Math, asked by skahmedsaud, 9 months ago

7. Sang Jae walks a certain distance due North and then the same
distance plus a further 7 km due East. If the final distance from the
starting point is 17 km, find the distances he walks North and East.​

Answers

Answered by SohamAgarwal
68

Answer:

8 km in North and 15 km in east.

Step-by-step explanation:

Let the distance she walks in North be x. As North is perpendicular to East direction, therefore her total journey can be represented in the form of a right angled triangle.

By pythagoras theorem,

(Base)^2 + (Height)^2 = (Hypotenuse)^2

(Distance traveled in North)^2+(Distance traveled in East)^2= (Displacement)^2

= (x)^2 + (x + 7)^2 = (17)^2

= x^2 + x^2 + 49 + 14x = 289

= 2x^2 + 14x = 240

Eliminating 2 from each side

 x^2 + 7x = 120

x^2 + 7x -120 = 0

x^2 -8x + 15x -120 = 0

x(x - 8) + 15(x - 8) = 0

(x + 15)(x - 8) = 0

x = -15, 8

However, distance traveled in North direction cannot be in negative so x is 8 km. As the distance traveled in East is 7 more than that traveled in North, so it will be 15 km.

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