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Maths Case study (4 Marks)

MODEL PAPER 1 (BASIC) · CBSE STD 10 (CBSE - English Medium). 3 questions.

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Case study (4 Marks)

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Question 14 Marks
Read the following text carefully and answer the questions that follow:
Basant and Vinod lives in a housing society in Dwarka, New Delhi. There are two building in their housing society. The first building is 8 meter tall. One day, both of them were just trying to guess the height of the other multi-storeyed building. Vinod said that it might be a 45 degree angle from the bottom of our building to the top of multi-storeyed building so the height of the building and distance from our building to this multi-storeyed building will be same. Then, both of them decided to estimate it using some trigonometric tools. Let's assume that the first angles of depression of the top and bottom of an 8 m tall building from top of a multi-storeyed building are $30^{\circ}$ and $45^{\circ}$, respectively.
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i. Now help Vinod and Basant to find the height of the multi-storeyed building.
ii. Also, find he distance between two buildings.
Answer
Let h is height of big building,here as per the diagram.
AE = CD = 8 m (Given)
BE = AB-AE = (h - 8) m
Let AC = DE = x
Also, $\angle F B D=\angle B D E=30^{\circ}$
$\angle F B C=\angle B C A=45^{\circ}$
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In $\triangle ACB , \angle A=90^{\circ}$
$\tan 45^{\circ}=\frac{A B}{A C}$
$\Rightarrow x = h , \ldots$ (i)
In $\triangle BDE , \angle E=90^{\circ}$
$\tan 30^{\circ}=\frac{B E}{E D}$
$\Rightarrow \quad x=\sqrt{3}(h-8)$.(ii)
From (i) and (ii), we get
$h=\sqrt{3} h-8 \sqrt{3}$
$h(\sqrt{ } 3-1)=8 \sqrt{ } 3$
$h=\frac{8 \sqrt{3}}{\sqrt{3}-1}=\frac{8 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{1}{2} \times(24+8 \sqrt{ } 3)=\frac{1}{2} \times(24+13.84)=18.92 m$
Hence height of the multistory building is 18.92 m and the distance between two buildings is 18.92 m.
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Question 24 Marks
Answer
i. Position of Neena = (3, 6)
Position of Karan = (6, 5)
Distance between Neena and Karan = $\sqrt{(6-3)^2+(5-6)^2}$
$=\sqrt{9+(-1)^2}$
$=\sqrt{10}$
ii. Co-ordinate of seat of Akash = 2, 3
iii.
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Co-ordinate of middle point = $\left(\frac{2+5}{2}, \frac{3+2}{2}\right)$
= 3.5, 2.5
OR
Binu = (5, 5); Karan = (6, 5)
Distance = $\sqrt{(6-5)^2+(5-2)^2}$
$=\sqrt{1+9}$
$=\sqrt{10}$
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Question 34 Marks
Read the following text carefully and answer the questions that follow:
Kamla and her husband were working in a factory in Seelampur, New Delhi. During the pandemic, they were asked to leave the job. As they have very limited resources to survive in a metro city, they decided to go back to their hometown in Himachal Pradesh. After a few months of struggle, they thought to grow roses in their fields and sell them to local vendors as roses have been always in demand. Their business started growing up and they hired many workers to manage their garden and do packaging of the flowers.
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In their garden bed, there are 23 rose plants in the first row, 21 are in the $2^{\text {nd }}, 19$ in $3^{\text {rd }}$ row and so on. There are 5 plants in the last row.
i. How many rows are there of rose plants?
ii. Also, find the total number of rose plants in the garden.
iii. How many plants are there in 6th row.
OR
If total number of plants are 80 in the garden, then find number of rows?
Answer
i. The number of rose plants in the $1^{\text {st }}, 2^{\text {nd }}, \ldots$ are $23,21,19, \ldots 5$
a = 23, d = 21 - 23 = - 2, $a_n=5$
$\therefore a_n=a+(n-1) d$
or, 5 = 23 + (n - 1)(-2)
or, 5 = 23 - 2n + 2
or, 5 = 25 - 2n
or, 2n = 20
or, n = 10
ii. Total number of rose plants in the flower bed,
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$S_{10}=\frac{10}{2}[2(23)+(10-1)(-2)]$
$S _{10}=5[46-20+2]$
$S_{10}=5(46-18)$
$S _{10}=5(28)$
$S _{10}=140$
iii. $a_n=a+(n-1) d$
$\Rightarrow a_6=23+5 \times(-2)$
$\Rightarrow a_6=13$
OR
$S _{ n }=80$
$S _{ n }=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 80=\frac{n}{2}[2 \times 23+(n-1) \times-2]$
$\Rightarrow 80=23 n-n^2+n$
$\Rightarrow n^2-24 n+80=0$
$\Rightarrow( n -4)( n -20)=0$
$\Rightarrow n=4$ or $n=20$
n = 20 not possible
$a_{20}=23+19 \times(-2)=-15$
Number of plants cannot be negative.
n = 4
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Case study (4 Marks) - Maths STD 10 Questions - Vidyadip