The area of a circle is 28pi cm^2. find the area of a circle with a radius half the length of the original circle.

The area of circle with radius half of original circle is 7 π  cm² .

Step-by-step explanation:

Given as :

The Area of original circle = 28 π square centimeter

Let The radius of original circle  = R

Let The area of circle with radius half of original circle = A square centimeters

Let The radius of circle with radius half of original circle = R'

Now, According to question

∵ Area of original circle =  π × radius²

So, 28 π cm² = π × R²

Or, R² =

Or, R² = 28 cm²

∴  R = cm

i.e R = 2 cm

So, The radius of original circle = R =  2 cm

Again

∵ The radius of circle with radius half of original circle = R'

So, R' =

i.e  R' =

∴  R' =   cm

So, The area of circle with radius half of original circle =  π × R'²

i.e A = π × R'²

Or, A = π ×() ²

or, A =  π × 7

So, The area of circle with radius half of original circle = 7 π  cm²

Hence The area of circle with radius half of original circle is 7 π  cm² . Answer

∠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.