Calculate distance

Question

Amy and Mel are on a treasure hunt. They start at the same point and run in straight lines to each clue. They follow different routes based on their clues, which are not the same.

Amy's first clue is 50m south-west of the starting point.

Mel reaches her first clue when she is 120m north-west of Amy's first clue.

Mel then makes a quarter turn clockwise and runs the same distance to her second clue as she ran to her first clue.

How far is Mel's second clue from the starting point, to the nearest metre?

Answer 1

Set up the points on a rectangular coordinate system where North and South are vertical and East and West are horizontal. Southwest is, from 0,0, diagonal in the third quadrant. Northwest is a diagonal from 0,0 in the second quadrant. Draw your lines and mark your distances. Read carefully as Mel's first direction is from Amy's first clue. Quarter turn is 90 degrees. Once there mark your distance. The answer is obtained using the Pythagorean theorem:

a² + b² = c², 'c' is always the hypotenuse. Once you square the legs, add the numbers together and then square root to find the distance requested.

Answer 2

First I make a sketch to visualize the changes, then I can tabulate the changes in x and y. First leg, from the origin to point A1, is SW so both x and y change by -50 cos 45 = -50 sin 45 = -50 √2 /2. Second leg to point M1 is NW from there with Δx = -120 √2 /2 and Δy = +120 √2 /2. Making a quarter turn clockwise, Mel is now running NE so Δx = +120 √2 /2 and Δy = +120 √2 /2 to point M2. Total Δx = [-50 -120 +120] √2 /2 = -25 √2. Total Δy = [-50 +120 +120] √2 /2 = 95 √2. Total distance from starting point: x² = 625*2 + 9025*2 = 19,300 and x = 139m.