# 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

OR

(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

units

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

Area of semicircle

units

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

[From (ii) (a) ]

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

= 2(14) + 7 + π r

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

OR

(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

OR

(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

OR

(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

OR

(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

Then,

Using

\$\Rightarrow 400(K + 1) = 800

⇒ 400 K + 400 = 800

⇒ 400 K = 400

K = 1

Therefor the value of K is  1.