Let the width of rectangle MATH equal w. The length of the rectangle is 3cm longer than the width. In rectangle GEOM the width is 1 more than w and the length is 5 less than twice w. If their perimeters are equal, find w.

7 cm

Step-by-step explanation:

The perimeter P of a rectangle is given as

p = 2 (l + b) where l and b are the length and width respectively

hence for a rectangle where  the length of the rectangle is 3cm longer than the width w

the perimeter

= 2 (w + 3 + w)

= 2(2w + 3)

In another rectangle where the width is 1 more than w and the length is 5 less than twice w, the perimeter

= 2(2w - 5 + w +1)

= 2 (3w - 4)

since the perimeters are equal, equate both equations

2(2w + 3) = 2 (3w - 4)

2w + 3 = 3w -4

collect like terms

3w - 2w = 4 + 3

w = 7cm

you answer is a

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.