# Pythagoras Perimeters

*Pythagoras Perimeters printable worksheet - proof sortingPythagoras Perimeters printable worksheet - whole problem*

If this right-angled triangle has a perimeter of $12$ units, it is possible to show that the area is $36-6c$ square units.

Can you find a way to prove it?

Once you've had a chance to think about it, click below to see a possible way to solve the problem, where the steps have been muddled up.

Can you put them in the correct order?

a) Squaring both sides: $a^2+2ab+b^2 = 144-24c+c^2$

b) So Area of the triangle $=36-6c$

c) $a+b=12-c$

d) So $2ab=144-24c$

e) Area of the triangle $= \frac{ab}{2}$

f) By Pythagoras' Theorem, $a^2+b^2=c^2$

g) $a+b+c=12$

h) Dividing by $2$: $ab=72-12c$

*Printable Version*

Can you adapt your method, or the method above, to prove that when the perimeter is $30$ units, the area is $225 - 15c$ square units?**Extension**

Can you find a general expression for the area of a right angled triangle with hypotenuse $c$ and perimeter $p$?

With thanks to *Don Steward*, whose ideas formed the basis of this problem.

What equations can you write down about the area and the perimeter of right angled triangles?

How could you rearrange and manipulate the equations?

Siddhant from Indus International School of Bangalore in India, Noraisha, Venus, Aditya, Ahn, Jemma and Minjun from West Island School in Hong Kong, Guruvignesh from England, Agathiyan from Hymers College in the UK, Vo from ISM in Vietnam and Sophie and Milly from Trinity Catholic High School in the UK wrote correct proofs which used the steps in the correct order. This is Jemma's proof and explanation:

g) $a + b + c =12$ What we know is that the perimeter of the triangle is twelve and the perimeter is all the sides added up. So that is $a$ plus $b$ plus $c$

c) $ a + b=12âˆ’c$ In order to get the correct order for Pythagoras theorem ($a^2 + b^2 =c^2$) you must put the $c$ onto the other side of the equal sign. The positive $c$ will then turn into a negative number.

a) $a^2+ 2ab +b^2=144-24c+c^2$ We now can square both sides of the equation so that we are able to simplify when we reach the next step.

So $(a+b)^2 = a^2+ 2ab +b^2$ and for the other side $12\times12=144$ (The twelve is from the perimeter) and the $24 = 12\times2$

f) $ a^2 + b^2 =c^2$ Now we can use the Pythagoras theorem

d) $2ab=144-24c$ When we get rid of the $a^2,b^2$ and $c^2$ it will just equal $2ab=144-24c$

h) $ab=72-12c$ We can then simplify the equation by dividing everything by two so then $2ab\div2=ab,$ $ 144\div2=72$ and $24c\div2=12c$

e) Area of triangle$= \frac{ab}2$ To find the area of a triangle the equation is $\frac{a\times b}2$

b) $36-6c$ Therefore we know $a$ and we know $b$ so the equation must be $36-6c$

The last step comes from dividing both sides of $ab=72-12c$ by $2$.

Adithya from Hymers College in the UK and Piyush, Anna, Vaibhavi, Nicholas, Ahan, Sahiti, Caitlin, Winston, Natalia, Stephanie and Wing Tung from West Island School in Hong Kong sent in proofs that included the steps in a slightly different order. These proofs didn't flow as neatly, but are still convincing. For example, this is Natalia's proof. Note that Natalia lists all of the information available near the beginning, which is sensible, but less easy to follow than using the information only when it is needed:

g) $a+b+c=12$

c) $a+b=12âˆ’c$

e) Area of the triangle $=\frac{ab}2$

f) By Pythagoras' Theorem, $a^2+b^2=c^2$

a) Squaring both sides: $a^2+2ab+b^2=144âˆ’24c+c^2$

d) So $2ab=144âˆ’24c$

h) Dividing by $2$: $ab=72âˆ’12c$

b) So Area of the triangle $=36âˆ’6c$

Minhaj from St Ivo School in England wrote a proof that didn't use the steps suggested. This is Minhaj's proof:

$a + b + c = 12$ (as perimeter is $12$ units)

