The area of a rectangular garden is 308 square meters. the width is 3 meters longer than one-half of the length. find the length and the width of the garden.

Length (L): L

width (w): 1/2L + 3

Area (A) = L x w

      308 = (L)(1/2L + 3)

      308 = (1/2)L² + 3L

      616 = L² + 6L

        0 = L² + 6L - 616

        0 = (L - 22)(L + 28)

0 = (L - 22)            0 = (L + 28)

  L = 22                   L = -28

Since length cannot be negative, disregard the -28

L = 22

width (w): 1/2L + 3   = 1/2(22) + 3   = 11 + 3   = 14

width = 14 meters, length = 22 meters

point a is the center of this circle.

the ratio of the lengths of (ef/bc) and (ad/ih) is 2: 1.

step-by-step explanation:

∠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.