Challenge 2: the rooftop sculpture — Maths challenges

Download the problem sheet (PDF, in Catalan)

The problem

The neighbours in my building have come up with the idea of setting up a chill-out area on the shared roof terrace. An artist who lives in the flat below mine has suggested a sculpture shaped like a transparent cone, which he wants to fill with coloured water.

I am really not sure the roof can take the weight once it is full of water, and it worries me: every summer I read about some roof that has collapsed under the weight of the water in inflatable pools people put up without a second thought.

Tap the height, the slant height or the radius to see what each one is.

The sculpture: a transparent cone 2 m high with a 3 m slant height.

Which concepts we will need

We will need to revise Pythagoras' theorem (it always turns up), similarity, some areas and volumes. We also need to remember:

$1\ \text{m}^3 = 1000$ L, and for water, $1$ L $= 1$ kg.
An average roof can take at most $350$ kg for every $\text{m}^2$ of surface.

About rounding. Calculations use the exact values ($x^2 = 5$, $x = \sqrt{5}$, $\pi$) and only the final result is rounded, to two decimal places. Rounding halfway through can change a result in the second decimal place.

1. The radius of the base

To work out the volume of the cone we need the radius of the base. When we do not know something, we use a letter: we will call it $x$.

See the solution

The height, the radius and the slant height form a right triangle, with the slant height as the hypotenuse. Pythagoras:

3^2 = 2^2 + x^2 \;\Longrightarrow\; 9 = 4 + x^2 \;\Longrightarrow\; x^2 = 5

x = \sqrt{5} \approx 2{,}24 \text{ m}

The radius of the base is $x = \sqrt{5} \approx 2{,}24$ m.

2. The volume of the cone

Work out the volume of the cone.

See the solution

A cone has a third of the volume of a cylinder with the same base and height. Since $x^2 = 5$, we can use the $5$ directly:

V = \dfrac{A_B \cdot h}{3} = \dfrac{\pi \cdot x^2 \cdot h}{3} = \dfrac{\pi \cdot 5 \cdot 2}{3} = \dfrac{10\pi}{3} \approx 10{,}47 \text{ m}^3

$V \approx 10{,}47\ \text{m}^3$

3. The area it takes up

Work out how much of the roof the sculpture takes up.

See the solution

What it takes up on the roof is the base, a circle of radius $x = \sqrt{5}$ m. Since $x^2 = 5$:

A_B = \pi \cdot x^2 = 5\pi \approx 15{,}71 \text{ m}^2

Be careful not to round the radius before squaring it: with $2{,}24^2$ you would get $15{,}76\ \text{m}^2$.

If we also want the surface of the sculpture (for example, to know how much material is needed), we add the lateral area $\pi \cdot x \cdot g$:

A_T = A_B + A_L = 5\pi + \pi \cdot \sqrt{5} \cdot 3 \approx 15{,}71 + 21{,}07 = 36{,}78 \text{ m}^2

It takes up $A_B \approx 15{,}71\ \text{m}^2$ of the roof.

4. The conversions

How many m³ of water does it take to fill the cone? How many litres is that? How many kilograms? What load can the roof take on the area the cone covers?

See the solution

Cubic metres

It is the volume of the cone, $\dfrac{10\pi}{3}\ \text{m}^3$:

$\approx 10{,}47\ \text{m}^3$

Litres

\dfrac{10\pi}{3} \text{ m}^3 \cdot \dfrac{1000 \text{ L}}{1 \text{ m}^3} = \dfrac{10\,000\pi}{3} \text{ L} \approx 10\,471{,}98 \text{ L}

$\approx 10\,471{,}98$ L

Kilograms

Since $1$ L of water weighs $1$ kg:

10\,471{,}98 \text{ L} \cdot \dfrac{1\text{ kg}}{1 \text{ L}} \approx 10\,471{,}98\text{ kg}

$\approx 10\,471{,}98$ kg

Load it can take

The roof takes $350$ kg per $\text{m}^2$, and the sculpture rests on $5\pi \approx 15{,}71\ \text{m}^2$:

5\pi \text{ m}^2 \cdot \dfrac{350\text{ kg}}{1 \text{ m}^2} = 1750\pi\text{ kg} \approx 5\,497{,}79\text{ kg}

That area can take at most about $5\,497{,}79$ kg.

5. Does the roof collapse or not?

With everything you have worked out, does the roof collapse or not?

See the solution

10\,471{,}98\text{ kg} > 5\,497{,}79\text{ kg}

The water weighs almost twice what that part of the roof can hold ($10\,471{,}98 : 5\,497{,}79 \approx 1{,}9$). And we have not yet counted the weight of the cone or of the people in the chill-out area.

Yes: with the cone full, the roof would collapse. The sculpture cannot go up there full of water.

6. The scale drawing

We want a scale drawing of the cone to present our conclusions at the residents' meeting. We have decided that each metre will be 2 cm. How do we do it?

See the solution

Each real metre is $2$ cm on the drawing:

\dfrac{\text{drawing}}{\text{real}} = \dfrac{2 \text{ cm}}{1 \text{ m}} = \dfrac{2 \text{ cm}}{100 \text{ cm}} = \dfrac{1}{50}

The scale is $1:50$. We convert each measurement:

\text{Height: } 2 \text{ m} \cdot \dfrac{2 \text{ cm}}{1 \text{ m}} = 4 \text{ cm}

\text{Radius: } \sqrt{5} \text{ m} \cdot \dfrac{2 \text{ cm}}{1 \text{ m}} = 2\sqrt{5} \text{ cm} \approx 4{,}47 \text{ cm}

\text{Slant height: } 3 \text{ m} \cdot \dfrac{2 \text{ cm}}{1 \text{ m}} = 6 \text{ cm}

Sketch at a scale of 1:50 (measurements are on paper).

To draw it: a base with a $4{,}47$ cm radius ($8{,}94$ cm across) and the apex $4$ cm above the centre. If it is done right, the slant edges are $6$ cm long, which is a good check.

Scale $1:50$: height $4$ cm, radius $4{,}47$ cm and slant height $6$ cm.