Thinking about Geometry

Finding triangles

Here are some triangles.

Diagram shows triangles A with sides 4,4,4, B with sides 3,4,5, C with sides 3,3,2 and D with sides 2,root 3,1

The diagrams are not to scale.

Can you find …

… a triangle similar to triangle A with perimeter $6$ ?

Triangle A has perimeter $4 + 4 + 4 = 12$ , so we want to scale by factor $6/12 = 1/2$ , so we should choose the triangle with side lengths $2$ , $2$ and $2$ .

… a triangle similar to triangle B with perimeter $6$ ?

Triangle B has perimeter $3 + 4 + 5 = 12$ (like triangle A), so we again use factor $1/2$ , so we should choose the triangle with side lengths $3/2$ , $2$ and $5/2$ .

… a triangle similar to triangle C with perimeter $6$ ?

Triangle C has perimeter $3 + 3 + 2 = 8$ , so we want to scale by factor $6/8 = 3/4$ . So we should choose the triangle with side lengths $9/4$ , $9/4$ and $3/2$ .

… a triangle similar to triangle D with perimeter $6$ ?

Triangle D has perimeter $1 + 2 + \sqrt{3} = 3 + \sqrt{3}$ , so we want to scale by factor $\frac{6}{3 + \sqrt{3}} = \frac{6(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = 3 - \sqrt{3}.$ So we should choose the triangle with side lengths $3 - \sqrt{3}$ , $6 - 2\sqrt{3}$ and $-3 + 3\sqrt{3}$ .

… a triangle similar to triangle A with area $10$ ?

We need to remember that when we scale the lengths of a triangle by a certain factor, we scale the area of the triangle by the square of that factor. So we must allow for that.

Triangle A has area $4\sqrt{3}$ , so we want to scale by factor $\sqrt{\frac{10}{4\sqrt{3}}} = \sqrt{\frac{5\sqrt{3}}{6}}.$ So we should choose the triangle with side lengths $4\sqrt{\frac{5\sqrt{3}}{6}}$ , $4\sqrt{\frac{5\sqrt{3}}{6}}$ and $4\sqrt{\frac{5\sqrt{3}}{6}}$ .

… a triangle similar to triangle B with area $10$ ?

Triangle B has area $6$ , so we want to scale by factor $\sqrt{\frac{10}{6}} = \frac{\sqrt{15}}{3}.$ So we should choose the triangle with side lengths $\sqrt{15}$ , $\frac{4\sqrt{15}}{3}$ and $\frac{5\sqrt{15}}{3}$ .

… a triangle similar to triangle C with area $10$ ?

Triangle C has area $2\sqrt{2}$ , so we want to scale by factor $\sqrt{\frac{10}{2\sqrt{2}}} = \sqrt{\frac{5\sqrt{2}}{2}}.$ So we should choose the triangle with side lengths $\sqrt{10\sqrt{2}}$ , $3\sqrt{\frac{5\sqrt{2}}{2}}$ , $3\sqrt{\frac{5\sqrt{2}}{2}}$ .

… a triangle similar to triangle D with area $10$ ?

Triangle D has area $\frac{\sqrt{3}}{2}$ , so we want to scale by factor $\sqrt{\frac{10}{\sqrt{3}/2}} = \sqrt{\frac{20}{\sqrt{3}}} = \frac{2\sqrt{5}}{\sqrt[4]{3}}.$ So we should choose the triangle with side lengths $\frac{2\sqrt{5}}{\sqrt[4]{3}}$ , $\frac{4\sqrt{5}}{\sqrt[4]{3}}$ and $\frac{2\sqrt{15}}{\sqrt[4]{3}}$ .

… a right-angled triangle similar to one of triangles A, B, C, D with area $12$ ?

Of the four given triangles, only triangles B and D are right-angled. We can check that using Pythagoras’s theorem, since the side lengths $3$ , $4$ and $5$ satisfy his equation, as do the side lengths $1$ , $\sqrt{3}$ and $2$ .

