In the same way we ask: can we find the area of a 4-gon if we only know the lengths of its sides? The figures below reveal the answer to this question. The problem (see the figure) is about a 4-gon whose sides are known to be 3, 3, 3, 3 (it is a ‘rhombus’). As you can see, the areas of the three figures are different. (We drew the figures using GeoGebra and found the areas using the ‘Area’ tool. Please try this exercise yourself, or by using four rods joined together at their ends.)

Step-by-step solution

Idea: A rhombus is a parallelogram, so its area is base × height. Fixing the sides fixes the base but not the height.

1. A rhombus of side 3 has area = base × height = 3 × h, where h is the distance between the parallel sides.1 mark
2. Square: h = 3, area 9. Second figure: area 8.01 means h = 8.01 ÷ 3 = 2.67. Third figure: h = 5.41 ÷ 3 ≈ 1.80. Same four sides, three different heights.1 mark
3. With four rods joined at their ends the frame is not rigid: pushing it changes the angles and the height, so the area can be anything from 9 down to almost 0. So the area of a 4-gon cannot be found from its sides alone; we also need an angle, a diagonal, or a property such as being cyclic.1 mark

Check: A rhombus of side 3 whose angle is 30° has height 1.5 and area 4.5, yet another value ✓.

Answer to write in the exam

Common mistakes that cost marks

• Multiplying side × side = 9 for every rhombus. That is only right for the square.
• Thinking a 4-gon is rigid like a triangle. Three sides fix a triangle; four sides do not fix a 4-gon.

How this can come in the exam

Try one yourself

A rhombus has side 5 cm and height 4 cm. Another has side 5 cm and height 2.5 cm. Find both areas.

More questions like this

• Verify Brahmagupta’s formula for the case of a rectangle.
• The formula states that if the sides of the cyclic 4-gon have lengths a, b, c, d, and the semi-perimeter s is s = 12(a + b + c + d), then: Area of 4-gon = √((s − a)(s − b)(s − c)(s − d)). Please check out the formula against other special cases.
• Verify Brahmagupta’s formula for the case of an isosceles trapezium.
• Here is another example. Compare the following identities from algebra: (a + b)² = a² + b² + 2ab, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. Can you see that the first identity is a special case of the second one (put c = 0 in the second identity), and the second identity is a generalisation of the first one?
• Try to work out why this method works. You will find that it is a geometrical translation of the formula (a + b2)² − (a − b2)² = ab.

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