Using Pythagoras' theorem,

$c = (a^2+b^2)^\frac12\\

\Rightarrow a + b + (a^2+b^2)^\frac12 = 12\\

\Rightarrow (a^2+b^2)^\frac12 = 12 - a - b\\

\Rightarrow a^2+b^2 = (12 - a - b)^2 = 144 - 24a - 24b + 2ab + a^2 + b^2\\

\Rightarrow 144 - 24a - 24b + 2ab = 0$

Dividing by $4$ gives $36 - 6a - 6b + \frac12 ab = 0\\

\Rightarrow\frac12 ab = 6 (a + b) - 36$

The area of the triangle is $\frac12 ab$ and $a + b = 12 - c$

$\Rightarrow\frac12 ab = 6 ( 12 - c ) -36 = 72 - 6c - 36 = 36 - 6c\\

\therefore\text{Area}= 36 - 6c$ square units

Adithya, Siddhant, Minhaj, Anna, Caitlin, Winston, Wing Tung, Minjun, Guruvignesh, Agathiyan, Sophie and Milly proved that a right-angled triangle whose perimeter is $30$ units has an area of $225-15c$ units. This is Milly's proof:

Amrit from Hymers College in the UK and Vo both proved this by first finding and proving the area in the general case (in the extension), and then substituting in $p=12$. This is Amrit's proof of the general case (extension) and also the second part.

A right angled triangle has sides $a,b$ (the base and height) and $c$ (the hypotenuse) which make up its perimeter $p$. Therefore,

$a+b+c=p$

Subtracting $c$ from both sides,

$a+b=p-c$

Squaring both sides,

$a^2+2ab+b^2=p^2-2pc+c^2$

By Pythagoras' theorem, $a^2+b^2=c^2$. Plugging $c^2$ in for $a^2+b^2$ on the LHS,

$2ab+c^2=p^2-2pc+c^2$

Subtracting $c^2$ from both sides,

$2ab=p^2-2pc$

Factorising,

$2ab=p(p-2c)$

Dividing both sides by $4$, we get that the Area of the triangle, $A$, is

$A=\dfrac{ab}2=\dfrac{p(p-2c)}4$

When the perimeter is $30$, the area of the triangle in terms of $c$ is

$\dfrac{30(30-2c)}4$

Factorising,

$\dfrac{60(15-c)}4$

Multiplying by $\dfrac{1/4}{1/4}$,

$15(15-c)$

Expanding the brackets, the area of the triangle is

$225-15c$

Well done to Shreya, Apurva and Aryan from Dhirubhai Ambani international School in India, and to Caitlin, Winston, Wing Tung, Minjun, Guruvignesh, Agathiyan, Matt, Sophie and Milly who also correctly found and proved the general expression for the area of a right-angled triangle. This is Guruvignesh's proof, which is slightly different:

$P = a+b+c$ (Equation for the perimeter of a right-angled triangle)

$P-c = a+b$ (Subtract $a$ from both sides)

$P^2-2Pc+c^2 = a^2+2ab+b^2$ (Square both sides)

$P^2-2Pc = a^2+2ab+b^2-c^2$ (Subtract $c^2$ from both sides)

$A = \frac{ab}2$ (Equation for the area of a triangle)

$4A = 2ab$ (Multiply both sides by $4$)

$P^2-2Pc = a^2+b^2+4A -c^2$ (Since, $2ab=4A$, we can substitute that into the previous

equation)

$a^2+b^2 = c^2$ (Pythagorean Theorem)

$P^2-2Pc = c^2-c^2+4A$ (Substitute the Pythagorean Theorem into the equation)

$P^2-2Pc = 4A$

$\dfrac{P^2-2Pc}4 = A$ (Divide both sides by $4$ to express $A$ in terms of $P$ and $c$)

*These printable resources may be useful: Pythagoras Perimeters Pythagoras Perimeters Proof Sorting*

### Why do this problem?

This problem offers a challenge in algebraic manipulation, bringing together students' knowledge of area, perimeter, and Pythagoras' Theorem. The Proof Sorter activity offers some scaffolding, so that students can see how to set out a proof, and then use the ideas to construct the proofs that follow on.

### Possible approach

"It is possible to show that a right-angled triangle with a perimeter of $12$ units has an area of $36-6c$ square units, where c is the length of the hypotenuse."

"Can you find a way to prove this?"

Give students some time to explore.

They may notice the special case of the $3,4,5$ triangle which satisfies the above conditions, but it is tricky to work out an efficient way to prove it holds for all right-angled triangles with a perimeter of $12$.

Once they have thought about the problem, offer them the Proof Sorter and invite them to put the statements in the correct order.

Once they are happy with the proof, challenge them to adapt their proof to find an expression, in terms of c, for the area of a right-angled triangle with a perimeter of $30$ units.

### Key questions

What equations can you write down about the area and the perimeter of right-angled triangles?

How could you rearrange and manipulate the equations?

### Possible support

Quadratic Patterns offers a good opportunity for students to practise manipulating algebraic expressions in a more accessible context.

### Possible extension

Challenge students to find a general expression, in terms of c, for the area of a right angled triangle with hypotenuse $c$ and perimeter $p$.