A rectangle has a perimeter of 80. if the length is four more than twice the width, what is the length of the rectangle

28

Step-by-step explanation:

l = length

w = width

p = perimeter

w = w

l + l + w + w = p (adding all the rectangle's sides together gets us our perimeter)

l = 2w + 4 (four more than twice the width)

Substitute 2w + 4 for 'l' in the original equation

(2w + 4) + (2w + 4) + w + w = 80

Solve equation normally:

6w + 8 = 80

6w = 72

w = 72/6 = 12

(w = 12)

Length = 2w + 4

Substitute 12 for 'w':

2(12) + 4 = l

24 + 4 = l

28 = l

The length is 28

∠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.