--- title: The Pythagorean Theorem subtitle: Proof, converse, distance, and applications author: Mathematics Department organization: Your School description: A compact lesson and practice set on the Pythagorean theorem, its converse, exact radicals, coordinate distance, and applications. keywords: [Pythagorean theorem, converse, right triangle, hypotenuse, distance formula, triangle classification] template: assignment paper: us-letter titlePage: false toc: false headingNumbers: false equationNumbers: false columns: 1 figurePlacement: none tableCaptionPosition: top header: Pythagorean Theorem | Proof and Applications footer: Student lesson and practice pdfProfile: tagged language: en-US --- # The Pythagorean theorem A **right triangle** has one $90^\circ$ angle. Its two perpendicular sides are the **legs**; the side opposite the right angle is the **hypotenuse**. The hypotenuse is always the longest side. :::figure{#fig:anatomy caption="A right triangle with legs a and b and hypotenuse c" alt="A right triangle has perpendicular legs labeled a and b, a right-angle marker between them, and the opposite sloping side labeled c as the hypotenuse" width=58 placement=none} ![Right triangle anatomy](right-triangle-anatomy.svg) ::: If the legs have lengths $a$ and $b$ and the hypotenuse has length $c$, then :::equation{#eq:theorem alt="a squared plus b squared equals c squared" numbered=false} \[ a^2+b^2=c^2. \] ::: This is an equality of **areas**: the area of a square built on the hypotenuse equals the sum of the areas of squares built on the legs. It applies only to right triangles. The letters do not matter, but the role of each side does: the hypotenuse must be the side alone on the right of the equation. > [!NOTE] > The theorem bears Pythagoras's name in much modern English usage, but right-triangle calculations were known in Old Babylonian mathematics long before his lifetime, and the *gougu* rule appears in ancient Chinese mathematics. Surviving evidence does not establish that Pythagoras personally discovered or first proved the result. Sources are listed in `references.bib`. ## Why the theorem is true In @fig:proof, four congruent right triangles, each with legs $a$ and $b$, fit inside a square of side $a+b$. Their four hypotenuses form a central quadrilateral with four equal sides of length $c$. At each central corner, the two acute angles of a right triangle meet; because those angles sum to $90^\circ$, every central angle is a right angle. The central quadrilateral is therefore a square. The outer square can now be measured in two ways. :::figure{#fig:proof caption="A rearrangement proof: four right triangles surround a square of side c inside a square of side a plus b" alt="Four congruent right triangles with legs a and b fill the corners of a large square; their hypotenuses form a tilted central square labeled c squared" width=62 placement=none} ![Area proof of the Pythagorean theorem](pythagorean-area-proof.svg) ::: The area of the outer square is $(a+b)^2$. The same region consists of four triangles of area $ab/2$ and the central square of area $c^2$. Hence $$ (a+b)^2=4\left(\frac{ab}{2}\right)+c^2. $$ Expanding and subtracting $2ab$ from both sides gives $$a^2+2ab+b^2=2ab+c^2\quad\Longrightarrow\quad a^2+b^2=c^2.$$ The proof explains the formula rather than merely checking it for a few triples. # Finding an unknown side For a missing hypotenuse, add the squares of the legs and take the positive square root. For a missing leg, subtract the known leg's square from the hypotenuse's square. A length is positive, so only the positive square root is used. :::table{#tbl:forms caption="The same theorem solved for each possible unknown side"} | Unknown | Formula | Condition | | --- | --- | --- | | Hypotenuse $c$ | $c=\sqrt{a^2+b^2}$ | $a$ and $b$ are the legs | | Leg $a$ | $a=\sqrt{c^2-b^2}$ | $c$ is the hypotenuse | | Leg $b$ | $b=\sqrt{c^2-a^2}$ | $c$ is the hypotenuse | ::: **Example 1.** A right triangle has legs $9$ cm and $12$ cm. Then $$c=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ cm}.$$ The answer is longer than either leg, as a hypotenuse must be. **Example 2.** A right triangle has hypotenuse $13$ m and one leg $5$ m. If the other leg is $x$, then $$x=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ m}.$$ Adding the squares here would produce a side longer than the stated hypotenuse, exposing the error. ::::keep **Example 3.** Legs of $7$ units and $11$ units give $$c=\sqrt{7^2+11^2}=\sqrt{170}\approx13.0.