Using the Pythagorean theorem and its converse. Includes the vocabulary of legs, hypotenuse and theorem, finding a hypotenuse from two legs, finding a leg from the hypotenuse, solving when the sides are given as expressions, using the converse to test whether a triangle is right-angled, and applying the theorem to diagonals.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
The Pythagorean Theorem and Its Converse
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-729 — the lesson these objectives are drawn from
Warm-up
Chapter 9 solved equations by taking square roots. This lesson supplies a geometric reason to do so, over and over.
Discussion prompt
If x squared equals a hundred and x is a length, what is x? Why is only one answer reported?
Hint: Lengths cannot be negative.
Answer:
\[ x^2 = 100 \;\Longrightarrow\; x = \pm 10, \text{ but as a length } x = 10 \]
The algebra gives both roots and the situation keeps one, exactly as with times in Lesson 9.2. Every calculation in this lesson ends the same way, so the rejection becomes routine rather than a special remark.
Concept
In a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. The hypotenuse is the side opposite the right angle and the other two sides are the legs.
Pythagorean theorem — If a triangle is a right triangle with legs of lengths a and b and hypotenuse of length c, then a squared plus b squared equals c squared.
A theorem is a statement that can be proved.
Figure (svg): A right triangle with its legs and hypotenuse labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-724
Section
Section 1
Concept
Given both legs, square each, add them, and take the positive square root to find the hypotenuse. Only the positive root is used, since a length cannot be negative.
\[ a^2 + b^2 = c^2 \]
The hypotenuse is opposite the right angle.
Figure (svg): Finding a hypotenuse from two legs
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-724 — Example 1, part a, and its Study Tip on the positive square root
Picture it
Squares add.
Figure (svg): Finding a hypotenuse from two legs
The hypotenuse always comes out longer than either leg, which is a quick sanity check on any answer. If it does not, the sides have probably been misidentified.
Worked example
This is Example 1, part a, from the textbook.
\[ \text{Given legs } a = 6 \text{ and } b = 8, \text{ find } c. \]
Write the theorem
Why: The general relationship.
\[ a ^{2} + b ^{2} = c ^{2} \]
Substitute
Why: Six and eight.
\[ 36 + 64 = c ^{2} \]
Add
Why: A hundred.
\[ c ^{2} = 100 \]
Take the positive root
Why: A length.
\[ c = 10 \]
Figure (svg): Finding a hypotenuse from two legs
\[ c = 10 \]
Verify: check the size
Why: Ten is longer than both six and eight, as a hypotenuse must be, and shorter than their sum of fourteen, as any side of a triangle must be. Both checks pass in a couple of seconds.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-724
Faded example
Three steps in order.
Fill in the blanks
6^2 + 8^2 = 100 \;\Longrightarrow\; c = 10
Why: The squares are added and then a single root is taken of the total. Taking roots of each square separately would simply return the original legs.
Worked example
Guided Practice 1 to 3.
\[ \text{Find } c \text{ for legs } 12 \text{ and } 5, \; 3 \text{ and } 4, \; 12 \text{ and } 16. \]
Take the first
Why: 144 plus 25.
\[ c ^{2} = 169, \; c = 13 \]
Take the second
Why: Nine plus sixteen.
\[ c ^{2} = 25, \; c = 5 \]
Take the third
Why: 144 plus 256.
\[ c ^{2} = 400, \; c = 20 \]
Note the pattern
Why: All came out whole.
Figure (svg): Finding a hypotenuse from two legs
\[ 13, \quad 5, \quad 20 \]
Verify: notice the third is a multiple
Why: Twelve, sixteen and twenty is four times three, four and five, so it is the same triangle enlarged. Multiplying every side of a right triangle by the same factor keeps it right-angled, which is worth knowing for spotting answers quickly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725
Trap
\[ c = 6 + 8 = 14 \]
Add the two legs to get the third side
Why: The theorem relates the three sides, so they were added.
