Case study (4 Marks)

Question 14 Marks

A 'circus' is a company of performers who put on shows of acrobats, clowns etc. to entertain people started around $250$ years back, in open fields, now generally performed in tents.
One such 'Circus Tent' is shown below.

The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are $9\ m$ and $30\ m$ respectively and height of conical part is $8\ m$ with same diameter as that of the cylindrical part, then find
$(1)$ The area of the canvas used in making the tent;
$(2)$ The cost of the canvas bought for the tent at the rate $Rs. 200\ \text{per} \ m^2, $ if $30 \ m^2$ canvas was wasted during stitching.

Answer

$1$. For cylinder,
height $=9 m$, diameter $=30\ m$
$\Rightarrow$ radius $=\frac{30}{2}=15\ m$.
For cone,
height $=8\ m,$ radius $=15\ m$
$\therefore$ slant height $, l =\sqrt{(8)^2+(15)^2}$
$ =\sqrt{64+225}$
$ =\sqrt{289}=17\ m$
Area of canvas required
$= \text{C.S.A}$ of cylinder $+ \text{C.S.A}$ of cone
$=2 \pi r h+r \pi l$
$=\pi r(2 h+l)$
$=\frac{22}{7} \times 15(2 \times 9+17)$
$=\frac{22}{7} \times 15 \times 35$
$=22 \times 15 \times 5=1650 \ m^2$.
$2$. The cost of the canvas $= ($Area of canvas required $+$ area of canvas wasted during stitching$) \times 200$
$=(1650+30) \times 200$
$=1680 \times 200$
$=\text { Rs. } 3,36,000$

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Question 24 Marks

A company deals in casting and moulding of metal on orders received from its clients.

The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part, then find
(1) The area of the canvas used in making the tent;
(2) The cost of the canvas bought for the tent at the rate Rs. 200 per sq. m, if 30 sq. m canvas was wasted during stitching.

Answer

a - Given, radius of cone, $r =21 cm$
height of cone, $h=28 cm$
Also, radius of hemisphere, $r =21 cm$
Volume of 50 toys
$=50 \times[$ Volume of sphere + Volume of cone $]$
$=50 \times\left(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\right)$
$=50 \times \frac{1}{3}\left(2 \times \frac{22}{7} \times(21)^3+\frac{22}{7}(21)^2 \times 28\right)$
$=\frac{50}{3} \times \frac{22}{7} \times(21)^2[2 \times 21+28]$
$=\frac{50 \times 22 \times 21 \times 21 \times 70}{7 \times 3}$
$=50 \times 22 \times 21 \times 70$
$=1617000 cm^3$
Hence, volume of 50 toys is $1617000 cm^3$
b -
$
\begin{aligned}
\frac{\text { Volume of hemisphere }}{\text { Volume of cone }} =\frac{\frac{2}{3} \pi r^3}{\frac{1}{3} \pi r^2 h} \\
=\frac{2 r}{h} \\
=\frac{2 \times 21}{28} \\
=\frac{21}{14}=\frac{3}{2}
\end{aligned}
$
Hence, the required ratio is $3: 2$

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Maths Case study (4 Marks)

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