Starting from point O on a flat school playground, a child walks 10 y

Question

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?

(A) 18
(B) 16
(C) 14
(D) 12
(E) 10

(A) 18 
 (B) 16 
 (C) 14 
 (D) 12 
 (E) 10

Let point O be the origin. 
 
moving 10 yards due north = (0,10). 
moving 6 yards due east = (6,10). 
moving 2 yards due south = (6,8) =Final point P.

The distance from origin O(0,0) to point P(6,8) = 10.

Even though I did it in this way, the best way is to not use the coordinates.

chetan2u · Answer

Work on the movement in east-west and north south Direction as all moves will be at 90 degree and finally we will have a Right angle triangle..
N-S = 10 UP and 2 down = 10-2 = 8....
E-W = 6..
HYP =  \sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10 
E

Abhishek009 · Answer

Distance.png Distance OP = \sqrt{8^2 + 6^2} =>10 Hence answer will be (E) 10

generis · Answer

iiiiii.png The distance in yards between points O and P is 10 The child creates a trapezoid with her walking pattern. The trapezoid can be divided into a rectangle and a right triangle 1) Find shape created by walking pattern. It will contain right angles (with compass points, "due " = 90 degrees) She walks 10 yards due north from point O to point A: OA = 10 Then 6 yards due east from A to B, which means a right angle at A. AB = 6 Then 2 yards due south from B to P. Another right angle. BP = 2 Connect P to O. She has created a trapezoid with her travel 2) Divide the shape into other shapes where distance between start and finish is easy to calculate Draw a line from P, PX on diagram, that is perpendicular to OA The new line divides trapezoid into two new shapes. Upper region is a rectangle. Lower region is right triangle Rectangle has length = 6 , width = 2 The rectangle's width, BP = AX = 2, divides original OA = 10 into two lengths AX and OX: (OA - AX) = OX (10 - 2) = 8 = OX Looking at leg lengths PX and OX: the bottom portion is a 3-4-5 right triangle 3) Find distance between begin and end point Find length of hypotenuse OP Rule: If a right triangle has two legs in ratio 3x:4x, it is a 3x-4x-5x triangle and the hypotenuse = 5x Leg PX = 6 Leg OX = (10 - 2) = 8 (3x: 4x: 5x) = (6: 8: 10 )* Distance between points O and P, in yards: 10 Answer (E) * Or use Pythagorean theorem: 6^2 + 8^2 = OP^2 36 + 64 = OP^2 100 = OP^2 OP = 10

ArnauG · Answer

To determine the distance between points O and P, we can use the concept of displacement.

The child first walks 10 yards due north, which means they move 10 yards in the positive y-direction.

Next, the child walks 6 yards due east, which means they move 6 yards in the positive x-direction.

Finally, the child walks 2 yards due south, which means they move 2 yards in the negative y-direction.

To find the total displacement, we can add the displacements in the x-direction and y-direction separately.

In the x-direction, the child moved 6 yards east.

In the y-direction, the child moved 10 yards north and then 2 yards south, resulting in a net movement of 10 - 2 = 8 yards north.

Thus, the total displacement in the x-direction is 6 yards and in the y-direction is 8 yards.

To find the distance between points O and P, we can use the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides in a right triangle.

Using the Pythagorean theorem, we have:

Distance^2 = (6 yards)^2 + (8 yards)^2 = 36 yards^2 + 64 yards^2 = 100 yards^2

Distance = 10 yards

Therefore, the distance between points O and P is 10 yards, corresponding to option (E).

bumpbot · Answer

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Starting from point O on a flat school playground, a child walks 10 y

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