Case study (4 Marks)

Question 14 Marks

Metallic silos are used by farmers for storing grains. Farmer Girdhar has decided to build a new metallic silo to store his harvested grains. It is in the shape of a cylinder mounted by a cone.
Dimensions of the conical part of a silo is as follows:
Radius of base $= 1.5 m$
Height $= 2 m$
Dimensions of the cylindrical part of a silo is as follows:
Radius $= 1.5 m$
Height $= 7 m$
On the basis of the above information answer the following questions.
$(i)$ Calculate the slant height of the conical part of one silo.
$(ii)$ Find the curved surface area of the conical part of one silo.
$(iii)(A)$ Find the cost of metal sheet used to make the curved cylindrical part of $1$ silo at the rate of $₹ 2000\ per\ m ^2$.
OR
$(iii) (B)$ Find the total capacity of one silo to store grains.

Answer

$(i)\ I =\sqrt{r^2+h^2}$
$=\sqrt{(1.5)^2+(2)^2}$
$=\sqrt{2.25+4}$
$=\sqrt{6.25}$
$=2.5 m$
$(ii)\ \text{CSA}$ of cone $=\Pi rl$
$=\frac{22}{7} \times 1.5 \times 2.5$
$=11.78 m^2$
$(iii)\ (A) \text{CSA}$ of cylinder $=2 \pi rh$
$=2 \times \frac{22}{7} \times 1.5 \times 7$
$=66 m^2$
Cost of metal sheet used $=66 \times 2000$
$=₹ 1,32,000$
OR
$(iii)\ (B)$ Volume of cylinder $=\Pi r^2 h$
$ =\frac{22}{7} \times(1.5)^2 \times 7$
$ =49.5 m^3$
Volume of cone $=\frac{1}{3} \Pi r^2 h$
$ =\frac{1}{3} \times \frac{22}{7} \times(1.5)^2 \times 2$
$ =4.71 m^3$
Total capacity $=49.5+4.71=54.21 m^3$

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Question 24 Marks

Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer.
Here the largest triangle is represented by $\triangle A B C$ and smallest one with shelf is represented by $\triangle D E F . P Q$ is parallel to $E F$.
$(i)$ Show that $\triangle DPQ \sim \triangle DEF$.
$(ii)$ If $DP =50 cm$ and $PE =70 cm$ then find $\frac{P Q}{E F}$.
$(iii) \ (A)$ If $2 A B=5 D E$ and $\triangle A B C \sim \triangle D E F$ then show that $\frac{\text { perimeter of } \triangle A B C}{\text { perimeter of } \triangle D E F}$ is constant.
OR
$(iii) \ (B)$ If $AM$ and $DN$ are medians of triangles $A B C$ and $D E F$ respectively then prove that $\triangle ABM \sim \triangle DEN$.

Answer

$(i) \ \angle DPQ =\angle DEF$
$\angle PDQ =\angle EDF $
$(ii)$ Therefore $\Delta DPQ \sim \Delta DEFD E=50+70=120 \ cm$
$\frac{D P}{D E}=\frac{P Q}{E F}$ Therefore $\frac{P Q}{E F}=\frac{50}{120}$ or $\frac{5}{12}$
$(iii) \ (A)\ \frac{A B}{D E} =\frac{5}{2}=\frac{B C}{E F}=\frac{A C}{D F}$
$ \Rightarrow AB=\frac{5}{2} DE$
$ \frac{\text { perimeter of } \triangle A B C}{\text { perimeter of } \triangle D E F}=\frac{\frac{5}{2}(D E+E F+F D)}{D E+E F+F D}=\frac{5}{2} \text { ( Constant) }$
OR

$\frac{A B}{D E}=\frac{B C}{E F}=\frac{B C / 2}{E F / 2}=\frac{B M}{E N}$
Also $\angle B =\angle E$
Therefore $\triangle ABM \sim \triangle DEN$

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Question 34 Marks

Ms. Sheela visited a store near her house and found that the glass jars are arranged one above the other in a specific pattern.
On the top layer there are $3$ jars. In the next layer there are $6$ jars. In the $3^{rd}$ layer from
the top there are $9$ jars and so on till the $8^{th}$ layer.
On the basis of the above situation answer the following questions.
$(i)$ Write an $A.P$ whose terms represent the number of jars in different layers starting
from top . Also, find the common difference.
$(ii)$ Is it possible to arrange $34$ jars in a layer if this pattern is continued? Justify your
answer.
$(iii) (A)$ If there are $' n\ '$ number of rows in a layer then find the expression for finding the total number of jars in terms of $n$.
Hence find $S_8$.
OR
$(iii) (B)$ The shopkeeper added $3$ jars in each layer. How many jars are there in the $5^{th}$ layer from the top?

Answer

$(i)$ First term $a=3, A.P$ is $3,6,9,12 \ldots \ldots, 24$
common difference $d=6-3=3$
$(ii) 34=3+(n-1) 3$
$\Rightarrow n =34 / 3=11 \frac{1}{3}$ which is not a positive integer.
Therefore, it is not possible to have $34$ jars in a layer if the given pattern is continued.
$(iii)(A)$
$S_n=\frac{n}{2}[2 \times 3+(n-1) 3]$
$=\frac{n}{2}[6+3 n-3]$
$=\frac{n}{2}[3+3 n]$
$=3 \frac{n}{2}[1+n]$
$S_8=3 \times \frac{8}{2}(1+8)$
$=108$
OR
$(B)$
$A.P$ will be $6,9,12, \ldots \ldots$
$\qquad a =6, d=3$
$ t_5=6 +(5-1) 3$
$ =6+12$
$ =18$

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Maths Case study (4 Marks)

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