## Triangle area

I’ve just seen the following image on spikedmath.com:

Some geometry questions.
Source: spikedmath.com

The second answers seem to be obviously the correct ones, right? Wrong.

According to Heron’s formula you can calculate a triangles area like this:

Let $$a, b, c$$ be the side lengths of the triangle.
$$s := \frac{a+b+c}{2}$$
$$T = \sqrt{s \cdot (s-a) \cdot (s-b) \cdot (s-c)}$$

So the area of the first triangle is
$$s_1 := \frac{16}{2} = 8$$
$$T_1 := \sqrt{8 \cdot (8-5) \cdot (8-5) \cdot (8-6)} = \sqrt{8 \cdot 3 \cdot 3 \cdot 2} = 3 \cdot 4 = 12$$

The area of the second one is
$$s_1 := \frac{18}{2} = 9$$
$$T_1 := \sqrt{9 \cdot (9-5) \cdot (9-5) \cdot (9-8)} = \sqrt{9 \cdot 4 \cdot 4 \cdot 1} = 3 \cdot 4 = 12$$

Both triangles have the same area!

When you draw it, it looks like this:

Both triangles in one picture

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### 6 Responses to “Triangle area”

1. Mike says:

Points for showing your working out, but -2 for incorrectly labelling your diagram.

• Martin Thoma says:

Thank you for your comment. I’ve corrected the typo.

2. Hsn says:

The diagram would be much more intuitive if each triangle is shown as a set of two triangle of 3x4x5 side to side (but the side varies)

• Martin Thoma says:

I’m sorry, I don’t understand what you mean. Could you make a scribble, upload it e.g. to http://imgur.com/ and post another comment, please?

• kmwho says:

You can divide both the (isosceles) triangles into two equal right triangles of 3x4x5, and depending on whether you place them side by side with common side as 4 or 3 you get the first or second triangle

3. Platty says:

Just draw two 3-4-5 triangles that share the side of length 4 and two 3-4-5 triangles that share the side of length 3. These triangles formed are equivalent to the 5-5-6 and 5-5-8 triangles, respectively, demonstrating that they have the same area.