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I) find, in terms of π, the curved surface area of a cone with circular base diameter 10cm and height 12cm.
Ii) if the cone is made of paper, and the paper is flattened out into the sector of a circle, what is the angle of the sector?
[tex]\pi[/tex]
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Sagot :

MidnightShowers

Answer:

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( i )

  • given - a cylinder with base diameter = 10 cm ! And , height = 12 cm

  • To find - CSA of cone

now ,

since diameter = 10 cm

[tex]\implies \: radius = \frac{10}{2} = 5 \: cm \\ [/tex]

now ,

[tex]slant \: height \: ( \: l \: ) = \sqrt{h {}^{2} + r { }^{2} } \\ \\ \implies \: \sqrt{(12) {}^{2} + (5) {}^{2} } \\ \\ \implies \: \sqrt{144 + 25} \\ \\ \implies \: \sqrt{169} \\ \\ \implies \: 13 \: cm[/tex]

now ,

[tex]CSA \: of \: cone = \pi \: \times r \times l \\ \\ \implies \: \pi \: \times 5 \times 13 \\ \\ \implies \: 65\pi \: cm {}^{2} [/tex]

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( ii )

now when we flatten the cone ,

the slant height becomes the radius of the sector , while the arc becomes the circumference of cone !

[tex]\therefore \: radius = l = 13 \: cm \\ \\ arc \: length = circumference = 2\pi \: r[/tex]

now ,

[tex]s = d \: \theta \: ( \: for \: a \: sector \: ) \\ \\ \therefore \: l\theta \: = 2\pi \: r \\ \\ \implies \: \theta = \frac{2\pi \: r}{l} \\ \\ \implies \: \theta = \frac{2 \times \pi \times 5}{13} \\ \\ \implies \: \theta \: = \frac{10\pi}{13} \\ \\ \implies \theta \: = \: 0.77 \pi [/tex]

hope helpful :D

reinhard10158

Step-by-step explanation:

i)

remember, the radius is half the diameter !

the overall surface area of a cone (base circle + "mantle", the lateral, curved surface) is

A = pi×r × (r + sqrt(h² + r²))

so, only the "mantle" or lateral curved stave area is

pi×r × sqrt(h² + r²) = pi×5 × sqrt(12² + 5²) = pi×5×sqrt(169) =

= pi×5×13 = 65×pi cm²

ii)

the length of the arc of the sector is the circumference of the circle, that was the base of the original cone.

that is

2×pi×r = 2×pi×5 = 10×pi cm

the total circumference of the large circle containing that sector is

2×pi×r' = 2×pi×sqrt(h² + r²)

as the radius r' of the circle containing the sector (representing the "mantle" of the cone) is the lateral height of the cone (the distance along its side from the base to the top). and that is per Pythagoras simply the square root of the sum of the square of the inner height and the square of the base radius.

in our case that is

2×pi×sqrt(12² + 5²) = 2×pi×13 = 26×pi cm

the angle of the sector is then 10/26 of the total of 360° for the whole circle :

360 × 10/26 = 138.4615385...°

or, if you need this in terms of pi too, remember that 360° are represented by 2×pi (the arc length of the standard circle).

2×pi × 10/26 = pi × 20/26 = pi × 10/13 =

= 0.769230769... × pi

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