The theorem adds the squares, not the sides. Fourteen would be the sum of the two legs, which is longer than any side of a triangle can be — a triangle with sides six, eight and fourteen would be flat.
\[ 6^2 + 8^2 = c^2 \;\Longrightarrow\; c = 10 \]
Square first, add, then take the root
Why: Three operations, in that order.
The answer must lie between the longer leg and the sum of the two, which fourteen fails.
Matching
Square, add, root.
Match the pairs
Why: Three of these are multiples of the three-four-five triangle and one is not. Recognising such families lets many answers be written down without any calculation.
Elimination
In a right triangle.
Eliminate the wrong options
How do you identify it?
Survives elimination: A
Why: Being opposite the right angle is what defines it, and being longest follows from that. Both descriptions are independent of how the triangle happens to be drawn.
Socratic
It follows from the theorem.
Discussion prompt
Explain why c is always greater than either leg. Then say what that gives you as a check.
Hint: Compare c squared with a squared.
Answer:
Since c squared equals a squared plus b squared and b squared is positive, c squared is larger than a squared — and larger squares mean larger lengths for positive quantities. The same argument applies to b, so the hypotenuse exceeds both legs.
As a check it is immediate: any answer for a hypotenuse that is not the largest of the three numbers is wrong, and any answer for a leg that exceeds the hypotenuse is wrong. That catches most substitution errors before any arithmetic is rechecked.
Section
Section 2
Concept
When the hypotenuse and one leg are known, substituting into the theorem and rearranging gives the other leg's square as a difference.
\[ b^2 = c^2 - a^2 \]
The answer is often irrational.
Figure (svg): Finding a leg from the hypotenuse and the other leg
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-724 — Example 1, part b
Picture it
The hypotenuse's square is the larger.
Figure (svg): Finding a leg from the hypotenuse and the other leg
Substituting into the theorem as written and then rearranging is safer than remembering a second formula. There is only one relationship to recall.
Worked example
This is Example 1, part b, from the textbook.
\[ \text{Given } a = 5 \text{ and } c = 6, \text{ find } b. \]
Write the theorem
Why: As always.
\[ a ^{2} + b ^{2} = c ^{2} \]
Substitute
Why: Five and six.
\[ 25 + b ^{2} = 36 \]
Rearrange
Why: Subtract twenty-five.
\[ b ^{2} = 11 \]
Take the positive root
Why: Not a perfect square.
\[ b = \sqrt{11} \approx 3.32 \]
Figure (svg): Finding a leg from the hypotenuse and the other leg
\[ b = \sqrt{11} \approx 3.32 \]
Verify: check the size
Why: The answer is less than the hypotenuse of six, as a leg must be, and the exact value is a radical rather than a whole number. Most real triangles give irrational sides; the tidy ones in exercises are chosen deliberately.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-724
Faded example
Subtract from the hypotenuse's square.
Fill in the blanks
25 + b^2 = 36 \;\Longrightarrow\; b^2 = 11 \;\Longrightarrow\; b = \sqrt11}
Why: Substituting first and rearranging afterwards means only one relationship has to be remembered. The subtraction emerges from the algebra rather than being decided in advance.
Worked example
Where the difference is a perfect square.
\[ \text{Given } a = 9 \text{ and } c = 15, \text{ find } b. \]
Substitute
Why: Eighty-one and two hundred and twenty-five.
\[ 81 + b ^{2} = 225 \]
Rearrange
Why: Subtract eighty-one.
\[ b ^{2} = 144 \]
Take the root
Why: A perfect square.
\[ b = 12 \]
Check the triangle
Why: Nine, twelve, fifteen.
\[ 3 \times (3, 4, 5) \]
Figure (svg): Finding a leg from the hypotenuse and the other leg
\[ b = 12 \]
Verify: recognise the family
Why: Nine, twelve and fifteen is three times the three-four-five triangle, which is why the answer came out whole. Spotting a familiar multiple often lets the answer be written down before the arithmetic is finished.
