Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Sanjay and his mother visited in a mall. He observes that three shops are situated at $P, Q, R$ as shown in the figure from where they have to purchase things according to their need. Distance between shop $P$ and $Q$ is $8 m$ and between shop $P$ and $R$ is $6 m$.
Considering $O$ as the center of the circles.

$i$. Find the Measure of $\angle QPR$.
$ii$. Find the radius of the circle.
$iii$. Find the Measure of $\angle QSR$.
OR
Find the area of $\Delta PQR$.

Answer

$i$. We know that angle in the semicircle $=90^{\circ}$
Here $QR$ is a diameter of circle and $\angle QPR$ is angle in semicircle.
Hence $\angle QPR =90^{\circ}$
$ii$. $\angle QPR =90^{\circ}$
$\Rightarrow QR ^2= PQ ^2+ PR ^2$
$\Rightarrow QR ^2=8^2+6^2$
$\Rightarrow QR =\sqrt{64+36}$
$\Rightarrow QR =10 m$
$iii.$ Measure of $\angle QSR =90^{\circ}$
Angles in the same segment are equal.
$\angle QSR$ and $\angle QPR$ are in the same segment.
OR
Area $\Delta PQR =\frac{1}{2} \times P Q \times P R$
$\Rightarrow$ Area $\Delta PQR =\frac{1}{2} \times 8 \times 6=24 m^2$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Ladli Scheme was launched by the Delhi Government in the year $2008$. This scheme helps to make women strong and will empower a girl child. This scheme was started in $2008$.
The expenses for the scheme are plotted in the following bar chart.

$i$. What are the total expenses from $2009$ to $2011$?
$ii$. What is the percentage of no of expenses in $2009-10$ over the expenses in $2010-11$?
$iii$. What is the percentage of minimum expenses over the maximum expenses in the period $2007-201$1?
OR
What is the difference of expenses in $2010-11$ and the expenses in $2006-09$?

Answer

$i.$ Expenses in $2009-10 = 9160$ Million
Expenses in $2010-11 = 10300$ Million
Total expenses from $2009$ to $2011$
$= 9160 + 10300$
$= 19460$ Million
$ii.$ Expenses in $2009-10 = 9160$ Million
Expenses in $2010-11 = 10300$ Million
Thus percentage of no of expenses in $2009-10$ over the expenses in $2010-11$
$=\frac{9160}{10300} \times 100$
$=88.93 \%$
$iii.$ The minimum expenses $($in $2007-08) = 5.4$ Million
The maximum expenses $($in $2010-11) = 10300$ Million
Thus percentage of no of minimum expenses over the maximum expenses
$=\frac{5.4}{10300} \times 100$
$=0.052 \%$
OR
The expenses in $2010-11 = 10300$ Million
The expenses in $2006-09 = 9060$ Million
The difference $= 10300 - 9060$ Million
$= 1240 $ Million

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Question 34 Marks

A bus stop is barricaded from the remaining part of the road, by using $50$ hollow cones made of recycled cardboard. Each cone has a base diameter of $40\ cm$ and a height of $1\ m.$

$i.$ Find the curved surface area of the cone.
$ii.$ If the outer side of each of the cones is to be painted and the cost of painting is $₹ 12$ per $m^2$, what will be the cost of painting all these cones ? $($Use $\pi=3.14$ and take $\sqrt{1.04}=1.02 )$

Answer

Diameter of cone $= 40 \ cm$
$\Rightarrow$ Radius of cone$\ ( r )=\frac{40}{2}$
$=20 \ cm$
$=\frac{20}{100} m$
$=0.2 m$
Height of cone $(h) = 1 m$
Slant height of cone $(l)=\sqrt{r^2+h^2}$
$=\sqrt{(0.2)^2+(1)^2}$
$=\sqrt{1.04} m$
Curved surface area of cone $=\pi r l$
$=3.14 \times 0.2 \times \sqrt{1.04}$
$=0.64056 m^2$
$\therefore$ Cost of painting $1\ m^2$ of a cone $= Rs.12$
$\therefore$ Cost of painting $0.64056\ m^2$ of a cone $=12 \times 0.64056= Rs. 7.68672$
$\therefore$ Cost of painting of $50$ such cones $=50 \times 7.68672= Rs. 384.34 ($approx.$)$

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Maths Case study (4 Marks)

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