Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Modern curricula include several problem$-$solving strategies. Teachers model the process, and students work independently to copy it. Sheela Maths teacher of class $9^{th}$ wants to explain the properties of parallelograms in a creative way, so she gave students colored paper in the shape of a quadrilateral and then ask the students to make a parallelogram from it by using paper folding.

$i.$ How can a parallelogram be formed by using paper folding?
$ii.$ If $\angle RSP =30^{\circ}$, then find $\angle RQP$.
$iii.$ If $\angle RSP =50^{\circ}$, then find $\angle SPQ$ ?
OR
If $SP =3 cm$, Find the $RQ.$

Answer

$i.$ By joining mid points of sides of a quadrilateral one can make parallelogram.
$S$ and $R$ are mid points of sides $A D$ and $C D$ of $\Delta A D C, P$ and $Q$ are mid points of sides $A B$ and $B C$ of $\Delta A B C$, then by mid point theorem $SR \| AC$ and $SR =\frac{1}{2} AC$ similarly $PQ \| AC$ and $PQ =\frac{1}{2} AC$.
Therefore $S R \| P Q$ and $S R=P Q$
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Hence $PQRS$ is parallelogram.
$ii. \ \angle RQP .=30^{\circ}$, Opposite angles of a parallelogram are equal.
$iii.$ Adjacent angles of a parallelogram are supplementary.
Thus, $\angle RSP +\angle SPQ =180^{\circ}$
$50^{\circ}+\angle SPQ =180^{\circ}$
$\angle SPQ =180^{\circ}-50^{\circ}$
$=130^{\circ}$
OR
$RQ = 3 \ cm$
Opposite side of a parallelogram are equal.

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Reeta was studying in the class $9^{th} C$ of St. Surya Public school, Mehrauli, New Delhi$-110030$
Once Ranjeet and his daughter Reeta were returning after attending teachers' parent meeting at Reeta's school.
As the home of Ranjeet was close to the school so they were coming by walking.
Reeta asked her father, "Daddy how old are you?"
Ranjeet said, "Sum of ages of both of us is $55$ years, After $10$ years my age will be double of you.

$i.$ What is the second equation formed?
$ii.$ What is the present age of Reeta in years?
$iii.$ What is the present age of Ranjeet in years?
OR
If the ratio of age of Reeta and her mother is $3 : 7$ then what is the age of Reeta's mother in years?

Answer

$i. x - 2y = 10$
$ii. x + y = 55 \ldots(i)$ and $x - 2y = 10 \ldots(ii)$
Subtracting $(ii)$ from $(i)$
$x+y-x+2 y=55-10$
$\Rightarrow 3 y=45$
$\Rightarrow y=15$
So present age of Reeta is $15$ years.
$iii. x + y = 55 \ldots(i)$ and $x - 2y = 10 \ldots(ii)$
Subtracting $(ii)$ from $(i)$
$x + y - x + 2y = 55 - 10$
$\Rightarrow 3 y=45$
$\Rightarrow y=15$
Put $y = 15$ in equation $(i)$
$x+y=55$
$\Rightarrow x+15=55$
$\Rightarrow x=55-15=40$
So Ranjeet's present age is $40$ years.
OR
Let Reeta;s mother age be $'z\ '.$
Given Reeta age : Her mother age $= 7 : 5$
We know that Reeta age $= 15$ years
$\frac{\text { Mother age }}{\text { Reeta age }}=\frac{7}{5}$
$\Rightarrow z =\frac{7}{3} \times y$
$\Rightarrow z =\frac{7}{3} \times 15$
$\Rightarrow$ Here Mother age $=35$ years
Hence Reeta's mother's age is $35$ years.

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
$11 \ 950-1000 5$ Vinod and Basant have an adventure tourism business in Rishikesh. They have a resort in Rishikesh but now they are planning to build some tent houses too.
The newly built tent house will have all the basic amenities and it will attract the young tourists coming for adventure. Their conical tent is $9 m$ high and the radius of its base is $12 m.$

$i.$ What is the cost of the canvas required to make it, if $1 m^2$ canvas costs $₹ 10$?
$ii.$ How many persons can be accommodated in the tent, if each person requires $2 m^2$ on the ground?
$iii.$ How many persons can be accommodated in the tent, if each person requires $15 m^3$ of space to breathe in?
OR
If each person requires $20 m^3$ of space to breathe in and $100$ person can be accommodated then what should be height of tent?

Answer

$i.$ We have,
$r =$ Radius of the base of the conical tent $=12 m$
$h =$ Height of the conical tent $=9 m$
$\therefore 1=$ Slant height of the conical tent $=\sqrt{r^2+h^2}$
$=\sqrt{12^2+9^2} m=\sqrt{225 m}=15$
Area of lateral surface $=\pi r l=\frac{22}{7} \times 12 \times 15 m^2=565.7 m^2$
$\therefore$ Total cost of canvas ₹ $(565.2 \times 10)=$ ₹ 5652
$ii.$ We have,
$r =$ Radius of the base of the conical tent $=12 m$
$h =$ Height of the conical tent $=9 m$
$\therefore 1=$ Slant height of the conical tent $=\sqrt{r^2+h^2}$
$=\sqrt{12^2+9^2} m=\sqrt{225 m}=15$
Area of the base of the conical tent $=\pi r ^2=\frac{22}{7} \times 12 \times 12 m^2=452.16 m^2$
Since each person requires $2$ sq . meters of floor area.
$\therefore$ Max. no. of persons who will have enough space on the ground $=\frac{452.16}{2}=226$
$iii.$ We have,
$r =$ Radius of the base of the conical tent $=12 m$
$h =$ Height of the conical tent $=9 m$.
$\therefore 1=$ Slant height of the conical tent $=\sqrt{r^2+h^2}$
$=\sqrt{12^2+9^2} m=\sqrt{225} m=15$
Volume of the conical tent $=\frac{1}{3} \times$ Area of the base $\times$ Height
$\Rightarrow$ Volume of the conical tent $=\frac{1}{3} \times 452.16 \times 9 m^3$
We have, Air space required person $=15 m^3$
$\therefore$ No. of persons who will have enough air space to breathe in $=\frac{1356.48}{15}=90$
Hence, $90$ persons can be accommodated.
OR
We have,
$r =$ Radius of the base of the conical tent $=12 m$
$h =$ Height of the conical tent $=9 m$.
$\therefore 1=$ Slant height of the conical tent $=\sqrt{r^2+h^2}$
$=\sqrt{12^2+9^2} m=\sqrt{225} m=15$
Let new height is $H$ and radius $= 12 m $
Each person requires $20 m^3$ of space to breathe
Thus volume of air required for $100$ persons $=20 \times 100=2000 m^3$
$2000=\frac{1}{3} \pi \times r^2 h$
$2000=\frac{1056 h}{7}$
$h=13.25 \ m$
Thus new height would be $13.25 \ m.$

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Maths Case study (4 Marks)

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