This morning I posted this about a lovley simple triangle puzzle I had found and explored. One of the responses came from Mrs Watson (@MrsJMWatson) who tweeted this reply:

The link is to this page from Nrich and contains this puzzle:

*Triangle T has sides 6,5,5 and Triangle Q has sides 8,5,5 What is the ratio Area T:Area Q? *

Now this is a great example if a “low barrier, high ceiling” problem. I think your first instinct are probably to go with what you are most familiar with. In today’s earlier problem the respondents who are used to working with the sine rule for area instinctively went for that, but others who are more used to working with other things went other ways.

For this problem my instinct was to use the cosine rule to find an angle, and then the sine rule for area ((1/2) absin(c)) to find the areas and simplify the ratios. My friend Steve, also a math teacher up to A level, actually worked through this, but didn’t need the sine rule for area as the ratio is easy to spot when you see the angles add up to 180 degrees (we’ll let him off for using degrees just this once).

I didn’t go down this route, I was thinking about generalising for all triangles and thought Heron’s Formula would be better for this.

I used it and found this lovely solution:

* T) a=5,b=5,c=6 s=8*

Area: (8.3.3.2)^1/2=12 U) a=5,b=5,c=8 s=9

Area:(9.4.4.1)^1/2 = 12

*
**So Area T: Area U = 12:12 = 1:1*

Discussing my method and answer with various people today I have been shocked at how many people (most it seems) haven’t heard of “Heron’s Formula”!

**Heron’s Formula**

Heron’s Formula is a fantastic piece of mathematics. I don’t know how I know about it. I have known about it so long it didn’t occur to me others wouldn’t. I guess I had assumed I learnt it at school, but if that was the case others would know about it. I’ve read and watched a lot about maths over the years, so I guess I must have picked it up from a book or show.

For those of you who don’t know, Heron’s Formula states:

*For a triangle with side lengths a,b and c *

*
*Area=(s(s-a)(s-b)(s-c))^1/2

*where s is the semiperimeter (ie s=(a+b+c)/2)*

A truly marvellous formula. It’s named after Hero of Alexandria, who along with this formula is credited with being the first person to envisage imaginary and complex numbers.

Hero’s own proof is pretty cool, and involves cyclic quadrilaterals and properties of right angles triangles. There are lots of other proofs too, my favourite is the trigonometric proof, which I think would have been what I ended up with if I had decided to generalise this problem using trig!

Later in the discussion Mrs Watson said the phrase “Pythagorean Triples“, and I instantly saw both triangles could be cut into 2 3,4,5 triangles. I think this is the nicest solution. When I checked back to another discussion I noticed that Andy (@andycav_25) had also had this realisation. His instinct was to draw a perpendicular height. I wondered if I would have gone down this route if I’d had paper to work on instead if working mentally?!

Finally, Mrs Watson mentioned that her year 7 class do it via construction. She didn’t elaborate much on this, but I imagine she means draw both triangles accurately, measure the height, work out the areas. This is, in itself, quite nice, and shows that pupils can tackle the problem using whatever tools they have at their disposal.

This is a lovely problem, I’d love to hear how your instincts would tackle it, how your students for if you try it, and if you previously knew about Heron’s Formula!

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## Area the Hero’s Way

This morning I posted this about a lovley simple triangle puzzle I had found and explored. One of the responses came from Mrs Watson (@MrsJMWatson) who tweeted this reply:

The link is to this page from Nrich and contains this puzzle:

Triangle T has sides 6,5,5 and Triangle Q has sides 8,5,5 What is the ratio Area T:Area Q?Now this is a great example if a “low barrier, high ceiling” problem. I think your first instinct are probably to go with what you are most familiar with. In today’s earlier problem the respondents who are used to working with the sine rule for area instinctively went for that, but others who are more used to working with other things went other ways.

For this problem my instinct was to use the cosine rule to find an angle, and then the sine rule for area ((1/2) absin(c)) to find the areas and simplify the ratios. My friend Steve, also a math teacher up to A level, actually worked through this, but didn’t need the sine rule for area as the ratio is easy to spot when you see the angles add up to 180 degrees (we’ll let him off for using degrees just this once).

I didn’t go down this route, I was thinking about generalising for all triangles and thought Heron’s Formula would be better for this.

I used it and found this lovely solution:

T) a=5,b=5,c=6 s=8Area: (8.3.3.2)^1/2=12 U) a=5,b=5,c=8 s=9

Area:(9.4.4.1)^1/2 = 12

So Area T: Area U = 12:12 = 1:1Discussing my method and answer with various people today I have been shocked at how many people (most it seems) haven’t heard of “Heron’s Formula”!

Heron’s FormulaHeron’s Formula is a fantastic piece of mathematics. I don’t know how I know about it. I have known about it so long it didn’t occur to me others wouldn’t. I guess I had assumed I learnt it at school, but if that was the case others would know about it. I’ve read and watched a lot about maths over the years, so I guess I must have picked it up from a book or show.

For those of you who don’t know, Heron’s Formula states:

For a triangle with side lengths a,b and cArea=(s(s-a)(s-b)(s-c))^1/2

where s is the semiperimeter (ie s=(a+b+c)/2)A truly marvellous formula. It’s named after Hero of Alexandria, who along with this formula is credited with being the first person to envisage imaginary and complex numbers.

Hero’s own proof is pretty cool, and involves cyclic quadrilaterals and properties of right angles triangles. There are lots of other proofs too, my favourite is the trigonometric proof, which I think would have been what I ended up with if I had decided to generalise this problem using trig!

Later in the discussion Mrs Watson said the phrase “Pythagorean Triples“, and I instantly saw both triangles could be cut into 2 3,4,5 triangles. I think this is the nicest solution. When I checked back to another discussion I noticed that Andy (@andycav_25) had also had this realisation. His instinct was to draw a perpendicular height. I wondered if I would have gone down this route if I’d had paper to work on instead if working mentally?!

Finally, Mrs Watson mentioned that her year 7 class do it via construction. She didn’t elaborate much on this, but I imagine she means draw both triangles accurately, measure the height, work out the areas. This is, in itself, quite nice, and shows that pupils can tackle the problem using whatever tools they have at their disposal.

This is a lovely problem, I’d love to hear how your instincts would tackle it, how your students for if you try it, and if you previously knew about Heron’s Formula!

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