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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?

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.