Question

Case study 2: Satellite Images : Satellite images are images of Earth collected by imaging satellites operated by governments and businesses around. In this view, his house is pointed out by a flag, which is situated at the point of intersection of and y-axes. If he goes 2 cm east and 3 cm north . Basus from the house, then he reaches to a grocery store. If how he goes 4 cm west and 6 cm south from the house, then he reaches to his office. If he goes 6 cm east and 8 cm south from the house, then he reaches to a food court. If he goes 6 cm west and 8 cm north from the house, he reaches to a his kid's school. Based on the above information, answer the following questions. B 46) What is the distance between grocery store and food court? (A) V137cm (B) V129cm (C) 8 15 cm V (D) 16V3 cm 47. What is the distance of the school from the house? (A) 10 cm (B) 15 cm (C) 20 cm (D) 25 cm 48) If the grocery store and office lie on a line, what is the ratio of distance of house from grocery store to that from office? (A) 2:1 (B) 3:1 (C)4:1 (D) 5:1 49) What is the ratio of distances of house from school to food court. (A) 1:1 (B) 2:1 (C) 3:1 (D) 4:1 56) What shape is formed by the coordinates of positions of school, grocery store, food court and office? (A) square (B) rectangle (C) rhombus (D) quadrilateral​

Answer 1

46) a) √137

47) a) 10cm

48) a) 2:1

49) a) 1:1

50) b) rectangle

Every answer is correct.

Answer 2

1) Distance from the grocery store to food court = √137 cm

2) Distance of school from house = 10 cm

3) Ratio of distance of house from groery store to that of office = 2:1

4) Ratio of distance of house from school to food court =  1:1

The above question can be solved using the pythagoras formula where the length of the hypotenuse in a right angled trirangle is equal to the square root of the sum of squares of the adjacent sides.

Solving accordingly,

1) The length of adjacent sides formed by the right angled triangle between grocery store and the food court are 11cm and 4cm.

so, Distance between grocery store and the food court = √(11²)+(4²)

= √121+16

=√137 cm

2) The length of adjacent sides formed by the right angled triangle between school and the house are 8cm and 6cm.

∴ Distance between school and the house = √(8²)+(6²)

= √64+36

= √100

= 10cm

3) Solving similarly,

Distance of house from grocery store = √(2²)+(3²)

= √4+9

= √13

and Distance of house from office = √(4²)+(6²)

= √16+36

= √52

Ratio of the two distances = √$\frac{52}{13}$

= √$\frac{4}{1}$

= 2:1

4) Similar to no (3)

Ratio of the two distances = $\frac{10}{10}$

= 1:1