Question
A paper cone has a base diameter of 8 and a height of 3cm make a sketch of the cone and hence use Pythagoras theorem to calculate the slant height
Answers
Cannot sketch on these posts.
radius = 4
4^2 + 3^2 = x^2
Solve for x.
radius = 4
4^2 + 3^2 = x^2
Solve for x.
Radius of base = 4 cm, height = 3 cm
Volume = (1/3)π(4^2)(3) = 16π cm^3
slant height --- s
s^2 = 4^2 + 3^2
s = 5 , (did your recognize the standard 3-4-5 right angled triangle?)
So the radius of the sector is 5 and the arclength is the circumference of the base of the cone.
Circumference of base = arc of sector = 8π cm
circumference of circle containing our sector = 10π cm, so the area of the sector is 4/5 the area of the big circle
area of sector = (4/5)π(5^2) = 20π cm^2
Of course I could have just used the formula
lateral area of cone = πrl, where r is the radius of the cone and l is the slant height
= π(4)(5) = 20π
for the sector angle:
sector-angle/360 = 8π/10π = 4/5
sector angle = 288°
Volume = (1/3)π(4^2)(3) = 16π cm^3
slant height --- s
s^2 = 4^2 + 3^2
s = 5 , (did your recognize the standard 3-4-5 right angled triangle?)
So the radius of the sector is 5 and the arclength is the circumference of the base of the cone.
Circumference of base = arc of sector = 8π cm
circumference of circle containing our sector = 10π cm, so the area of the sector is 4/5 the area of the big circle
area of sector = (4/5)π(5^2) = 20π cm^2
Of course I could have just used the formula
lateral area of cone = πrl, where r is the radius of the cone and l is the slant height
= π(4)(5) = 20π
for the sector angle:
sector-angle/360 = 8π/10π = 4/5
sector angle = 288°