Kylie needs to pack her baton for a
color-guard competition. The baton
is 38 inches long. She has a rectangular
box with a base of 13 inches by 35
inches and a height of 13 inches.
Part A
Could the baton lie flat on a diagonal
along the base of the box ? Explain.
Part B
Could the baton fit along the interior
diagonal of the box? Explain.

Step-by-step explanation:

For A, we would need to use the Pythagorean theorem. We have 13^2+35^2=38^2. Is this true? No It is not, 13^2 + 35^2 = 1394. The square root of that is about 37.3, so it does not fit.

For B, we would need to use the Pythagorean theorem again, but in a different way . We would need to find 13^2 + 35^2, and then square root that. That would be our base of our right triangle. The height would be 13, and we use the Pythagorean theorem again! We get 169+1394=1563. The square root of 1563 is about 39.5, so it does fit.

mental cunning

explanation:

he is often referred to as a man of twists and turns

∠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.