Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Two trees are standing on flat ground. The angle of elevation of the top of Both the trees from a point $X$ on the ground is $60^{\circ}$. If the horizontal distance between $X$ and the smaller tree is $8\ m$ and the distance of the top of the two trees is $20\ m$.

$i$. Calculate the distance between the point $X$ and the top of the smaller tree.
$ii.$ Calculate the horizontal distance between the two trees.
$iii$. Find the height of big tree.
OR
Find the height of small tree.

Answer

$i$. In $\triangle D C X$
$\tan 60^{\circ}=\frac{D C}{C X}$
$\sqrt{3}=\frac{D C}{8}$
$DC =8 \sqrt{3}\ m$
$\text { DX }=\sqrt{D C^2+C X^2}$
$=\sqrt{(8 \sqrt{3})^2+8^2}$
$=\sqrt{192+64}$
$=\sqrt{256}$
$=16\ m$
Hence, distance between $X$ and top of smaller tree is $16 \ m$.
$ii$. In $\triangle BAX$
$\cos 60^{\circ}=\frac{A X}{B X}$
$\frac{1}{2}=\frac{A C+8}{36}$
$36=2 A C+16$
$20 = 2\ AC$
$\frac{20}{2}=10 AC$
$AC = 10$
$\therefore$ horizontal distance between both trees is $10\ m$ .
$iii$. Height of big tree $= AB$
$\therefore$ In $\triangle B A X$
$\tan 60^{\circ}=\frac{A B}{A X}=\frac{A B}{18}$
$AB =18 \sqrt{3}\ m$
OR
Height of small tree $= CD$
In $\triangle CDX$
$\tan 60^{\circ}=\frac{C D}{C X}$
$\sqrt{3}=\frac{C D}{8}$
$CD =8 \sqrt{3} \ m$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is.
The left-right $($horizontal$)$ direction is commonly called $X-$ axis.
The up $-$ down $($vertical$)$ direction is commonly called $Y-$ axis.
In Green Park, New Delhi Suresh is having a rectangular plot $\text{ABCD}$ as shown in the following figure. Sapling of Gulmohar is planted on the boundary at a distance of $1 m$ from each other. In the plot, Suresh builds his house in the rectangular area $\text{PQRS}$. In the remaining part of plot, Suresh wants to plant grass.

$i$. Find the coordinates of the midpoints of the diagonal $QS$
$ii$. Find the length and breadth of rectangle $\text{PQRS}$?
$iii$. Find Area of rectangle $\text{PQRS}$.
OR
Find the diagonal of rectangle.

Answer

$i.$

Middle point of $QS =\left(\frac{10+3}{2}, \frac{6+2}{2}\right)$
$= (6.5, 4)$
$ii. $ Length $= RS. = \sqrt{(10-3)^2+(2-2)^2}$
$RS. =\sqrt{7^2+0}$
$RS. = 7 m$
Breadth $= RQ = \sqrt{(10-10)^2+(2-6)^2}$
$=\sqrt{0+16}$
$=4 m$
$iii$. Area of rectangle $= 1 \times b$
$=7 \times 4$
$=28 m^2$
OR
Diagonal $=\sqrt{l^2+b^2}$
$=\sqrt{7^2+4^2}$
$=\sqrt{49+16}$
$=\sqrt{65}$

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
Saving money is a good habit and it should be inculcated in children right from the beginning. Rehan’s mother brought a piggy bank for Rehan and puts one ₹ 5 coin of her savings in the piggy bank on the first day. She increases his savings by one ₹ 5 coin daily.

Based on the above information, answer the following questions:
i. How many coins were added to the piggy bank on $8^{\text {th }}$ day?
ii. How much money will be there in the piggy bank after 8 days?
iii. a. If the piggy bank can hold one hundred twenty ₹ 5 coins in all find the number of days she can contribute to put ₹ 5 coins into it.
OR
b. Find the total money saved, when the piggy bank is full.

Answer

i. 8 coins
ii. Money in the piggy bank day wise 5, 10, 15, 20 ...
Money after 8 days = ₹ 180
iii. a. We can have at most 120 coins.
$\frac{n}{2}[2(1)+(n-1) 1]=120$
$n^2+n-240=0$
Solving for n, we get, n = 15 as $n \neq-16$
$\therefore$ Number of days $=15$

OR

b. Total money saved $=120 \times 5=₹ 600$

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Maths Case study (4 Marks)

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