Trap
\[ b^2 = 5^2 + 6^2 = 61 \]
Add the two given squares
Why: The theorem adds squares, so the two given ones were added.
The theorem adds the two legs' squares to give the hypotenuse's. Here six is the hypotenuse, so its square is the total and the known leg's square must be subtracted from it.
\[ b^2 = 6^2 - 5^2 = 11 \]
Substitute into the theorem first, then rearrange
Why: Rather than deciding the operation in advance.
An answer larger than the hypotenuse is impossible, which catches this immediately.
Sorting
Which side is unknown?
Sort into buckets
Sort each situation by the operation needed.
The operation is decided entirely by which side is missing. Substituting into the theorem before rearranging makes that decision automatic rather than something to remember.
Prediction
Compared with the hypotenuse.
Predict first
What must be true of any leg's length?
Correct: It is less than the hypotenuse.
\[ b^2 = c^2 - a^2 < c^2 \;\Longrightarrow\; b < c \]
Why: Since the hypotenuse's square is the sum of both legs' squares, each leg's square is less than it, and so is each leg. That gives an immediate check: an answer for a leg that exceeds the given hypotenuse means the sides were mixed up, most likely by adding when a subtraction was needed.
Socratic
The rearranged form could be memorised.
Discussion prompt
Say why writing the theorem out and substituting is safer than recalling a formula for a leg. Then say what would have to be memorised otherwise.
Hint: How many arrangements are there?
Answer:
There is only one relationship to remember, and substituting the known values into it makes the required rearrangement obvious. The alternative is to recall which square is subtracted from which, and that is exactly the detail people reverse under pressure.
Memorising separate formulas would mean holding three versions — one for each side being unknown — and choosing between them correctly. One equation plus ordinary rearranging replaces all three, and it also makes the check available, since an impossible answer shows up as a negative square.
Section
Section 3
Concept
When the sides are described in terms of an unknown, substituting into the theorem produces a quadratic equation. Solve it and reject any negative solution as a length.
All of Chapter 10's factoring applies.
Figure (svg): A right triangle whose sides are given as expressions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725 — Example 2, Use the Pythagorean Theorem
Picture it
The same theorem, a longer equation.
Figure (svg): A right triangle whose sides are given as expressions
Squaring the bracket is where the extra work lies, and it is Lesson 10.3's pattern. The rest is factoring and rejecting a negative length.
Worked example
This is Example 2 from the textbook.
\[ \text{One leg is } 3 \text{ longer than the other and the hypotenuse is } 15. \text{ Find the legs.} \]
Label the sides
Why: Let x be the shorter leg.
\[ x \text{ and } x + 3 \]
Substitute
Why: Into the theorem.
\[ x ^{2} + (x + 3) ^{2} = 225 \]
Expand and simplify
Why: The bracket squares out.
\[ 2 x ^{2} + 6 x - 216 = 0 \]
Factor and solve
Why: Reject the negative.
\[ x = 9 \]
Figure (svg): A right triangle whose sides are given as expressions
\[ 9 \text{ and } 12 \]
Verify: check with the theorem
Why: Eighty-one plus a hundred and forty-four is two hundred and twenty-five, which is fifteen squared. The legs also differ by three as required, so both conditions of the problem are met.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725
Faded example
Three terms, not two.
Fill in the blanks
(x + 3)^2 = x^2 + 6x + 9
Why: The middle term is twice the product of the two terms, which is the part most often dropped. Losing it changes the equation and the answer.
Worked example
The rejection, made explicit.
\[ \text{Why is } x = -12 \text{ discarded?} \]
Note what x represents
Why: The shorter leg.
Consider the value
Why: Negative twelve.
Check the algebra
Why: It does solve the equation.
Reject it
Why: The situation forbids it.