Let’s find a triangle similar to B with area $12$ . And we can do that using the same approach as earlier.

Triangle B has area $6$ , so we want to scale by factor $\sqrt{\frac{12}{6}} = \sqrt{2}.$ So we should choose the triangle with side lengths $3\sqrt{2}$ , $4\sqrt{2}$ and $5\sqrt{2}$ .

… an isosceles triangle similar to one of triangles A, B, C, D with area $12$ ?

Of the four given triangles, only triangles A and C are isosceles. We could choose either, but let’s choose C and save A for later since it’s equilateral too.

Triangle C has area $2\sqrt{2}$ , so we want to scale by factor $\sqrt{\frac{12}{2\sqrt{2}}} = \sqrt{3\sqrt{2}}.$ So we should choose the triangle with side lengths $2\sqrt{3\sqrt{2}}$ , $3\sqrt{3\sqrt{2}}$ and $3\sqrt{3\sqrt{2}}$ .

… an equilateral triangle similar to one of triangles A, B, C, D with area $12$ ?

Our only option here is triangle A.

Triangle A has area $4\sqrt{3}$ , so we want to scale by factor $\sqrt{\frac{12}{4\sqrt{3}}} = \sqrt[4]{3}.$ So we should choose the triangle with side lengths $4\sqrt[4]{3}$ , $4\sqrt[4]{3}$ and $4\sqrt[4]{3}$ .

… a scalene triangle similar to one of triangles A, B, C, D with area $12$ ?

Here we can choose between triangles B and D. Since we’ve already done B, let’s pick D.

Triangle D has area $\frac{\sqrt{3}}{2}$ , so we want to scale by factor $\sqrt{\frac{12}{\sqrt{3}/2}} = \sqrt{8\sqrt{3}} = 2\sqrt{2\sqrt{3}}.$ So we should choose the triangle with side lengths $2\sqrt{2\sqrt{3}}$ , $4\sqrt{2\sqrt{3}}$ and $2\sqrt{6\sqrt{3}}$ .

… a right-angled triangle with area $12$ that is not similar to any of triangles A, B, C, D?

We have lots of choice here, we just have to make sure that our right-angled triangle isn’t similar to triangle B or D(it definitely won’t be similar to either of the others).

Let’s try to be lazy when constructing an example. An easy way to make a triangle have area $12$ is to give it base $12$ and height $2$ , for example, and then we can use Pythagoras to finish off.

Choose the triangle with side lengths $2$ , $12$ and $\sqrt{2^2 + 12^2} = \sqrt{4 + 144} = \sqrt{148} = 2\sqrt{37}$ . The side lengths satisfy Pythagoras’s equation so the triangle is right-angled, and its area is $12$ . It is easy to check that it is not similar to triangle B or triangle D, by looking at ratios of two sides (e.g. $\frac{3}{4} \neq \frac{2}{12}$ ).

… an isosceles triangle with area $12$ that is not similar to any of triangles A, B, C, D?

We need to avoid both A and C here. It should be quite straightforward to build an isosceles triangle with area $12$ , by picking the base and height and then finding the missing side length.

Choose the isosceles triangle with base $2$ (as one side of the triangle) and height $12$ . The other sides have length $\sqrt{1^2 + 12^2} = \sqrt{145}$ . This is clearly isosceles, clearly has area $12$ , and cannot be similar to triangle C (since they have the same base but different heights).

… an equilateral triangle with area $12$ that is not similar to any of triangles A, B, C, D?

This is a bit tricky…

Every equilateral triangle is similar to every other equilateral triangle! So any equilateral triangle will be similar to A, so there is no triangle that meets the criteria.

… a scalene triangle with area $12$ that is not similar to any of triangles A, B, C, D?

There are lots and lots of examples here. Again, a nice approach might be to specify the base and height, since they determine the area, and then just to make sure that the triangle is not isosceles or right-angled. The hardest part might be calculating the side lengths!

Thinking about Geometry

Finding triangles

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