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
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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