Converse of Pythagoras theoremIB MYP Maths Extended: Revision notes

Pythagoras' theorem says that in a right-angled triangle with hypotenuse $c$, $a^2+b^2=c^2$. Its converse goes the other way: if the sides of a triangle satisfy $a^2+b^2=c^2$, the triangle is right-angled, and the right angle is opposite the longest side $c$. To test a triangle: 1. Identify the longest side as $c$. 2. Work out $a^2+b^2$ and $c^2$. 3. Compare them. Example: sides $9,12,15$. $9^2+12^2=81+144=225=15^2$, so the triangle is right-angled, with the right angle opposite the $15$ cm side. Its area is $\frac12\times9\times12=54$ cm$^2$.

If $a^2+b^2\ne c^2$, compare the two values (with $c$ the longest side):

• $c^2=a^2+b^2$: right-angled
• $c^2<a^2+b^2$: acute-angled (all angles less than $90^\circ$)
• $c^2>a^2+b^2$: obtuse-angled (one angle greater than $90^\circ$) Examples: $8,15,16$: $64+225=289>256$, so acute. $5,6,9$: $25+36=61<81$, so obtuse. For three lengths to make a triangle at all, the two shorter sides must add up to more than the longest side. The lengths $2,3,6$ cannot form a triangle because $2+3<6$.

Triangles can also be named by how many sides are equal.

• Equilateral: all three sides equal (and every angle is $60^\circ$, so it is acute).
• Isosceles: exactly two sides equal.
• Scalene: all three sides different. Use both names together. Sides $5,5,8$: two equal sides, so isosceles, and $5^2+5^2=50<64$, so obtuse. Sides $13,13,10$: isosceles and acute, since $169<169+100$. Sides $6.1,8.0,10.1$ are scalene.

A Pythagorean triple is a set of whole numbers $(a,b,c)$ with $a^2+b^2=c^2$. Common ones are $(3,4,5)$, $(5,12,13)$, $(8,15,17)$ and $(7,24,25)$. Multiplying a triple by any whole number $k$ gives another, because $(ka)^2+(kb)^2=k^2(a^2+b^2)=(kc)^2$. So $(6,8,10)$, $(9,12,15)$ and $(12,16,20)$ all work. Spotting a triple, or a multiple of one, lets you recognise a right-angled triangle quickly.

Builders and carpenters check corners are square by measuring. A frame with sides $2.4$ m and $1.8$ m has a square corner only if the diagonal is exactly $3.0$ m, since $2.4^2+1.8^2=9=3.0^2$. A diagonal of $3.1$ m gives $9.61>9$, so the corner is wider than $90^\circ$. Real measurements are rounded. If lengths are given to the nearest $0.1$ m, each could be $0.05$ m out. A triangle with sides $6.1,8.0,10.1$ gives $101.21<102.01$, which looks slightly obtuse, but a true longest side of $10.06$ m (within the rounding) would give a right angle exactly. When the two sides of the comparison are this close, you cannot be sure, so re-measure or check with a $3$-$4$-$5$ rope.