Government council of a local public development authority

Case Study Question Class 10

Government council of a local public development authority of Dehradun to build an adventurous playground on the top of a hill, which will have adequate space for parking.

After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats. [CBSE 2023]

Based on the above information, answer the following question :

(i) What is the total perimeter of the parking area ? 1

(ii) (a) What is the total area of parking and the two quadrants ? 2

(b) What is the ratio of area of playground to the area of parking area ? 2

(iii) Find the cost of fencing the playground and parking area at the rate of Rs 2 per units. 1

Solution :

(i)Radius of the parking area (r)= 7/2

Perimeter of the parking area = Circumference of the semicircle of parking area + Diameter of semicircular parking area

= π r + 2 r

$= \pi \times (\frac{7}{2}) + 2(\frac{7}{2})$

$= \frac{22}{7}\times \frac{7}{2} + 7$

$= 11 + 7 =18$ units

(ii) (a) Radius of semicircle = 7/2 units

Radius of quadrant = 2 units

Area of semicircle $= \dfrac{\pi r^2}{2}$

$= \dfrac{\frac{22}{7}\times (\frac{7}{2})^2}{2}$

$= \dfrac{\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}}{2}$

$= \dfrac{11\times 7}{2\times 2}$

$= \dfrac{77}{4} = 19.25$ units

Area of quadrant $= \dfrac{1}{4} \pi r^2$

$= \dfrac{1}{4}\times \dfrac{22}{7}\times (2)^2$

$= \dfrac{22}{7} = 3.14$

Area of parking field and two quadrant = Area of Parking + 2(Area of quadrant)

= 19.25 + 2(3.14)

= 19.25 + 6.28

= 25.53 units

OR

(b) Length of playground = 14 units

Breadth of playground = 7 units

The ratio of area of playground to the area of parking area

$The ratio = \dfrac{\text{Area of play ground}}{\text{Area of parking}}$

$= \dfrac{14 \times 7}{\frac{77}{4}}$ [From (ii) (a) ]

$= \dfrac{14 \times 4}{11}$

$= \dfrac{56}{11} = 56 : 11$

(iii) Perimeter of the playground and parking area = 2(Length of playground) + Breadth of playground + Circumference of Semicircle

= 2(14) + 7 + π r

$= 28 + 7 + \dfrac{22}{7}\times \dfrac{7}{2}$

$= 35 + 11 = 46$ units

Cost of fencing per unit = Rs 2

Cost of fencing = 2 × Perimeter of the playground and parking area

= 2(46) = Rs 92

Case Study 2

Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey Rs x per student and Cricket Rs y per student. School ‘P’ decided to award a total of Rs 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award Rs 7,370 for the two games 4 and 3 students respectively.

Based on the given information, answer the following questions:

(i) Represent the following information algebraically (in terms of x and y). 1

(ii) (a) What is the prize amount for hockey ? 2

(b) Prize amount on which game is more and by how much ? 2

(iii) What will be the total prize amount if there are 2 students each from two games ? 1 [CBSE 2023]

Solution :

Given,

Award prizes to their students for games of Hockey = Rs x per student

Award prizes to their students for games of Cricket = Rs y per student

For School ‘P’

5 x + 4 y = 9,500 ……….(i)

For school ‘Q’

4 x + 3y = 7,370 ……….(ii)

(ii) (a) From question (i)

Solving equation (i) and (ii)

5 x + 4 y = 9,500 ……….(i)

4 x + 3y = 7,370 ……….(ii)

Multiplying 3 in (i) and 4 in (ii) and substracting (i) to (ii)

3(5 x + 4 y) – 4(4 x + 3 y) = 3×9500 – 4×7370

⇒ 15 x + 12 y – 16 x – 12 y = 28500 – 29480

⇒ – x = – 980

⇒ x = 980

Putting this value in eq (i)

5 x + 4 y = 9,500

⇒ 5(980) + 4y = 9500

⇒ 4900 + 4 y = 9500

⇒ 4 y = 9500 – 4900

⇒ 4 y = 4600

⇒ y = 1150

The prize amount for hockey = Rs 980

(ii)(b) The prize amount for hockey = Rs 980

and The prize amount for Cricket = Rs 1150

The prize of Cricket is more than the Hockey

Tmore amount of Cricket = 1150 – 980

= Rs 170

(iii) Total prize of 2 student of Hockey and 2 student of Cricket

= 2 x + 2y

= 2(980) + 2(1150)

= 1960 + 2300

= Rs 4,260

Case Study 3

Jagdish has a field which is in the shape of right angle triangle ΔAQC. He want to leave a space in the form of square PQRS inside the field of growing wheat and remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.

Based on the above information, answer the following questions :

(i) Taking O as origin, coordinate of P are (-200, 0) and of Q are (200, 0). PQRS being a square, what are the coordinates of R and S ? 1

(ii) (a) What is the area of square PQRS ? 2

(ii) (b) What is the length of diagonal PR in square PQRS ? 2

(iii) If S divides CA in the ratio K:1, what is the value of K, where point A is (200, 800) ? 1

[CBSE 2023]

Solution:

(i) Taking O as origin, the cordinate of P(-200, 0) and Q(200, 0). PQRS is a square

From the graph

Coordinate of R is (200, 400) and the coordinate of S is (-200, 400)

(ii)(a) Side of a square PQRS = PQ = PO + OQ

⇒ PQ = 200 + 200 = 400 units

Area of square(PQRS) = (Side)²

= (400)² = 160000 sq units

(ii)(b) Length of the diagonal of square(PR) = √2×side

= √2×400 = 400√2 units

(iii) Given point A(200, 800) and Point S is (-200, 400)

Here Point C is (-600, 0)

For calculation of K we have to use Section formula

Since, S divides CA in K : 1 Then

$S(-200, 400) = (X, Y), C(-600, 0) = (x_1, y_1), A(200, 800) = (x_2, y_2)$

Then,

$X = \dfrac{Kx_2 + x_1}{K + 1}, Y = \dfrac{Ky_2 + y_1}{K + 1}$

Using $Y = \dfrac{Ky_2 + y_1}{K + 1}$

$400 =\dfrac{K\times 800 + 0}{K + 1}$

$\Rightarrow 400(K + 1) = 800

⇒ 400 K + 400 = 800

⇒ 400 K = 400

⇒ K = 1

Therefor the value of K is 1.

Case Study Question Class 10

Solution :

OR

Case Study 2

Solution :

Case Study 3

Solution:

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