Drew is going to his friend’s house after school and wants to track how far he walks. he first walks 1.25 miles north, then 0.75 miles east, and finally 0.25 miles south to get to his friend’s house. when drew leaves, his friend tells him that there’s a shortcut through the park to walk straight back to his house, 1.25 miles.

a rectangular shape of park split into a standard triangle and a polygon

how many miles does drew save by taking the shortcut instead of retracing the same route back home?

Drew saves a distance of by taking the shortcut.

Step-by-step explanation:

Here is an image of the routes.

We have drawn it assigning the signs NSEW where N=North,S=South ,E=East and W=West.

Now the total distance covered by Drew = OA+AB+BC

We see by joining C with O it has a shortcut.

And it forms a right angled triangle.

, at D.

By using Pythagoras theorem we can find the shortcut distance.

OC is the hypotenuse and

So Drew saves of distance following the advice of his friend and opting for the shortcut route.

sorry, i can't understand your question. will later if you describe more clearly!

: )

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.