The Central Idea
In a right triangle, let the two legs have lengths \(a\) and \(b\), and let the hypotenuse have length \(c\).
\[ \boxed{a^2+b^2=c^2} \]This equation is the Pythagorean Theorem. It says that the square of the longest side is exactly the sum of the squares of the two perpendicular sides.
The Theorem Applies to Right Triangles
The Pythagorean Theorem is not a formula for every triangle. Its defining condition is a right angle.
If a triangle has a \(90^\circ\) angle, then the two sides forming that angle are the legs, and the side opposite the right angle is the hypotenuse.
Only after identifying those roles should we write
\[ a^2+b^2=c^2. \]Identifying the Hypotenuse
The hypotenuse is always opposite the right angle. It is also the longest side of a right triangle.
The symbols \(a\), \(b\), and \(c\) are conventional, not mandatory. What matters is that the quantity placed in the \(c\)-position represents the hypotenuse.
For example, if the legs have lengths \(x\) and \(7\), and the hypotenuse has length \(12\), then the correct setup is
\[ x^2+7^2=12^2. \]Why Squares Appear
The theorem can be interpreted geometrically. Construct a square outward on each side of a right triangle. The areas of the squares on the legs are \(a^2\) and \(b^2\), while the area of the square on the hypotenuse is \(c^2\).
The theorem states that
\[ \boxed{\text{area on leg 1}+\text{area on leg 2}=\text{area on hypotenuse}}. \]So the relationship is fundamentally an equality of areas, not simply a rule for adding side lengths.
A Simple Numerical Example
A right triangle has legs of lengths \(3\) and \(4\). Let the hypotenuse be \(c\).
\[ 3^2+4^2=c^2. \]Evaluate the squares:
\[ 9+16=c^2. \] \[ 25=c^2. \]Because a side length is positive,
\[ \boxed{c=5}. \]Finding the Hypotenuse
When both legs are known, add their squares and take the positive square root.
If the legs have lengths \(6\) and \(8\), then
\[ 6^2+8^2=c^2. \] \[ 36+64=c^2. \] \[ 100=c^2. \] \[ \boxed{c=10}. \]Finding a Missing Leg
If the hypotenuse and one leg are known, isolate the square of the missing leg by subtraction.
Suppose the hypotenuse is \(13\) and one leg is \(5\). Let the missing leg be \(b\).
\[ 5^2+b^2=13^2. \] \[ 25+b^2=169. \] \[ b^2=144. \] \[ \boxed{b=12}. \]Side Lengths Are Positive
Algebraically, the equation
\[ x^2=144 \]has two real solutions:
\[ x=12\qquad\text{or}\qquad x=-12. \]But a geometric side length cannot be negative. Therefore, when solving for a length, we take the positive value:
\[ \boxed{x=12}. \]Exact Radical Answers
Not every right triangle has integer side lengths. Suppose the legs have lengths \(5\) and \(7\).
\[ 5^2+7^2=c^2. \] \[ 25+49=c^2. \] \[ 74=c^2. \]Therefore,
\[ \boxed{c=\sqrt{74}}. \]Since \(74=2\cdot37\) contains no perfect-square factor greater than \(1\), \(\sqrt{74}\) is already in simplified exact form.
Simplifying Radical Lengths
Suppose a right triangle has legs \(6\) and \(12\).
\[ c^2=6^2+12^2=36+144=180. \]Thus,
\[ c=\sqrt{180}. \]Factor out the perfect square \(36\):
\[ \sqrt{180}=\sqrt{36\cdot5}=6\sqrt5. \]Therefore,
\[ \boxed{c=6\sqrt5}. \]Exact Form and Decimal Approximation
An exact radical and a decimal approximation serve different purposes.
For example,
\[ \sqrt{74}\approx8.6023. \]The radical \(\sqrt{74}\) is exact. The decimal is approximate because it has been rounded.
Unless a problem requests a decimal measurement, retaining the simplified radical usually preserves more information.
Estimating an Irrational Length
Even before using a calculator, nearby perfect squares give useful bounds. Since
\[ 64<74<81, \]taking positive square roots gives
\[ 8<\sqrt{74}<9. \]This provides a quick reasonableness check for any decimal approximation.