$$ $\sqrt{170}$ is the exact value. The decimal is an approximation to the nearest tenth. Round only at the end; early rounding can make a correct result drift. :::: > [!TIP] > Before calculating, circle the right angle and mark the opposite side. Then estimate: the hypotenuse must exceed either leg but be less than the sum of the legs. # The converse and triangle classification The theorem says that a right triangle satisfies $a^2+b^2=c^2$. Its **converse** also holds: > [!IMPORTANT] > If the side lengths of a triangle satisfy $a^2+b^2=c^2$, with $c$ the longest side, then the triangle is right. More generally, sort the lengths so $a\le b\le c$ and compare $a^2+b^2$ with $c^2$. First verify $a+b>c$; otherwise the lengths do not form a triangle. :::table{#tbl:classification caption="Classifying a triangle from its three side lengths"} | Comparison | Classification | Meaning for the largest angle | | --- | --- | --- | | $a^2+b^2=c^2$ | right | exactly $90^\circ$ | | $a^2+b^2>c^2$ | acute | less than $90^\circ$ | | $a^2+b^2100=10^2,$$ so the triangle is acute. For $5,7,10$, $25+49<100$, so the triangle is obtuse. **Example 6.** The lengths $3,4,8$ fail the triangle inequality because $3+4<8$. They form no triangle, so acute/right/obtuse classification is not available. # Distance in the coordinate plane Let $P(x_1,y_1)$ and $Q(x_2,y_2)$. Horizontal and vertical changes form perpendicular legs of lengths $|x_2-x_1|$ and $|y_2-y_1|$. The segment $PQ$ is the hypotenuse, so :::equation{#eq:distance alt="the distance from P to Q equals the square root of the quantity x two minus x one squared plus the quantity y two minus y one squared" numbered=false} \[ PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \] ::: :::figure{#fig:distance caption="Horizontal and vertical changes make a right triangle whose hypotenuse is the distance between two points" alt="A coordinate grid shows points P and Q connected by a sloping segment, with horizontal change delta x and vertical change delta y forming a right triangle" width=55 placement=none} ![Coordinate distance as a right triangle](coordinate-distance.svg) ::: **Example 7.** Between $P(-2,1)$ and $Q(4,9)$, the changes are $6$ and $8$. Therefore $$PQ=\sqrt{6^2+8^2}=\sqrt{100}=10.$$ Changing the subtraction order changes both signs but not the squared values. # Applications **Example 8: ladder.** A $13$ ft ladder reaches $12$ ft up a vertical wall. The ladder is the hypotenuse. If its foot is $x$ ft from the wall, then $$x^2+12^2=13^2\quad\Longrightarrow\quad x=\sqrt{25}=5\text{ ft}.$$ **Example 9: rectangular shortcut.** A rectangular field is $90$ m by $120$ m. Its diagonal is $$d=\sqrt{90^2+120^2}=\sqrt{22500}=150\text{ m}.$$ The units in $a^2+b^2$ are square meters; taking the square root returns a length in meters. # Practice Give exact answers unless a decimal is requested. Show the equation that identifies the hypotenuse. ::::regions{tracks="1 1" gutter=18} :::region **A. Unknown lengths** 1. In @fig:anatomy, name the two legs and the hypotenuse. 2. Find the hypotenuse when the legs are $6$ cm and $8$ cm. 3. Find the hypotenuse when the legs are $8$ m and $15$ m. 4. Find the hypotenuse when the legs are $7$ units and $9$ units. Give an exact value and a decimal to the nearest tenth. 5. Find the missing leg when $c=13$ in and the other leg is $5$ in. 6. Find the missing leg when $c=25$ ft and the other leg is $7$ ft. 7. Find the missing leg when $c=10$ units and the other leg is $6$ units. 8. Find the diagonal of a $9$ cm by $12$ cm rectangle. **B. Converse and classification** For 9-13, first decide whether the lengths form a triangle. If they do, classify it as acute, right, or obtuse. 9. $9,12,15$ 10. $8,10,13$ 11. $9,12,14$ 12. $3,4,8$ 13. $7,24,25$ ::: :::region **C. Coordinates and applications** 14. Find the distance between $(1,2)$ and $(7,10)$. 15. Find the distance between $(-3,4)$ and $(5,-11)$. 16. Find the exact diagonal of a square with side $7$ cm, then approximate it to the nearest tenth. 17. A $17$ m guy wire is attached $15$ m above level ground. How far is its anchor from the pole? 18. A rectangular screen is $48$ cm wide and $36$ cm high. Find its diagonal. 19. A baseball diamond is a square with side $90$ ft. Find the distance from home plate to second base exactly and to the nearest foot. **D. Explain and correct** 20. A student finds a missing leg from $c=17$ and $a=8$ by writing $b=\sqrt{17-8}=3$. Identify the error and give the correct length. 21. A student says a $5,6,7$ triangle is right because the sides are consecutive after skipping one integer. Test and classify the triangle. 22. Explain why $a^2+b^2=c^2$ must be checked with the longest side as $c$. 23. Construct a right triangle whose hypotenuse is irrational. State its exact length. 24. **Exit ticket.** A triangle has side lengths $12,35,37$. Classify it, then find the distance between $(-5,2)$ and $(7,37)$ without starting a new calculation. ::: :::: ## Working record For each calculation, record the longest side, the equation before taking a square root, the exact value, and a reasonableness check. ________________________________________________________________________