Figure (svg): A right triangle whose sides are given as expressions
\[ x > 0 \]
Verify: see what it would describe
Why: At x equal to negative twelve the other leg would be negative nine, and squaring both does give two hundred and twenty-five — so the equation is satisfied. The rejection is a statement about lengths rather than about the algebra, exactly as with times in Chapter 9.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725
Error analysis
The student substituted expressions into the theorem.
Annotate
On: \( \begin{aligned} x^2 + (x + 3)^2 &= 15^2 \\ x^2 + x^2 + 9 &= 225 \\ 2x^2 &= 216 \end{aligned} \)
This is the squaring-a-sum error from Chapter 10 appearing in a geometric setting. Whenever a bracket is squared, writing the three-term expansion out in full is what prevents it.
Faded example
The algebra gives two roots.
Fill in the blanks
The solutions are 9 and -12, but a length cannot be negative, so the shorter leg is 9.
Why: Both values satisfy the equation and only one describes a triangle. That rejection is a modelling judgement rather than an algebraic one, and it should be stated rather than done silently.
Elimination
Legs x and x + 3, hypotenuse 15.
Eliminate the wrong options
Which equation follows from the theorem?
Survives elimination: A
Why: Every side is squared, including the hypotenuse, and the bracket is squared as a whole. Options C and D each get one of those two points wrong.
Socratic
The theorem is just three squares.
Discussion prompt
Explain why substituting expressions gives a quadratic equation. Then say what that means about how many solutions to expect and to check.
Hint: What is the highest power of x?
Answer:
Each side is squared, so an expression linear in x becomes quadratic, and adding two such squares leaves a quadratic. That is why the theorem, which is not itself about quadratics, produces one as soon as the sides are unknown.
So up to two solutions should be expected, and both must be examined — not for algebraic correctness but for whether they describe a possible triangle. In practice one is almost always negative and rejected, which makes the pattern predictable once you have seen it two or three times.
Section
Section 4
Concept
The converse of the Pythagorean theorem says that if the sum of the squares of two sides equals the square of the third, then the triangle is a right triangle. It is used to test a triangle rather than to measure one.
\[ a^2 + b^2 = c^2 \;\Longrightarrow\; \text{right triangle} \]
The theorem and its converse run in opposite directions.
Figure (svg): Two columns distinguishing the theorem from its converse
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-729 — the converse named in the lesson title and key words
Picture it
Given the angle, or given the sides.
Figure (svg): Two columns distinguishing the theorem from its converse
The theorem assumes a right angle and concludes a relationship; the converse assumes the relationship and concludes a right angle. Keeping the two apart is what makes each usable.
Worked example
Applying the converse.
\[ \text{Are the triangles with sides } 9, 12, 15; \; 5, 6, 8; \text{ and } 8, 15, 17 \text{ right-angled?} \]
Take the first
Why: Eighty-one plus a hundred and forty-four.
\[ 225 = 15^2 \;\checkmark \]
Take the second
Why: Twenty-five plus thirty-six.
\[ 61 \ne 64 \]
Take the third
Why: Sixty-four plus two hundred and twenty-five.
\[ 289 = 17^2 \;\checkmark \]
Report
Why: Two are right-angled.
Figure (svg): Two columns distinguishing the theorem from its converse
\[ 9, 12, 15 \;\checkmark \quad 5, 6, 8 \;\times \quad 8, 15, 17 \;\checkmark \]
Verify: check the longest side was used as c
Why: In each case the largest number was squared on its own, since only the longest side can be the hypotenuse. Testing with a shorter side as c would give a false negative every time.
Sorting
Square the two shorter sides and compare.
Sort into buckets
Sort each set of side lengths by whether it forms a right triangle.
Two of the three right triangles are multiples of three-four-five and the third is the eight-fifteen-seventeen family. Both are worth recognising, since they recur constantly in exercises.
