As part of summer camp, Henry goes on a treasure hunt. He starts at the base of a tree and walks 180 feet due north. He then turns and walks 80 feet due east. He turns again and walks 80 feet due south . How far is Henry from the treel Round your answer to the nearest foot.

      Henry is 128 feet far from the tree.

Given in the question,

From the figure attached,

AB = CD = 80 feet

AO = AD + OD

180 = 80 + OD

OD = 100 feet

To get the distance from the last location of Henry and base of the tree, apply Pythagoras Theorem in ΔODC.

OC² = OD² + CD²

OC² = (100)² + (80)²

OC =

OC = 128.06 feet

OC ≈ 128 feet

     Therefore, Henry is 128 feet far from the tree.

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128.06 ft

Step-by-step explanation:

First Henry walks 180 ft North

Then 80 feet East and after that 80 ft south

So he is at a direction of North east from the starting Point

Vertically he is at a distance of 100 ft from starting point

and Horizontally at a distance of 80 ft from starting point

So, net distance from starting point is

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)