ABCD, EFGH, and HIJK are squares.
AABCD = 400 cm2, O is the center of the circle, O ∈
FE
,
FG
and
BC
tangent to the circle (see the picture on the right).

Find the length of
AM
.

40/3 cm

Step-by-step explanation:

FE : DC

1 : 3

Since area ABCD = 400

s² = 400

s = 20

FE = ⅓(20) = 20/3 cm

AM = DC - FE

AM = 20 - 20/3 = 40/3

∠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.