Pythagorean Triples
A Pythagorean triple is a set of three positive integers \((a,b,c)\) satisfying
\[ a^2+b^2=c^2. \]Common examples include
\[ (3,4,5),\qquad(5,12,13),\qquad(8,15,17),\qquad(7,24,25). \]Recognizing common triples can make calculations much faster, but the theorem itself remains the underlying justification.
Scaled Pythagorean Triples
If \((a,b,c)\) is a Pythagorean triple, multiplying all three values by the same positive factor produces another right-triangle side set.
Starting from \((3,4,5)\), multiply by \(4\):
\[ (12,16,20). \]Check:
\[ 12^2+16^2=144+256=400=20^2. \]This scaling behavior is consistent with the similarity of all triangles obtained by uniformly enlarging or reducing a given right triangle.
Connection to Similarity
The Pythagorean relationship can be proved using the similarity ideas from the previous lesson.
Draw an altitude from the right-angle vertex to the hypotenuse. The original right triangle is divided into two smaller right triangles. Each smaller triangle is similar to the original triangle and to the other smaller triangle by AA similarity.
Those proportional relationships imply that each leg squared equals the hypotenuse times the adjacent segment of the hypotenuse. Adding the two relationships yields
\[ a^2+b^2=c^2. \]This is one of many proofs of the theorem and shows that the relationship is deeply connected to Euclidean similarity.
A Similarity Proof in Symbols
Let the altitude to the hypotenuse divide the hypotenuse \(c\) into segments of lengths \(p\) and \(q\), so that
\[ p+q=c. \]Similarity gives the leg relationships
\[ a^2=cp \]and
\[ b^2=cq. \]Add them:
\[ a^2+b^2=cp+cq=c(p+q). \]Since \(p+q=c\),
\[ \boxed{a^2+b^2=c^2}. \]The Converse of the Pythagorean Theorem
The original theorem begins with a right triangle and concludes that its side lengths satisfy a square relationship.
The converse reverses the direction:
If a triangle has side lengths \(a\), \(b\), and \(c\), where \(c\) is the longest side, and
\[ a^2+b^2=c^2, \]then the triangle is a right triangle.
Testing Whether a Triangle Is Right
Consider side lengths \(9\), \(12\), and \(15\). The longest side is \(15\), so compare
\[ 9^2+12^2 \]with
\[ 15^2. \]We have
\[ 81+144=225 \]and
\[ 15^2=225. \]Therefore,
\[ \boxed{9^2+12^2=15^2}, \]so the triangle is right by the converse of the Pythagorean Theorem.
The Longest Side Must Play the Hypotenuse Role
When using the converse, first identify the longest side and place its square alone on the right side of the comparison.
For side lengths \(6\), \(8\), and \(10\), the correct test is
\[ 6^2+8^2\stackrel{?}{=}10^2. \]Testing a different arrangement can produce an irrelevant inequality and obscure the geometric meaning of the theorem.
Beyond Equality: Acute and Obtuse Comparisons
The Pythagorean comparison also helps classify a triangle by its largest angle. Let \(c\) be the longest side.
- If \(a^2+b^2=c^2\), the triangle is right.
- If \(a^2+b^2>c^2\), the triangle is acute.
- If \(a^2+b^2<c^2\), the triangle is obtuse.
This comparison extends the side-angle ordering relationship developed in the triangle-classification lesson.
Example: Acute, Right, or Obtuse?
Classify a triangle with side lengths \(5\), \(6\), and \(8\). The longest side is \(8\).
\[ 5^2+6^2=25+36=61 \]while
\[ 8^2=64. \]Since
\[ 61<64, \]the angle opposite the side of length \(8\) is greater than \(90^\circ\). Therefore,
\[ \boxed{\text{the triangle is obtuse}}. \]First Verify That the Side Lengths Form a Triangle
A Pythagorean comparison does not replace the triangle inequality. Before classifying three proposed lengths as acute, right, or obtuse, they must actually form a triangle.
For example, \(2\), \(3\), and \(6\) do not form a triangle because
\[ 2+3\not>6. \]Therefore, no triangle-angle classification applies to that set of lengths.
Right Triangles in the Coordinate Plane
Horizontal and vertical changes naturally form perpendicular legs. Suppose two points are
\[ A(x_1,y_1) \]and
\[ B(x_2,y_2). \]The horizontal change has magnitude
\[ |x_2-x_1|, \]and the vertical change has magnitude
\[ |y_2-y_1|. \]These form the legs of a right triangle whose hypotenuse is the straight-line distance from \(A\) to \(B\).
