Sort all of the following triangle side lengths into if each set would form a right, acute, or obtuse triangle.

  acute, acute, obtuse*, right, obtuse

Step-by-step explanation:

The Pythagorean theorem is used as a reference for classifying triangles from their side lengths. If the sum of the squares of the short sides is equal to the square of the long side, then the triangle is a right triangle. If the long side is too short, then triangle is acute. If the long side is too long, the triangle is obtuse.

__

The math can be tedious, so it is convenient to let a spreadsheet do it. The results are attached.

The first two triangles are acute; the 4th one is a right triangle, and the remaining two are obtuse.

_____

Additional comment

* Note that triangle #3, with side lengths 2, 4, 6, is classified as "obtuse" because the largest angle is greater than 90°. In fact, with those side lengths, the largest angle is 180°. Some would argue it isn't a triangle at all because its area is 0 and it looks like a line segment.

(Our spreadsheet doesn't care if the triangle inequality is not satisifed. It is only concerned with the Pythagorean theorem.)

news channels,news papers,and the internet

∠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.