Question

SECTION - 11
Case study based questions are compulsory. Attempt any four sub parts of each question.
Each subpart carries 1 mark.
17
Aravind goes for a picnic from his school to a park in his town. The plan of
the town is given below. If his house is situated at (100, 450), the school at
(-400,150) and park entrance at (300,-250) and the coordinates are in
metres, assuming the distances are covered in straight lines, answer the
following questions;
500?
4 (100,450)
409
300
s(-430,150)
200
100
1000 400
2034
P (300,- 250)
-300
30
1
1
-5001
What distance does Aravind cover to reach his school from his home?
(a)1100m
(b) 100 V61m
(c) 600 m
(d) 300 V2m
What is the distance between the school and park?
(a)1100m
(b) 100 V65m
(c) 100v5 m
(d) 100 Vām
Aravind went to the same park the next day with his family from his home.
What is the distance travelled?
(a)300m
(b) 100 V65m
(0710015 m
(d) 100 V53m
1
SHI)
1
(iv)
There is an ice cream parlour Q in the way between his house and park
which divides the distance HP in the ratio 2:3. What are its coordinates?
(a)(180,170) (b) (200,100) (c)(400, 700) d) (100,-350)
1
(v)
There is a stationary R in the midway between his house and school. What
are its coordinates?
(a)(-100,200) (b) (200,100) (c)(- 150,300) d) (900,425)​

Answer 1

Answers:

(I) 100√34 ( Answer is not in the options).

(ii) 100√65

(iii) 100√53

(iv) (180,170)

(v) (-150, 300)

Steps:

(i) Distance formula-

= √((100+400)^2 + (450-150)^2)

= √ ( 500^2 + 300^2)

=√(250000+900000

= √340000 = 100√34

(ii) Distance formula-

=√((300+400)^2 + (-250-150)^2)

=√(490000+160000)

=√650000 =100√65

(iii) Distance formula-

=√((100-300)^2+(450+250)^2)

=√(40000+490000)

=√530000 =100√53

(iv) Section formula-

x= (2*300+3*100)/2+3

y= (2*250+3*450)/2+3

x= 180

y= 170

coordinates= (180,170)

(v) Mid-point formula-

x= (100-400)/2

y= (450+150)/2

x= -150

y= 300

coordinates= (-150,300)

= Hope I was able to help you .

Answer 2

Given : Arvind goes from School to Park

House ( 100 , 450)

School ( - 400 , 150)

Park entrance (300 , -250)

To Find : Distance Arvind cover from home to school

Solution:

Distance Arvind cover from home to school

House ( 100 , 450)  School ( - 400 , 150)

Distance covered = √(-400 - 100)² + (150 - (450))²

= √500² + 600²

= 100√25 +36

= 100√61  m

Distance  from School to Park

School ( - 400 , 150) Park   (300 , -250)

Distance covered = √(-400 - 300)² + (150 - (-250))²

= √700² + 400²

= 100√49 + 16

= 100√65 m

Distance  from home to park

House ( 100 , 450)  Park   (300 , -250)

Distance covered = √(100 - 300)² + (450 - (-250))²

= √(-200)² + 700²

= 100√4 + 49

= 100√53 m

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