Deriving the Distance Formula
Let the distance between \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be \(d\). By the Pythagorean Theorem,
\[ (x_2-x_1)^2+(y_2-y_1)^2=d^2. \]Taking the positive square root gives
\[ \boxed{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}. \]The coordinate distance formula is therefore not an unrelated formula. It is the Pythagorean relationship written in coordinate language.
Example: Coordinate Distance
Find the distance between \((1,2)\) and \((7,10)\).
The horizontal change is
\[ 7-1=6, \]and the vertical change is
\[ 10-2=8. \]Therefore,
\[ d=\sqrt{6^2+8^2}=\sqrt{100}=\boxed{10}. \]The Order of Subtraction Does Not Affect Distance
In the distance formula, we could write \(x_2-x_1\) or \(x_1-x_2\), provided we use the corresponding coordinates consistently.
Squaring removes the sign:
\[ (x_2-x_1)^2=(x_1-x_2)^2. \]The same is true for the vertical difference. Distance is nonnegative and does not depend on which point is named first.
Diagonal of a Rectangle
A rectangle's length and width meet at a right angle. Its diagonal therefore forms the hypotenuse of a right triangle.
If a rectangle has length \(12\) and width \(5\), then its diagonal \(d\) satisfies
\[ d^2=12^2+5^2=144+25=169. \]Thus,
\[ \boxed{d=13}. \]Diagonal of a Square
If a square has side length \(s\), its diagonal \(d\) satisfies
\[ s^2+s^2=d^2. \]Therefore,
\[ 2s^2=d^2 \]and
\[ \boxed{d=s\sqrt2}. \]This is an important example of an exact irrational length arising naturally from Euclidean geometry.
A Three-Dimensional Extension
The same relationship can be applied in stages to perpendicular directions in three-dimensional space.
A rectangular box with side lengths \(a\), \(b\), and \(c\) has space diagonal \(d\). First, the diagonal of the base is
\[ \sqrt{a^2+b^2}. \]Then use that base diagonal and the third dimension as perpendicular legs:
\[ d^2=(a^2+b^2)+c^2. \]Hence,
\[ \boxed{d=\sqrt{a^2+b^2+c^2}}. \]Units in the Pythagorean Relationship
The theorem squares lengths during the calculation. If \(a\), \(b\), and \(c\) are measured in meters, then
\[ a^2, \quad b^2, \quad c^2 \]are expressed in square meters during the intermediate area relationship. Taking the square root returns the final side length to meters.
Compatible units should be established before applying the theorem.
Example with Unit Conversion
A right triangle has one leg of \(3\) meters and another leg of \(400\) centimeters.
Convert first:
\[ 400\text{ cm}=4\text{ m}. \]Then
\[ c^2=3^2+4^2=25, \]so
\[ \boxed{c=5\text{ m}}. \]Reasonableness: The Hypotenuse Must Be Longest
A computed hypotenuse should always be longer than either leg.
If the legs are \(9\) and \(12\), then a claimed hypotenuse of \(10\) is immediately impossible because \(10<12\).
The exact theorem confirms this:
\[ c=\sqrt{9^2+12^2}=\sqrt{225}=15. \]Reasonableness: Triangle Inequality Still Applies
In a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\), we must also have
\[ c<a+b. \]For a \(3\)-\(4\)-\(5\) triangle,
\[ 5<3+4. \]The Pythagorean relationship and the triangle inequality describe different structural requirements, and a valid right triangle satisfies both.
Common Error: Adding the Side Lengths
If the legs are \(3\) and \(4\), the hypotenuse is not
\[ 3+4=7. \]The theorem relates squares of lengths:
\[ c=\sqrt{3^2+4^2}=\sqrt{25}=5. \]Common Error: Forgetting the Final Square Root
Suppose
\[ c^2=169. \]The side length is not \(169\). The equation gives the square of the length. We still need
\[ c=\sqrt{169}=\boxed{13}. \]Common Error: Treating the Hypotenuse Like a Leg
Suppose the hypotenuse is \(10\) and one leg is \(6\). The missing leg is not found from
\[ 10^2+6^2. \]Instead, the hypotenuse square must equal the sum of the leg squares:
\[ 6^2+b^2=10^2. \]Therefore,
\[ b^2=100-36=64 \]and
\[ \boxed{b=8}. \]Common Error: Using the Theorem Without a Right Angle
Knowing three side lengths does not automatically justify
\[ a^2+b^2=c^2. \]For a general triangle, the equality may fail because the included angle between the two shorter sides need not be \(90^\circ\).
