The volume of a cube is 7.14 cubic meters. what is the cross-sectional area that is parallel to one of its faces?
a =

A = 3.706 square meters

Step-by-step explanation:

the length of a cube has equal side.

therefore, the volume of a cube is given by S³

V = S³ = 7.14

where S is the length of a side

the surface area of a cube is = 6S²

where the area a aside is calculated, then it is multiply by 6.

if S³ = 7.14

S = ∛(7.14) = 1.925 m

What is the cross-sectional area that is parallel to one of its faces?

this is saying we should calculate for the cross-sectional area of one face. All faces in a cube is equal and parallel to each other.

the crossectional  area of one side of a cube is = S² = 1.925² = 3.706 square meters

A = 3.706 square meters

∠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.