Metal End Cone
August 5th, 2008
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Metal End Cone
Grade 9 Geometry Math help?
A seed hopper is in the shape of a cylinder with a diameter of 3.5m and a height of 4.7m. Both ends of the cylinder are capped with cone 2.1m high.
1) Find the number of square metres of sheet metal required to build the hopper.
2) Find the total volume of the hopper.
Please explain thanks 10 Points
Hi !
1)
I assume it looks something like
.._________
/................
._________/
Area of 2 cones
=2 * area of one cone
=2 * pi * r * l
where l is the slanting height of the cone.
l can be found by the pythagorean theorem.
l^2
= r^2 + h^2
= (d/2)^2 +h^2 ; (h=height of cone)
= (3.5/2)^2 + (2.1)^2
= 7.4725
l=√7.4725 =2.734 m
Hence,
Area of 2 cones
=2 * pi * (3.5/2) * 2.734
=30.057m^2
Area of the cylinder
= 2 * pi * r * h; (h=height of cylinder)
= 2 * pi * (3.5/2) * 4.7
= 51.68m^2
Total area of the hopper
= 30.057m^2 + 51.68m^2
= 81.74m^2
2) Total volume of the hopper
= volume of cylinder + volume of cone
Volume of cylinder
= pi * r^2 * h; (h=height of cylinder)
= pi * (3.5/2)^2 * 4.7
= 45.22m^3
Volume of 2 cones
= 2 * (1/3) * pi * r^2 * l
= (2/3) * pi * (3.5/2)^2 * 2.734
= 17.53m^3
Total volume
= 45.22m^3 + 17.53m^3
= 62.75m^3
Hope the above clarifies!
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Forging Pine Cones
Some good mathematical problems?
Question 1 : A given quantity of metal is to be cut into half cylinder with a rectangular base and semicircular
ends. If the total surface area is to be a minimum, then the ratio of the length of the
cylinder to the diameter of its semi-circular ends is
Options are
a)π : (2 + π)
b)π : (1 + π)
c)2π : (3 + π)
d)3π : (8 + 4π)
Question 2 : An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a
horizontal table. Inside this is placed
a solid metallic right circular cone, the diameter of whose base is 3.5 cm and height 8 cm. If
the cone is replaced by another cone, whose height is 1.75 cm and base radius is 2 cm then
the drop in the water level is
Options are :
a)0.62 cm
b)0.48 cm
c)0.88 cm
d)1 cm
I would be thanksful if you could provide me detailed solution of these problem
q1
A = [2πr(r+h)]/2 + 2rh
= πr(r+h) + 2rh
= πr^2 + (π+2)rh
dA/dr = 2πr + (π+2)h = 0
h = 2πr / (π+2)
= πd / (π+2)
h/d = π / (π+2)
h : d = π : π+2
answer = option a)
q2
differed volume
= π[8*3.5^2 - 1.75*(2*2)^2] / 12
= 70π / 12
= 35π / 6
differed volume = (base area)(height)
35π/6 = [π(7/2)^2]h
h = [35π/6]*[4/49π]
= 10/21
= 0.47619
= 0.48 cm
answer = option b)
=
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