Exercise 13 Page 602

We want to draw and label two cones with different dimensions but the same volume. Let $r_{1}$ and $r_{2}$ be their radii. Also, $h_{1}$ and $h_{2}$ will be the heights of the corresponding cones. We can see this in the graph.

A green cone with height h one and base radius r one, accompanied by an orange cone with height h two and base radius r two

Now, let's remember the formula for the volume of a cone. In the formula

r

is the radius of the base of the cone, and

h

is its height.

V = \frac{1}{3} π r^{2} h

We can use this formula to write the volumes of the cones

V_{1}

and

V_{2} .

V_{1} V_{2} = \frac{1}{3} π r_{1}^{2} h_{1} = \frac{1}{3} π r_{2}^{2} h_{2}

For the volumes to be equal, the following equation needs to be true.

V_{1} \frac{1}{3} π r_{1}^{2} h_{1} = V_{2} = \frac{1}{3} π r_{2}^{2} h_{2}

Let's transform the last equality further. This means dividing both sides by

\frac{1}{3} π .

\frac{1}{3} π r_{1}^{2} h_{1} r_{1}^{2} h_{1} = \frac{1}{3} π r_{2}^{2} h_{2} ⇓ = r_{2}^{2} h_{2}

We see that for the volumes to be equal the last equality needs to be true.

V_{1} = V_{2} \Leftrightarrow r_{1}^{2} h_{1} = r_{2}^{2} h_{2}

There are infinitely many pairs of numbers

(r_{1}, h_{1})

and

(r_{2}, h_{2})

that satisfy the equation. Let's then narrow it down to natural numbers. We can start by choosing the radii, for example

r_{1} = 6

and

r_{2} = 3 .

Let's substitute them into the equation and simplify. This will give us an equation for

h_{1}

and

h_{2} .

(6)^{2} h_{1} = (3)^{2} h_{2} ⇓ 36 h_{1} = 9 h_{2}

Now we need to pick such values of

h_{1}

and

h_{2}

that the equation will be true. One of the possible combinations is

h_{1} = 2

and

h_{2} = 8 .

We can check that it satisfies the equation.

36 h_{1} = 9 h_{2}

SubstituteII

$h_{1} = 2$ , $h_{2} = 8$

36 (2) = 9 (8)

Multiply

72 = 72 ✓

This gives us the dimensions of our cones. We can add them onto the graph.

Note that this is just a sample solution.

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