Use the Pythagorean Theorem when the triangle is already known to be right, or use its converse when testing whether the side lengths force a right triangle.
Common Error: Rounding Too Early
If an intermediate result contains a radical, retain the exact value until the final step whenever possible.
Early rounding can accumulate error, especially when the result will be used in another calculation.
For example, keep
\[ c=6\sqrt5 \]during subsequent exact work rather than immediately replacing it with a shortened decimal.
A Reliable Procedure for Finding a Missing Side
- Confirm that the triangle is right.
- Identify the hypotenuse, opposite the right angle.
- Assign the hypotenuse to the \(c\)-role.
- Write \(a^2+b^2=c^2\).
- Substitute the known side lengths.
- Isolate the square of the unknown side.
- Take the positive square root.
- Simplify the radical if possible.
- Approximate only if the problem requires it.
- Check that the result is geometrically reasonable.
A Reliable Procedure for Testing a Triangle
- Check that the three lengths satisfy the triangle inequality.
- Identify the longest side and call its length \(c\).
- Call the other two side lengths \(a\) and \(b\).
- Compute \(a^2+b^2\).
- Compute \(c^2\).
- Compare the two quantities.
- If they are equal, the triangle is right.
- If \(a^2+b^2>c^2\), the triangle is acute.
- If \(a^2+b^2<c^2\), the triangle is obtuse.
Example: Full Missing-Side Workflow
A right triangle has hypotenuse \(17\) and one leg \(8\). Find the other leg.
Let the missing leg be \(x\). Since \(17\) is the hypotenuse,
\[ x^2+8^2=17^2. \]Evaluate:
\[ x^2+64=289. \]Subtract \(64\):
\[ x^2=225. \]Take the positive square root:
\[ \boxed{x=15}. \]The result is less than the hypotenuse and forms the known triple \((8,15,17)\), providing two quick checks.
Example: Full Converse Workflow
Determine whether side lengths \(10\), \(24\), and \(26\) form a right triangle.
First, they satisfy the triangle inequality. The longest side is \(26\). Compare
\[ 10^2+24^2 \]with
\[ 26^2. \]Compute:
\[ 100+576=676 \]and
\[ 26^2=676. \]Since the values are equal,
\[ \boxed{\text{the triangle is right}}. \]Example: Applied Length
A ladder reaches \(12\) feet up a vertical wall while its base is \(5\) feet from the wall. Assuming the wall and ground are perpendicular, the ladder is the hypotenuse of a right triangle.
\[ L^2=12^2+5^2=144+25=169. \]Therefore,
\[ \boxed{L=13\text{ ft}}. \]Practice 1: Find the Hypotenuse
A right triangle has legs of lengths \(9\) and \(12\). Find the hypotenuse.
Practice 2: Find a Missing Leg
A right triangle has hypotenuse \(25\) and one leg \(7\). Find the other leg.
Practice 3: Exact Radical
A right triangle has legs of lengths \(4\) and \(9\). Find the hypotenuse in exact form.
Practice 4: Simplify the Radical
A right triangle has legs of lengths \(10\) and \(10\). Find the hypotenuse in simplified exact form.
Practice 5: Converse
Determine whether side lengths \(12\), \(35\), and \(37\) form a right triangle.
Practice 6: Classify by Pythagorean Comparison
A triangle has side lengths \(7\), \(8\), and \(10\). Classify it as acute, right, or obtuse.
Practice 7: Check Triangle Existence First
Can side lengths \(4\), \(5\), and \(10\) form an acute, right, or obtuse triangle? Explain.
Practice 8: Coordinate Distance
Find the distance between \((-2,3)\) and \((4,11)\).
Practice 9: Rectangle Diagonal
A rectangle is \(20\) centimeters long and \(21\) centimeters wide. Find its diagonal.
Practice 10: Square Diagonal
Find the exact diagonal length of a square whose side length is \(11\).
