Work out the surface area of this solid quarter cylinder in terms of pi. r 10cm. h 16​

9514 1404 393

  (320 +130π) cm²

Step-by-step explanation:

The perimeter of the base will be the sum of two radii and the arc length of a quarter circle:

  P = 2r +r(π/2) = r(2+π/2)

For a radius of 10 cm, the perimeter of the base is ...

  P = (10 cm)(2+π/2) = (20+5π) cm

The lateral area of the quarter-cylinder is the product of this perimeter and the height:

  LA = Ph = ((20 +5π) cm)(16 cm) = (320 +80π) cm²

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The total base area is the area of a half-circle of radius 10 cm, so is ...

  BA = 1/2πr² = (1/2)π(10 cm)² = 50π cm²

The total surface area is the sum of the base area and the lateral area:

  SA = BA +LA = (50π +(320 +80π)) cm² = (320 +130π) cm²

∠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.