Question

The semicircle shown is folded to form a right cylinder cone so that the arc PQ becomes the circumference of the base. Given that the diameter of the semicircle, PQ, is 10cm.
Find:
(i) the diameter if the base,
(ii) the curved surface area of the cone.

Answer 1

Answer: The diameter of the base of cone is 5 cm.

The curved surface area of the cone is 39.27 $cm^{2}$.

Step-by-step explanation:

The concept used behind the following problem is that the

1. The circumference of the semicircle becomes the circumference of the base of the right cylinder cone.

2. Radius of the circle becomes the slant height of the right cylindrical cone.

i) For the diameter of the base

Circumference of the semicircle = πR or 0.5πD

where R is the radius of the semicircle and D is the diameter of the semicircle.

Diameter of the given semicircle = 10 cm

Circumference of semicircle = 5π cm.

Circumference of the right cylinder cone =  2πr or πd

We know that,

Circumference of the semicircle = Circumference of the base of the cone

5π = πd

d = 5 cm

r = 2.5 cm

So, the diameter of the base of the right cylindrical cone is 5 cm.

ii) For the curved surface area of the cone

Radius of the semicircle = Diameter ÷ 2 = 5 cm

So, Slant height (l) of the right cylinder cone = 5 cm

We know that,

The Curved Surface Area of the right cylindrical cone = π×r ×l

where r= radius of the cone and l= slant height of the cone.

C.S.A of the cone = π×2.5×5

                             = 39.2699 $cm^{2}$

So, the curved surface area of the cone is 39.27 $cm^{2}$.

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References

1. https://brainly.com/question/27155435

2. https://brainly.com/question/10466463