We are asked to draw a with a that is between 50 and 100 square units.
Before we do that, let's remember the formula for the . The surface area S.A. of a cone with
ℓ and
r is given as follows.
S.A.=πrℓ+πr2
We should choose such values of
r and
ℓ that the expression on the right-hand side has a value between
50 and
100.
50<πrℓ+πr2<100
There are infinitely many pairs of numbers that satisfy the . We should choose just one. Since it is easier to focus on just one , let's say that the slant height will always be twice the radius.
ℓ=2r
We can now substitute
2r for
ℓ in the expression and simplify.
50<πrℓ+πr2<100
50<πr(2r)+πr2<100
50<2⋅π⋅r⋅r+πr2<100
50<2⋅π⋅r2+πr2<100
50<2πr2+πr2<100
50<3πr2<100
Now there is just one variable for us to consider. See that we need to choose a value of
r that satisfies the last inequality.
50<3πr2<100
Notice that
r=3 satisfies the condition. We can check that using a calculator. Since the radius of our cone is
3 units, this means that the slant height is
2(3)=6. Let's now find the surface area of the cone.
S.A.=πrℓ+πr2
S.A.=π(3)(6)+π(3)2
S.A.=π(3)(6)+π(9)
S.A.=18π+9π
S.A.=27π
S.A.=84.823001…
S.A.≈84.82
The volume of the cone is about
84.82 units3. Since this more than
50 and less than
100, the dimensions are correct. Last, let's draw the cone.
Note that this is just a sample solution. The slant height does not need to be twice the radius. Also, the radius does not have to be a . We could choose any two numbers r and ℓ that satisfy the inequalities.