The Pythagorean Theorem
Learning objectives
- State and prove the Pythagorean theorem.
- Use it to find missing sides and distances.
- Use the converse to recognise right triangles.
- Tell acute and obtuse triangles apart from the squares of their sides.
Right triangles
In a right triangle the side opposite the right angle is the hypotenuse; it is the longest side. The other two sides are the legs.
In a right triangle with legs \(a,b\) and hypotenuse \(c\):
\[a^2+b^2=c^2\]
Show proof
Place four copies of the triangle inside a square of side \(a+b\), so that the four hypotenuses form a tilted square of side \(c\) in the middle. The big square has area \((a+b)^2\). It consists of four triangles of area \(\frac{ab}{2}\) each and the middle square of area \(c^2\):
\[(a+b)^2=4\cdot\frac{ab}{2}+c^2\]
Expanding gives \(a^2+2ab+b^2=2ab+c^2\), so \(a^2+b^2=c^2\). ∎
Find the hypotenuse of a right triangle with legs 6 and 8.
\[c^2=6^2+8^2=36+64=100,\qquad c=10\]
The hypotenuse is 13 and one leg is 5. Find the other leg.
\[b^2=13^2-5^2=169-25=144,\qquad b=12\]
Whole numbers with \(a^2+b^2=c^2\) are called Pythagorean triples: \((3,4,5)\), \((5,12,13)\), \((8,15,17)\), \((7,24,25)\).
Explore on the grid
Drag the vertices. The page lists the squares of the sides and compares the two smaller ones added together with the largest. Can you make another right triangle?
If the sides of a triangle satisfy \(a^2+b^2=c^2\), then the angle opposite \(c\) is a right angle.
Show proof
Build a right triangle with legs \(a\) and \(b\). By the Pythagorean theorem its hypotenuse satisfies \(h^2=a^2+b^2=c^2\), so \(h=c\). The two triangles have three equal sides, so they are congruent by SSS, and the angle opposite \(c\) is a right angle. ∎
Let \(c\) be the longest side. If \(a^2+b^2>c^2\) the triangle is acute, and if \(a^2+b^2<c^2\) it is obtuse. The grid tool above shows this with the signs > and <.
Is a triangle with sides 7, 24, 25 a right triangle? What about 6, 7, 10?
We have \(7^2+24^2=49+576=625=25^2\), so 7, 24, 25 is right-angled. For 6, 7, 10: \(6^2+7^2=85<100=10^2\), so that triangle is obtuse.
Applications
Find the distance between \((1,2)\) and \((7,10)\).
The horizontal difference is 6 and the vertical difference is 8, so the distance is \(\sqrt{6^2+8^2}=\sqrt{100}=10\).
A 5 m ladder reaches 4 m up a wall. How far is its foot from the wall? By the theorem, \(d^2=5^2-4^2=9\), so \(d=3\) m.
Exercises
Find the hypotenuse of a right triangle with legs 9 and 12.
Show solution
\(c^2=81+144=225\), so \(c=15\).
A right triangle has hypotenuse 17 and one leg 8. Find the other leg.
Show solution
\(b^2=17^2-8^2=289-64=225\), so \(b=15\).
Is a triangle with sides 8, 15, 17 right-angled?
Show solution
Yes: \(8^2+15^2=64+225=289=17^2\).
Find the distance between \((-2,3)\) and \((4,-5)\).
Show solution
The differences are 6 and 8, so the distance is \(\sqrt{36+64}=10\).
Show that the diagonal of a square with side \(s\) is \(s\sqrt2\).
Show solution
The diagonal is the hypotenuse of a right triangle with both legs \(s\), so \(d^2=s^2+s^2=2s^2\) and \(d=s\sqrt2\).
A right triangle has legs \(x\) and \(x+1\) and hypotenuse 5. Find \(x\).
Show solution
From \(x^2+(x+1)^2=25\) we get \(2x^2+2x-24=0\), i.e. \(x^2+x-12=0\), so \((x+4)(x-3)=0\). The positive solution is \(x=3\).
Summary
- In a right triangle \(a^2+b^2=c^2\), where \(c\) is the hypotenuse.
- The converse holds: if \(a^2+b^2=c^2\), the triangle is right-angled.
- With \(c\) the longest side, \(a^2+b^2>c^2\) means acute and \(a^2+b^2<c^2\) means obtuse.
- The theorem gives distances between points and diagonals of rectangles.