Worked example
The step that decides the test.
\[ \text{For sides } 8, 15 \text{ and } 17, \text{ which must be the hypotenuse?} \]
Recall the size rule
Why: The hypotenuse is longest.
\[ c \text{ is the largest} \]
Pick the largest
Why: Seventeen.
\[ c = 17 \]
Square the other two
Why: Sixty-four plus two hundred and twenty-five.
\[ 289 \]
Compare
Why: Seventeen squared.
\[ 289 \;\checkmark \]
Figure (svg): Two columns distinguishing the theorem from its converse
\[ c = 17 \]
Verify: try it the wrong way round
Why: Using fifteen as the hypotenuse would give sixty-four plus two hundred and eighty-nine against two hundred and twenty-five, which fails — and would wrongly suggest the triangle is not right-angled. Identifying the longest side first is essential to the test.
Trap
The sides are 5, 6 and 8, so by the Pythagorean theorem 25 plus 36 equals 64.
Apply the theorem to any triangle
Why: The theorem relates three sides, and here are three sides.
The theorem applies only to right triangles, and this one has not been shown to be one. In fact twenty-five plus thirty-six is sixty-one rather than sixty-four, so by the converse it is not right-angled at all.
Test with the converse: 25 + 36 = 61, which is not 64, so it is not a right triangle.
Use the theorem when the right angle is known and the converse when it is not
Why: Direction matters.
The theorem never claims anything about triangles that are not right-angled.
Faded example
The longest side is c.
Fill in the blanks
8^2 + 15^2 = 64 + 225 = 289, \text289 17^2 = ___, \text___
Why: The two sums agree exactly, so the converse concludes the triangle is right-angled. Had they differed by even one, the conclusion would have been the opposite.
Elimination
Compare the directions.
Eliminate the wrong options
Which one is the converse of the Pythagorean theorem?
Survives elimination: A
Why: A converse swaps the hypothesis and the conclusion, so it starts from the side relationship and concludes the right angle. That is what makes it a test rather than a measuring tool.
Socratic
It looks like the same fact.
Discussion prompt
Explain why a converse does not automatically follow from a statement. Then give an everyday example where a statement is true but its converse is not.
Hint: Swapping the two halves changes the claim.
Answer:
A statement and its converse make different claims, so proving one says nothing about the other — the theorem says a right angle guarantees the relationship, and the converse says the relationship guarantees a right angle. Both happen to be true here, but that is a fact requiring its own proof.
An everyday case: if it is raining then the ground is wet, but the ground being wet does not mean it is raining, since it may have been hosed. Direction matters in mathematical statements for exactly the same reason, which is why the converse is named and stated rather than assumed.
Section
Section 5
Concept
A diagonal splits a rectangle into two right triangles, so the theorem gives its length. For a square, the diagonal is the side length times the square root of two.
\[ d = s\sqrt{2} \quad \text{for a square of side } s \]
The ratio holds whatever the side length.
Figure (svg): The diagonal of a square found with the theorem
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725 — Example 3, on the diagonal of a square board
Picture it
The theorem applies inside a square.
Figure (svg): The diagonal of a square found with the theorem
The shape contains no obvious triangle until the diagonal is drawn. Drawing it is what makes the theorem available.
Worked example
This is Example 3 from the textbook.
\[ \text{A square board is } 2 \text{ feet by } 2 \text{ feet. Find its diagonal.} \]
Draw the diagonal
Why: It forms a right triangle.
\[ \text{legs of } 2\text{ and } 2 \]
Substitute
Why: Both legs are two.
\[ c ^{2} = 4 + 4 \]
Simplify
Why: Eight.
\[ c ^{2} = 8 \]
Take the root
Why: Two root two.
\[ c \approx 2.8 \text{ ft} \]
Figure (svg): The diagonal of a square found with the theorem
\[ c = \sqrt{8} = 2\sqrt{2} \approx 2.8 \text{ feet} \]
Verify: check against the sides
Why: The diagonal of 2.8 feet is longer than either two-foot side and shorter than their sum of four, as it must be. Simplifying the radical to two root two also shows the ratio to the side directly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 725-725
Faded example
Both legs are the side.
