Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Vijay lives in a flat in a multi-story building. Initially, his driving was rough so his father keeps eye on his driving. Once he drives from his house to Faridabad. His father was standing on the top of the building at point A as shown in the figure. At point $C$, the angle of depression of a car from the building was $60^{\circ}$. After accelerating $20 m$ from point $C ,$ Vijay stops at point $D$ to buy ice cream and the angle of depression changed to $30^{\circ}$

$i.$ Find the value of $x. (1)$
$ii.$ Find the height of the building $AB. (1)$
$iii.$ Find the distance between top of the building and a car at position $D$ ? $(2)$
OR
Find the distance between top of the building and a car at position $C$? $(2)$

Answer

$i$. The above figure can be redrawn as shown below:

From the figure,
$\text { let } A B=h \text { and } B C=x$
$\text { In } \triangle A B C,$
$\tan 60=\frac{A B}{B C}=\frac{h}{x}$
$\sqrt{3}=\frac{h}{x}$
$h=\sqrt{3} x \ldots(i)$
$\text { In } \triangle A B D,$
$\tan 30=\frac{A B}{B D}=\frac{h}{x+20}$
$\frac{1}{\sqrt{3}}=\frac{\sqrt{3} x}{x+20}[\text { using (i) }]$
$x+20=3 x$
$x=10 m$
$ii$. The above figure can be redrawn as shown below:

Height of the building, $h =\sqrt{3} x =10 \sqrt{3}=17.32 m$
$iii$. The above figure can be redrawn as shown below:

Distance from top of the building to point $D$ .
$\text { In } \triangle ABD$
$\sin 30^{\circ}=\frac{A B}{A D}$
$\Rightarrow A D=\frac{A B}{\sin 30^0}$
$\Rightarrow A D=\frac{10 \sqrt{3}}{\frac{1}{2}}$
$\Rightarrow AD=20 \sqrt{3} m$
OR
The above figure can be redrawn as shown below:

Distance from top of the building to point $C$ is
$\text { In } \triangle ABC$
$\sin 60^{\circ}=\frac{A B}{A C}$
$\Rightarrow A C=\frac{A B}{\sin 60^0}$
$\Rightarrow A C=\frac{10 \sqrt{3}}{\frac{\sqrt{3}}{2}}$
$\Rightarrow AD=20 m$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
There are two routes to travel from source A to destination B by bus. First bus reaches at B via point C and second bus reaches from A to B directly. The position of A, B and C are represented in the following graph: Based on the above information, answer the following questions.

i. If the fare for the second bus is ₹ 15 / km, then what will be the fare to reach to the destination by this bus? (1)
ii. What is the distance between $A$ and $B$ ? (1)
iii. What is the distance between A and C? (2)
OR
If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B? (2)

Answer

i. Distance travelled by second bus $=7.2 km$
$\therefore$ Total fare $=7.2 \times 15=$ ₹ 08
ii. Required distance $=\sqrt{(2+2)^2+(3+3)^2}$
$=\sqrt{4^2+6^2}=\sqrt{16+36}=2 \sqrt{13} km \approx 7.2 km$
iii. Required distance $=\sqrt{(3+2)^2+(2+3)^2}$
$=\sqrt{5^2+5^2}=5 \sqrt{2} km$
OR
Distance between B and C
$=\sqrt{(3-2)^2+(2-3)^2}=\sqrt{1+1}=\sqrt{2} km$
Thus, distance travelled by first bus to reach to $B$
$=AC+CB=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} km \approx 8.48 km$
and distance travelled by second bus to reach to $B$
$=AB=2 \sqrt{13} km=7.2 km$
$\therefore$ Distance of first bus is greater than distance of the cond bus, therefore second bus should be chosen.

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Your friend Varun wants to participate in a $200\ m$ race. He can currently run that distance in $51$ seconds and with each day of practice it takes him $2$ seconds less. He wants to do in $31$ seconds.

$i.$ Write first four terms are in $AP$ for the given situations. $(1)$
$ii$. What is the minimum number of days he needs to practice till his goal is achieved? $(1)$
$iii$. How many second takes after $5^{th}$ days? $(2)$
$OR$
Out of $41, 30, 37$ and $39$ which term is not in the $AP$ of the above given situation? $(2)$

Answer

$\text { i. } 51,49,47, \ldots 31 \text { AP }$
$d=-2$
First $4$ terms of $AP$ are $: 51, 49, 47,45 \ldots$
$ii. 51,49,47, \ldots 31 AP$
$d=-2$
$t_n=a+(n-1) d$
$31=51+(n-1)(-2)$
$31=51-2 n+2$
$31=53-2 n$
$31-53=-2 n$
$-22=-2 n$
$n=11$
i.e.$,$ he acheived his goal in $11$ days.
$iii. 51,49,47, \ldots 31 AP$
$d=-2$
$t_6=a+(n-1) d$
$=51+(6-1)(-2)$
$=51+(-10)$
$=41$ sec
OR
The given $AP$ is
$51,49,47,45,43,41,39,37,35,33,31,29$
$\therefore 30$ is not in the $AP.$

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Maths Case study (4 Marks)

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