Question

from point x ankit walks 112m east to reach at point y .from point y ankit walks 15m towards north to reach point z . what is the straight line distance postion when he started betwwen position when he started and his postion now?

117m
113m
123m
127m

Answer 1

Answer:

${ \huge\fbox \pink {A}\fbox \blue {N} \fbox \purple {S} \fbox \green{W} \fbox \red {E} \fbox \orange {R}}$

Step-by-step explanation:

113 M

Answer 2

Given: From point x ankit walks 112m east to reach at point y. Then from point y he walks 15m towards north to reach point z.

To find: The straight line distance between the position when he started and his postion now

Solution: W can understand from the given question that the points x, y and z form a right angled triangle where xy is the base and yz is the perpendicular.

The required distance which we need to find is the hypotenuse zx of this triangle.

Here, xy = 112 m and yz = 15 m.

Using Pythagoras' theorem,

(zx)² = (xy)² + (yz)²

⇒ (zx)² = (112)² + (15)²

⇒ (zx)² = 12544 + 225

⇒ (zx)² = 12769

⇒ zx = 113 m

Therefore, the distance of the straight line between the position when he started and his postion now is 113 m

Answer: 113 m