Practice 11: Three-Dimensional Extension
Find the space diagonal of a rectangular box with side lengths \(2\), \(3\), and \(6\).
Practice 12: Identify the Error
A student says that a right triangle with hypotenuse \(13\) and one leg \(12\) has missing leg
\[ \sqrt{13^2+12^2}=\sqrt{313}. \]Explain the error and find the correct length.
Practice Answers
Practice 1
\[ c^2=9^2+12^2=81+144=225. \] \[ \boxed{c=15}. \]Practice 2
\[ x^2+7^2=25^2. \] \[ x^2=625-49=576. \] \[ \boxed{x=24}. \]Practice 3
\[ c^2=4^2+9^2=16+81=97. \]Since \(97\) has no perfect-square factor greater than \(1\),
\[ \boxed{c=\sqrt{97}}. \]Practice 4
\[ c^2=10^2+10^2=200. \] \[ c=\sqrt{200}=\sqrt{100\cdot2}=\boxed{10\sqrt2}. \]Practice 5
The longest side is \(37\):
\[ 12^2+35^2=144+1225=1369 \]and
\[ 37^2=1369. \]Therefore,
\[ \boxed{\text{the triangle is right}}. \]Practice 6
The longest side is \(10\). Compare:
\[ 7^2+8^2=49+64=113 \]and
\[ 10^2=100. \]Since \(113>100\),
\[ \boxed{\text{the triangle is acute}}. \]Practice 7
The two shorter lengths fail the triangle inequality:
\[ 4+5=9<10. \]Therefore,
\[ \boxed{\text{the lengths do not form a triangle}}. \]Acute, right, and obtuse classifications do not apply.
Practice 8
\[ d=\sqrt{(4-(-2))^2+(11-3)^2}. \] \[ d=\sqrt{6^2+8^2}=\sqrt{100}=\boxed{10}. \]Practice 9
\[ d^2=20^2+21^2=400+441=841. \] \[ \boxed{d=29\text{ cm}}. \]Practice 10
\[ d^2=11^2+11^2=242. \] \[ d=\sqrt{242}=\sqrt{121\cdot2}=\boxed{11\sqrt2}. \]Practice 11
\[ d=\sqrt{2^2+3^2+6^2} =\sqrt{4+9+36} =\sqrt{49} =\boxed{7}. \]Practice 12
The student added the square of the known leg to the square of the hypotenuse. Because \(13\) is the hypotenuse, its square belongs alone on the right side:
\[ x^2+12^2=13^2. \] \[ x^2=169-144=25. \] \[ \boxed{x=5}. \]What to Remember
- The Pythagorean Theorem applies to right triangles.
- The hypotenuse is opposite the right angle and is the triangle's longest side.
- With legs \(a\) and \(b\) and hypotenuse \(c\), \(a^2+b^2=c^2\).
- To find a hypotenuse, add the squares of the legs and take the positive square root.
- To find a missing leg, subtract the known leg square from the hypotenuse square.
- Geometric side lengths are positive.
- Keep exact radical form when appropriate and simplify perfect-square factors.
- Pythagorean triples are integer solutions such as \((3,4,5)\) and \((5,12,13)\).
- The converse tests whether three valid triangle side lengths form a right triangle.
- When testing, the longest side must occupy the hypotenuse role.
- For longest side \(c\), compare \(a^2+b^2\) with \(c^2\) to distinguish acute, right, and obtuse triangles.
- Check the triangle inequality before classifying proposed side lengths.
- The coordinate distance formula is a direct application of the Pythagorean relationship.
- Rectangle, square, and box diagonals can be found by constructing right triangles.
- Use compatible measurement units before applying the theorem.
- Use exact work as long as practical and round only when the context requires an approximation.
The central relationship is
\[ \boxed{a^2+b^2=c^2}. \]Its coordinate form is
\[ \boxed{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}. \]Where We Go Next
The Pythagorean relationship reveals a fundamental metric structure inside right triangles. The next lesson expands the Euclidean picture from triangles to polygons and studies how interior and exterior angle totals are organized by the number of sides.
The next lesson, Polygons and Angle Sums, develops polygon vocabulary, triangulation, interior-angle sums, regular-polygon angles, exterior-angle sums, and the relationship between local angles and the overall structure of a polygon.