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Starting from point O on a flat school playground, a child walks 10 yards due north, then 6 yards due east, and then 2 yards due south, arriving at point P. How far apart , in yards, are points O and P?

How do you get the conclusion?

2007-04-09 20:00:03 · 5 answers · asked by Judo 1 in Science & Mathematics Mathematics

5 answers

pythagorean theorem:
c^2 = a^2 + b^2

try to draw it in your paper so you'll get what i mean..

hope you get the picture ^_^

top = side b = 6 yards
************* _ _ _ _ _ _
all in all*****|************|
10yards****|_ _ _ _ _ _| 2 yards
*************|************/
*************|**********/
*************|********/
*************|******/
****side a =|****/ side C = ?
*****8 yards|**/
*************|*/
*************|/


left side is 10 yards
top is 6 yards
right side is 2 yards

you'll get a triangle with one leg = 8 yards and the other leg - 6yards...

to get the distance from O to P use pythagorean theorem in solving side C...

c^2 = 8^2 + 6^2
c^2 = 64 + 36
c^2 = 100
c = 10 yards

hope this helps

2007-04-09 20:11:17 · answer #1 · answered by kishimii 2 · 2 0

Point O is 10 yards from Point P.

Draw it out... 10 yards north, 6 yards east and 2 yards south. Make a triangle from point P by drawing a straight line going west intersecting the 10 yard line. You have a right triangle. 8 yards by 6 yards, solve for the hypotenuse using Pythagorean's Theorem.

A^2 + B^2 = C^2
8^2 + 6^2 = C^2
C= 10 yards

2007-04-09 20:16:06 · answer #2 · answered by shagohod77 2 · 0 0

walking 10 yards north then 2 yards south is like walking 8 yards north. so you end up with 8 yards north and 6 yards east, which looks like a right triangle with sides 8 and 6
so use the pythagorean thereom.
U get 6^2+8^2=x^2
36+64=x^2
100=x^2
x=10 yards

2007-04-09 20:12:56 · answer #3 · answered by Sam 5 · 0 0

The resultant vector is 8 yards north and 6 yards east.
By Pythagoras, (OP)² = 6² + 8² and OP is 10 yards.

2007-04-09 21:21:43 · answer #4 · answered by Zax 3 · 0 0

Let's look at the coordinates of the points the childs stops at each step:

O - start point (x0, y0)
10 yards to the north (x0 + 10, y0)
2 yards due east (x0 + 10, y0 + 6)
P - 2 yards due south (x0 + 10 - 2, y0+6) = (x0 + 8, y0 + 6)

The distance from O to P is = sqrt( {[(x0 + 8) - x0]^2 +
+ [(y0 + 6) - y0]^2} ) = sqrt(8^2 + 6^2) = sqrt(64 + 36) =
= sqrt(100) = 10

2007-04-09 20:11:22 · answer #5 · answered by Amit Y 5 · 0 0

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