Polygon wxyz has vertices w (-1.5, 1.5)x(6, 1.5) y(6, -4.5) and z (-1.5, -4.5 )is wxyz a rectangle

WXYZ is a rectangle

Step-by-step explanation:

∵ W = (-1.5 , 1.5) , X = (6 , 1.5)

∴ WX is a horizontal segment ⇒ same y-coordinates

∴ WX = 6 - -1.5 = 6 + 1.5 = 7.5 units ⇒ (1)

∵ Y = (6 , -4.5) , X = (6 , 1.5)

∴ XY is a vertical segment ⇒ same x-coordinates

∴ XY = 1.5 - -4.5 = 1.5 + 4.5 = 6 units ⇒ (2)

From (1) and (2) WX ⊥ XY

∵ Y = (6 , -4.5) , Z = (-1.5 , -4.5)

∴ YZ is a horizontal segment ⇒ same y-coordinates

∴ YZ = 6 - -1.5 = 6 + 1.5 = 7.5 units ⇒ (3)

From (1) and (3)

∴ WX // YZ and WX = YZ

∴ WXYZ is a rectangle ⇒ 2 opposite sides are parallel and equal

  and 2 adjacent sides are ⊥

a. -0.3 > -0.33

∠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.