The owner of the Rancho Grande has 3,044 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions (in yd) of the largest area he can enclose? shorter side? longer side? what is the area?

Step-by-step explanation:

Area of the rectangular piece A = xy

x is the length

y is the width

Perimeter =2x+2y

Given

Perimeter = 3044yds

3044 = 2x+2y

1522 = x+y

x = 1522-y

Substitute into the formula for area

A = xy

A = (1522-y)y

A = 1522y-y²

To maximize the area, dA/dy must be zero

dA/dy = 1522-2y

0 = 1522-2y

2y = 1522

y = 1522/2

y = 761

Since x+y = 1522

X+761 = 1522

X = 761

Dimension is 761yd by 761yd

Area = 761×761 = 579121yd²

∠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.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6