I) Yellow colour chart paper required to make one penholder a) 616cm2 (b) 154cm2 (c) 308cm2 (d) 539cm2

(ii) Blue colour chart paper required to make one penholder
(a) 308cm2 (b) 616cm2 (c) 539cm2 (d) 154cm2

(iii) TSA of one pen holder
(a) 385cm2 (b) 770cm2 (c) 700cm2 (d) 1232cm2

(iv) Total requirement of yellow colour chart paper
(a) 30800cm2 (b) 3850cm2 (c) 5390cm2 (d) 7700cm2

(v) Total requirement of blue colour chart paper
(a) 15400cm2 (b) 7700cm2 (c) 30800cm2 (d) 61600cm2

step-by-step explanation:

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∠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.