Fill in the blanks
d^2 = s^2 + s^2 = 2s^2 \;\Longrightarrow\; d = s\sqrt2}
Why: Both legs being equal makes the sum twice one square, and the root of that is the side times the root of two. The ratio is independent of the side length.
Worked example
Generalising from the board to a diamond.
\[ \text{A baseball diamond is a square with } 90 \text{ foot sides. Find the diagonal.} \]
Substitute
Why: Both legs are ninety.
\[ d ^{2} = 8100 + 8100 \]
Simplify
Why: Sixteen thousand two hundred.
\[ d ^{2} = 16 \, 200 \]
Take the root
Why: Ninety root two.
\[ d = 90\sqrt{2} \]
Evaluate
Why: Ninety times 1.414.
\[ \approx 127 \text{ feet} \]
Figure (svg): A baseball diamond with its diagonal marked
\[ d = 90\sqrt{2} \approx 127 \text{ feet} \]
Verify: check the ratio against the board
Why: The board's diagonal was two root two for a side of two, and the diamond's is ninety root two for a side of ninety. The ratio of root two holds at every size, which is why it is worth learning as a fact rather than recomputing.
Trap
\[ d = 2 \times 90 = 180 \text{ feet} \]
Double the side to cross the square
Why: Going across seems like going along two sides.
Going along two sides really is a hundred and eighty feet, but the diagonal cuts the corner. It is only about a hundred and twenty-seven feet, which is why runners take it.
\[ d = 90\sqrt{2} \approx 127 \text{ feet} \]
Use the theorem rather than adding the sides
Why: The diagonal is a hypotenuse.
The diagonal must exceed one side and fall short of two, which the correct answer does.
Prediction
Crossing a square rather than going round two sides.
Predict first
What fraction of the two-side route is the diagonal?
Correct: About 71 per cent.
\[ \dfrac{\sqrt{2}}{2} \approx 0.707 \]
Why: The diagonal is the side times the root of two, about 1.414 sides, while two sides is exactly two — so the ratio is about 0.707, or roughly seventy-one per cent. On a baseball diamond that is a hundred and twenty-seven feet against a hundred and eighty, a saving of about fifty-three feet. The ratio is the same for any square, which is why it is worth remembering as a number rather than recomputing.
Sorting
A right angle must be present or constructible.
Sort into buckets
Sort each situation by whether the Pythagorean theorem applies.
Rectangles and squares contain right angles by definition and a wall meets the ground at one. A general triangle or a slanted parallelogram does not, and for those a different relationship would be needed.
Socratic
It could be recomputed each time.
Discussion prompt
Say why the diagonal-to-side ratio of a square is worth remembering. Then name another shape whose proportions are worth knowing for the same reason.
Hint: How often do squares appear?
Answer:
Squares are everywhere — rooms, boards, tiles, city blocks, baseball diamonds — so the question of how far it is across one recurs constantly. Knowing that the answer is about 1.41 times the side lets it be estimated instantly, and it also gives a check on any computed answer.
The other worth knowing is the three-four-five triangle and its multiples, which appear so often in problems that recognising six-eight-ten or nine-twelve-fifteen saves the whole calculation. Builders use the same fact in reverse to construct right angles on site, measuring three and four units along two edges and checking that the diagonal is five.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| The theorem | The converse | |
|---|---|---|
| You are given | a right triangle | three side lengths |
| You conclude | the relationship between the squares | the triangle is right-angled |
| Used to | find a missing side | test whether a triangle has a right angle |
Both are true and they are different statements. Which one you need depends on whether the right angle is given or is the thing being asked about.
Pattern
To use the Pythagorean theorem on any problem, these five moves cover it.
