f3 area and volume.

Question 17 part A

Volume of cone = 1/3 * π * r^2 * h
where
r = radius
h= height

605π=1/3 * π * 11^2 * h
****SOLVE H****
h= 15cm

Question 17 part B
Total surface area = area of circle + area of curved surface area
=(π*r^2) + (πrL)

To find L, we will use Pythagoras' theorem:
a^2 + b^2= c^2
11^2 + 15^2 = c^2
346 = c^2
c= 18.60107524cm (THIS IS THE SLANT HEIGHT)

Total surface area = (π*r^2) + (πrL)
=(π*11^2) + (π(11)(18.60107524))
=1022.939726cm^2
=1022.9cm^2 (to 1 dp)

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Second Question
a) i) Circumference of a circle = 2πr
28=2πr
r = 4.456338407cm

ii) Use Pythagoras' theorem:
a^2 + b^2= c^2
where a= radius
b= VO (height)
c= slant height
4.456338407^2 + b^2 = 12^2
b^2=124.141048
b=11.14186017cm

b)Curved surface area of cone formula = πrL
=π *4.456338407 *12
=168cm^2

c) Volume of cone = 1/3 * π * r^2 * h
=1/3 * π * 4.456338407^2 * 15
=311.9436885cm^3