Solutions to Assignment 5
1. If we have a box, we are referring to a right rectangular prism. So, the surface area is
SA = 2AB + P h = 2lw + (2l + 2w)h.
If each dimension is doubled, then we have
2(2l)(2w) + 2((2l) + 2w))(2h)
= 4(2lw) + 4(2l + 2w)h
= 4SA
If each dimension is tripled, then we have
2(3l)(3w) + 2((3l) + 3w))(3h)
= 9(2lw) + 9(2l + 2w)h
= 9SA
We can generalize this to that if we increase each dimension by a factor of n, the total surface
area increases by a factor of n2 .
2. The lateral surface area of a cone is SAl = 21 Cl.
If l is tripled, then SAl = 12 C(3l), so our lateral surface area would be three times larger.
If the radius is tripled only, then we would have SAl = 12 (2π(3r)l, so again, we would have
tripled the lateral surface area.
In both cases, this is because the dimension being changed is taken only to the first power in
the formula.
3.
Based on our picture, we can see that the slant height l = 1.5 m. We also need to know the
240◦
2
radius of the base. Now, we can see that the central angle is 240◦ , so we have 260
◦ = 3
of a circle in this lateral face. If we had the whole circle, the circumference would be
C = 2π(1.5) = 3π m. But, since we have only 32 of a circle, the circumference of this
base is 2π m.
Now, if the lateral face has an arc of 2π m, then the circumference of the base must also be
2π m. So, we have
CB = 2π = 2πr ⇒ r = 1 m
Now we have all of the measurements we need.
The lateral surface area of the cone:
1
SAl = Cl
2
1
= 2π(1)(1.5)
2
= 1.5π m2
The total surface area:
SA = 1.5π m2 + AB
= 1.5π m2 + π(r)2
= 1.5π m2 + π(1)2
= 1.5π m2 + π m2
= 2.5π m2
4.
This figure is called a frustum. To find the area, we will find the lateral surface area of the
whole cone and subtract that of the top cone. Then we will add in the area of the bases.
Consider the cross section of the whole cone.
x
60 cm
40 cm
100 cm
Because both triangles have right angles and the top angle is in both, we have similar triangles
by AA. Therefore, can set up the proportion
x
x + 40
=
60
100
100x = 60x + 2400
40x = 2400
x = 60 cm
So, we have isosceles triangles here. Since they are also right triangles, we know at both
√ congruent angles are 45◦ and so we have that
the
hypotenuse
of
the
large
triangle
is
100
2 cm
√
and the that for the small triangle is 60 2 cm.
The lateral surface area of the small cone would therefore be
√
√
1
1
SAl = Cl = (2π(60))60 2 = 3600 2π cm2
2
2
and for the large cone, we have
√
√
1
1
SAl = Cl = (2π(100))100 2 = 10000 2π cm2
2
2
This means that
√ the lateral surface area of the frustum is the difference between these two,
which is 6400 2 cm2 .
Therefore, the total surface area for the frustum is this lateral surface area plus the area of
the bases.
√
SA = 6400 2 + π(100)2 + π(60)2
| {z }
large base
| {z }
small base
√
= (6400 2 + 13600)π cm2
5.
We need another cross section here to find all of the values we need.
40 cm
r
30 cm
25 cm
As with the last example, we have AA similarity because both are right triangles and the two
triangles share the top angle. Therefore, we can set up a proportion to find the radius we
need.
40
10
=
25
r
40r = 25
25
r=
cm
4
Now that we have the radius of the base of the cylinder, we can find the lateral surface area.
SAl = 2πrh = 2π
25
(30) = 375π cm2
4
6. We will assume we have a perfect cylinder for the can and that the tennis balls are perfect
spheres. So let’s consider the cross section of the can.
r
If the radius of the balls is r units, the length of the base of the can is 2r units and the height
is 6r units. Using this, we have all of the information we need to answer the questions.
The circumference of the can is 2πr and 2πr > 6r, so the circumference is greater.
Now we look at the surface areas. For the can, we have
SAl = 2πrh
= 2πr(6r)
= 12πr2
And, since there are 3 tennis balls, we have
3SAB = 3(4πr2 ) = 12πr2
So, the lateral surface area is exactly the same as that of the three tennis balls.