Case study (4 Marks)

Question 14 Marks

Partha, a software engineer, lives in Jerusalem for his work. He lives in the most convenient area of the city from where bank, hospital, post office and supermarket can be easily accessed. In the graph, the bank is plotted as A(9, 5), hospital as B(-3,1) and supermarket as C(5,5) such that A, B, C form a triangle.

Based on the above given information, answer the following questions:
(i) Find the distance between the bank and the hospital.
(ii) In between the bank and the supermarket, there is a post office plotted at E which is their mid-point. Find the coordinates of E.
(iii) (a) In between the hospital and the supermarket, there is a bus stop plotted as D, which is their mid-point. If Partha wants to reach the bus stand from the bank, then how much distance does he need to cover ?
OR
(b) P and Q are two different garment shops lying between the bank and the hospital, such that $BP = PQ = QA$. If the coordinates of P and Q are $(1, a)$ and $(b, 3)$ respectively, then find the values of ' $a$ ' and ' $b$ '.

Answer

(i) Distance between bank and hospital $=\sqrt{(-3-9)^2+(-1-5)^2}$
$=\sqrt{180}$ units or $6 \sqrt{5}$ units
(ii) Coordinates of E are $\left(\frac{9+5}{2}, \frac{5+(-5)}{2}\right)=(7,0)$
(iii) (a) Coordinates of D are $\left(\frac{-3+5}{2}, \frac{-1+(-5)}{2}\right)=(1,-3)$
Distance Partha need to cover $=\sqrt{(9-1)^2+(5-(-3))^2}$
$=\sqrt{128}$ units or $8 \sqrt{2}$ units
OR
(iii) (b) P is mid-point of BQ
$
\therefore a=\frac{-1+3}{2}=1
$
Q is mid-point of PA
$
\therefore b=\frac{1+9}{2}=5
$

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Question 24 Marks

A school has decided to plant some endangered trees on $51^{\text {st }}$ World Environment Day in the nearest park. They have decided to plant those trees in few concentric circular rows such that each succeeding row has $20$ more trees than the previous one. The first circular row has $50$ trees.

Based on the above given information, answer the following questions:
$(i)$ How many trees will be planted in the $10^{\text {th }}$ row?
$(ii)$ How many more trees will be planted in the $8^{\text {th }}$ row than in the $5^{\text {th }}$ row ?
$(iii)\ (a)$ If $3200$ trees are to be planted in the park, then how many rows are required ?
OR
$(b)$ If $3200$ trees are to be planted in the park, then how many trees are still left to be planted after the $11^{\text {th }}$ row ?

Answer

Here $a=50$ and $d=20$
$(i)$ Number of trees planted in $10^{\text {th }}$
row $=a_{10}=50+9 \times 20$
$=230$
$(ii) a_8-a_5=3 \times 20=60$
$(iii) \ (a)$ Let $S_n=3200$
$\Rightarrow \frac{n}{2}[2 \times 50+(n-1) \times 20]=3200$
$\Rightarrow n^2+4 n-320=0$
$\Rightarrow(n+20)(n-16)=0$
$n \neq-20$
$\therefore n=16$
Hence, required number of rows are $16$
OR
$(iii)\ (b)$ Required number of trees $=S_n-S_{11}$
$=3200-\frac{11}{2}[2 \times 50+10 \times 20]$
$=1550$
Hence, number of trees left are $1550$

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Question 34 Marks

Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely $X$ and $Y$ have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point $O$ is fixed in front of the fire engine. The ladder inclined at an angle $60^{\circ}$ to the horizontal is leaning against the wall of the terrace $($top$)$ of the building $Y$. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace $($top$)$ of the opposite building $X$ at an angle of $45^{\circ}$ with the ground. Both the buildings along with the foot of the ladder, fixed at $' O \ '$ are in a straight line.

Based on the above given information, answer the following questions :
$(i)$ Find the length of the ladder.
$(ii)$ Find the distance of the building $Y$ from point $'O\ '$, i.e. $OA$.
$(iii)\ (a)$ Find the horizontal distance between the two buildings.
OR
$(b)$ Find the height of the building $X$.

Answer

$(i)$ In $\triangle OAP$,
$\frac{ OP }{12 \sqrt{3}}=\operatorname{coesc} 60^{\circ}=\frac{2}{\sqrt{3}}$
$\Rightarrow OP =24 m$
$\therefore$ Length of ladder is $24 m$
$(ii)$ In $\triangle OAP$,
$\frac{OA}{12 \sqrt{3}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}}$
$\Rightarrow O A=12 m$
$\therefore$ the distance of the building $Y$ from point $O$
i e., $OA$ is $12 m$
$\text { (iii) (a) } O P=O R=24 m$
$\therefore \text { In } \triangle O C R,$
$\frac{\text { OC }}{24}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$\Rightarrow O C=12 \sqrt{2} m$
$\therefore$ distance between two buildings $ =OA+OC$
$ =(12+12 \sqrt{2}) m$ or $ 12(1+\sqrt{2}) m$
OR
$\text { (iii) (b) } O P=O R=24 m$
$\therefore In \triangle O C R,$
$\frac{RC}{24}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\Rightarrow RC=12 \sqrt{2} m$
$\therefore$ height of building $X$ is $12 \sqrt{2} m$

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Maths Case study (4 Marks)

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