Question

The given figure shows the distances covered by Ariv from
place A to place H. His direction of movement from A to
B, C to D, E to F and G to H is due North and that
from B to C. D to E and F to G is due East. Find the
aerial distance between A and H.
​

Answer 1

Answer:

your answer is in image

Answer 2

Answer:

The distance between A and H is 65 km.

Step-by-step explanation:

Let O be a point,

Such that,

AO = AB + CD + EF + GH,

OH = BC + DE + FG

By the diagram,

AB = CD = EF = GH = 14 km,

BC = DE = FG = 11 km,

Thus, AO = 14 + 14 + 14 + 14 = 56,

OH = 11 + 11 + 11 = 33

By the Pythagoras theorem,

$AH^2 = AO^2 + OH^2$

$=56^2 + 33^2$

$=3136+1089$

$=4225$

$\implies AH = \sqrt{4225}=65$

Hence, the distance between A and H is 65 km.