Math, asked by jayanthdfcslic, 1 year ago

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?


Devinitza: is any further clues provided

Answers

Answered by yasummu
2
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

yasummu: The answer from By pythagoras theorem is 
yasummu: 10^2 = x^2+(x+2)^2            => 100 = x^2 +x^2 +4+4x     => 100 = 2x^2 +4x+4   => 0 = 2x^2 + 4x +4 -100    => 2x^2 + 4x -96    => 2(x^2 +2x+ 48) = 0      => x^2 + 2x +48 = 0   => x^2 + 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
yasummu: Therefore distance travelled by Manisha = x =6km   and  by sandeep = x+2 = 6+2 =8
yasummu: sorry for this discomfort but if you will please mark it as brainliest
Answered by lahari2001
0
by using pythagoras we can do it

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