Manisha and sandeep started moving from thesame point in North and East direction respectively. manisha got tired after some times .she talked to sandeep on phone and came to know that he has tarvelled 2 k.m more than her . if distance betwwen them at that very moment is 10 k.m then find the distance travelled by them separetaly?
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With the given information we get a Right angled triangle ABC
AB is distance of manisha travelling to the north
AC is distance of Sandeep towards the east
Given that distance between them is 10km
BC is the hypotenuse = 10km
Let distance covered by Manisha be x km
then distance covered by Sandeep = x+2 km
⇒ AB = x
and AC = x+2
By Pythagoras Theorem
BC² = AB² + AC²
⇒ 10² = x² +(x+2)²
⇒ 100 = x² +x² +4 + 4x
⇒ 100 = 2x² +4x +4
⇒ 2x² +4x +4 - 100 = 0
⇒ 2x² + 4x - 96 = 0
⇒ 2(x² + 2x - 48) = 0
⇒ x² + 2x - 48 = 0
⇒ x² - 8x +6x - 48 = 0
⇒ x(x -8) + 6(x - 8) = 0
⇒ (x - 8) (x+6) = 0
⇒ (x-8) = 0 or (x+6) =0
⇒ x = 8 or x = - 6
Therefore distance travelled by Manisha = x = 8km [-6 can not be considered as distance can not be negative]
Distance covered by Sandeep = x+2 = 8+2 = 10km
AB is distance of manisha travelling to the north
AC is distance of Sandeep towards the east
Given that distance between them is 10km
BC is the hypotenuse = 10km
Let distance covered by Manisha be x km
then distance covered by Sandeep = x+2 km
⇒ AB = x
and AC = x+2
By Pythagoras Theorem
BC² = AB² + AC²
⇒ 10² = x² +(x+2)²
⇒ 100 = x² +x² +4 + 4x
⇒ 100 = 2x² +4x +4
⇒ 2x² +4x +4 - 100 = 0
⇒ 2x² + 4x - 96 = 0
⇒ 2(x² + 2x - 48) = 0
⇒ x² + 2x - 48 = 0
⇒ x² - 8x +6x - 48 = 0
⇒ x(x -8) + 6(x - 8) = 0
⇒ (x - 8) (x+6) = 0
⇒ (x-8) = 0 or (x+6) =0
⇒ x = 8 or x = - 6
Therefore distance travelled by Manisha = x = 8km [-6 can not be considered as distance can not be negative]
Distance covered by Sandeep = x+2 = 8+2 = 10km
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by using pythagoras we can do it
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