$$\quad \displaystyle\dfrac{h_1}{h} = \displaystyle\dfrac{3}{2}$$
$$ \Rightarrow h_1 = \displaystyle\dfrac{3}{2}h$$
$$\dfrac{r_1}{r_2} = \displaystyle\dfrac{h_1}{h_1-h} = \displaystyle\dfrac{\displaystyle\dfrac{3}{2}h}{\displaystyle\dfrac{1}{2}h} = 3$$
Volume of the cone AXY=$$ \displaystyle\dfrac{1}{3}\pi r_2^2(h_1 - h)$$
                                  $$= \displaystyle\dfrac{1}{3}\pi r_2^2(\displaystyle\dfrac{3}{2}h - h)$$
                                  $$= \displaystyle\dfrac{1}{6}\pi r_2^2h$$
VolumeoffrustumXYBC=$$\dfrac { 1 }{ 3 } \pi h(r_{ 1 }^{ 2 }+r_{ 2 }^{ 2 }+r_{ 1 }r_{ 2 })$$
                                       =$$\dfrac { 1 }{ 3 } \pi h(9r_{ 2 }^{ 2 }+r_{ 2 }^{ 2 }+3r_{ 2 }^{ 2 })$$
                                        =$$\dfrac { 1 }{ 3 } \pi h(13r_{ 2 }^{ 2 })$$
$$\displaystyle\dfrac{\mbox{Volume of cone AXY}}{\mbox{Volume of cone XYBC}} = \displaystyle\dfrac{\displaystyle\dfrac{1}{6}\pi r_2^2h}{\displaystyle\dfrac{13}{3}\pi r_2^2h}$$
                                           $$ = \displaystyle\dfrac{1}{26}$$
The ratio between the volume of the cone AXY and the remaining portion BCXY is $$1:26$$

Let ABC be the right triangle, right angled triangle at A, whose sides AB and AC measures 6 cm and 8 cm respectively.
$$\therefore$$ Hypotenuse $$BC=\sqrt {(15)^2+(20)^2}$$
$$=\sqrt {225}=25 cm$$
AO(A'O) is the radius of the common base of the double cone fanned by revolving the triangle around BC.
Area of $$\Delta ABC=\dfrac {1}{2}\times 15\times 20=\dfrac {1}{2}BC\times AO$$.
$$\Rightarrow 150=\dfrac {1}{2}\times 25\times AO$$
$$\Rightarrow AO=12 cm$$
Volume of the double cone
$$=\dfrac {1}{3}\pi \times AO^2\times BO+\dfrac {1}{3}\pi \times AO^2\times CO$$
$$=\dfrac {1}{3}\pi \times AO^2(BO+CO)$$
$$=\dfrac {1}{3}\pi \times AO^2\times BC$$
$$=\dfrac {1}{3}\times 3.14\times 12\times 12\times 25=3768 cm^3$$
Surface area of the double cone
$$=\pi \times AO\times AB+\pi \times AO\times AC$$
$$=\pi \times AO(AB+AC)$$
$$=3.14\times 12\times (15+20)$$
$$=3.14\times 12\times 35=1318.8 cm^2$$

Given, Radius of cylindrical container $$=6 cm$$
Height of cylindrical container $$=15 cm$$

The tent has two parts 
(I) a cylindrical part and (II) a conical part 
(I) curved surface area of cylindrical portion 
$$\displaystyle =2\pi rh$$
$$\displaystyle 2\times \frac{22}{7}\times \frac{105}{2}\times 6m^{2}$$
=1980 $$\displaystyle m^{2}$$
(II) Curved sufrace area of conical portion 
$$\displaystyle \pi rl$$
$$\displaystyle \frac{22}{7}\times \frac{105}{2}\times 50m^{2}$$
$$\displaystyle =8250m^{2}$$
$$\displaystyle \therefore $$ Total area of canvas required to build the tent 
=Curved surface area of conical portion 
$$\displaystyle =1980m^{2}+8250m^{2}=10230m^{2}$$

It is given that
$$OM = 8, OP = 4, and OB = 10$$
$$\displaystyle \therefore BM^{2}=OB^{2}=OM^{2}$$
$$\displaystyle =100-64=36$$
$$\displaystyle \Rightarrow PM=6$$ so that $$PC = 3$$
Now volume of cone OAB
$$\displaystyle =\frac{1}{3}.\pi .6^{2}.8=96\pi $$
Volume of cone OCD $$\displaystyle =\frac{1}{3}.\pi .3^{2}.4=12\pi $$
$$\displaystyle \therefore $$ Volume of frustum $$\displaystyle ABCD=96\pi -12\pi =84\pi $$
Reqd. ratio $$\displaystyle =\frac{96\pi }{84\pi }=\frac{8}{7}\Rightarrow 8:7$$

Radius of wire$$=\dfrac{0.4}{2}=0.2\ cm$$
Radius of sphere$$=\dfrac{6}{2}=3\ m$$
Volume of wire (cyl) $$=$$ Vol of sphere 
$$\displaystyle \Rightarrow \pi \times \left ( 0.2 \right )^{2}\times h=\dfrac{4}{3}\times \pi \times 3^{3}$$
$$\displaystyle\Rightarrow h=900\ cm = 9\ m $$

Length of the bigger cuboid $$AF = (5 + 4)\ cm = 9\ cm$$