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Problem of the Week
Problem B and Solution
How to Net a Balloon

Problem

In \(1783\), the Montgolfier brothers launched the first hot air balloon flight in history, using a balloon that they created out of fabric and paper. The flight was short but successful. For an art exhibit, Vijay used cardboard to create a model of a balloon inspired by the Montgolfier brothers’ balloon. Vijay’s model, as well as its net, are shown below. Note that these diagrams are not drawn to scale.

Vijay’s model.A net made up of four rectangles, four smaller triangles, and four larger triangles. The rectangles all have base 3 m and height 2 m and are placed side by side with their 2 m heights touching. A smaller triangle is placed above each rectangle with the top edge of the rectangle aligned with the base of the triangle. These triangles all have base 3 m and height 2 m. A larger triangle is placed below each rectangle with the bottom edge of the rectangle aligned with the base of the triangle. These triangles all have base 3 m and height 4 m.

  1. Calculate the total area of cardboard Vijay used in his model. This is also called the surface area of Vijay’s model.

  2. Suppose you have a sheet of paper measuring \(90\) cm by \(60\) cm. Draw a net for a balloon that you could make using this sheet of paper. Write the dimensions for each shape on your net.

Solution

  1. To calculate the total area we notice that the net has four triangles on top, four rectangles, and four triangles on the bottom. We will calculate the area of each of these shapes separately and then add them together. \[\begin{aligned} \text{Area of one top triangle} &= \frac{1}{2}\times \text{base} \times \text{height}\\ &= \frac{1}{2} \times 3 \times 2\\ &= \frac{1}{2} \times 6 \\ &= 3~\text{m}^2 \end{aligned}\] Since there are four top triangles, the total area is \(4 \times 3 = 12~\text{m}^2\). \[\begin{aligned} \text{Area of one rectangle} &= \text{length} \times \text{width}\\ &= 3 \times 2 \\ &= 6~\text{m}^2 \end{aligned}\] Since there are four rectangles, the total area is \(4 \times 6 = 24~\text{m}^2\). \[\begin{aligned} \text{Area of one bottom triangle} &= \frac{1}{2}\times \text{base} \times \text{height}\\ &= \frac{1}{2} \times 3 \times 4\\ &= \frac{1}{2} \times 12 \\ &= 6~\text{m}^2 \end{aligned}\] Since there are four bottom triangles, the total area is \(4 \times 6 = 24~\text{m}^2\). Therefore, the total area of cardboard used is \(12+24+24=60~\text{m}^2\).

  2. There are many possible nets. Here is one.

    A net made up of four rectangles with base 22.5 cm and height 20 cm placed in a row, with triangles with base 22.5 cm and height 20 cm placed along the top edge and the bottom edge of each rectangle. The net has width 90 cm and height 60 cm.