Step one decides everything else, since putting a leg where the hypotenuse belongs turns an addition into a subtraction and gives a plausible wrong answer.
Check
Squares add.
Check your understanding
A right triangle has legs 6 and 8. What is the hypotenuse?
Answer: A
Why: Thirty-six plus sixty-four is a hundred, whose positive square root is ten.
Check
Subtract when a leg is unknown.
Check your understanding
A right triangle has leg 5 and hypotenuse 6. What is the other leg?
Answer: A
Why: Twenty-five plus the unknown square is thirty-six, so that square is eleven and the leg is its root.
Check
Use the converse.
Check your understanding
Is a triangle with sides 5, 6 and 8 a right triangle?
Answer: A
Why: The squares of the two shorter sides total sixty-one, which is not sixty-four, so the converse does not apply and the triangle is not right-angled.
Real world
This is the baseball question from the lesson opener. A standard diamond is a square with 90 foot sides, and the distance from home plate to second base is its diagonal.
Discussion prompt
Find that distance exactly and to the nearest foot, and say how much shorter it is than running along the two base paths.
Hint: The diagonal is the hypotenuse of a right triangle.
Answer:
\[ d^2 = 90^2 + 90^2 = 16\,200 \;\Longrightarrow\; d = 90\sqrt{2} \approx 127 \text{ feet} \]
Running from home to first to second covers a hundred and eighty feet, so the direct throw across the diamond saves about fifty-three feet.
The exact answer is ninety root two, which is worth writing down because it shows the general rule: a square's diagonal is always its side times the root of two, about 1.41 times, whatever the size. That ratio is why a catcher's throw to second is so much shorter than the runner's path, and it is the same fact that made the two-foot board's diagonal two root two feet.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A right triangle has one leg 5 and hypotenuse 6. What is the other leg?
Correct: root 11, about 3.32.
\[ 5^2 + b^2 = 6^2 \;\Longrightarrow\; b^2 = 11 \]
Why: Six is the hypotenuse, so its square is the total: twenty-five plus the unknown square equals thirty-six, giving eleven. The first option adds the two given squares, which would be right if both given lengths were legs, but a leg can never exceed the hypotenuse and 7.81 is larger than six — that size check alone rejects it in a second. The third option subtracts the sides instead of their squares, and the fourth stops before taking the root. Substituting into the theorem as written and rearranging afterwards makes the subtraction emerge from the algebra rather than being a decision to remember.
Explain it
They added the two given squares when the hypotenuse was one of the given sides.
Discussion prompt
In no more than four sentences, explain how to tell which operation is needed. Then give them a check on the answer's size.
Hint: Which side is opposite the right angle?
Answer:
A usable answer: find the hypotenuse first — it is opposite the right angle and is always the longest side. If it is the unknown you add the two squares, and if it is given you subtract the known leg's square from it.
The check is on size. A leg must come out shorter than the hypotenuse, so if your answer for a leg is bigger than the hypotenuse you added when you should have subtracted.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The hypotenuse is fixed by looking for the side opposite the right angle, which is also the longest. The operation is fixed by substituting into the theorem before rearranging. Algebraic sides are fixed by expanding the squared bracket in full and rejecting negative lengths. The converse is fixed by asking whether the right angle is given or is what you are trying to establish. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page draw a right triangle, mark the right angle, label the two legs and the hypotenuse, and write the theorem beneath it. Underneath, work two problems side by side: one finding a hypotenuse from two legs and one finding a leg from the hypotenuse, writing the substitution before the rearrangement in both so the different operations emerge rather than being chosen. In the middle, take a triangle whose sides are given as expressions, sketch and label it, substitute, expand the squared bracket in full with all three terms, solve the quadratic and strike through the negative solution with a reason. Beneath that, write the theorem and its converse side by side with arrows showing which direction each runs, then test three sets of side lengths with the converse, marking which side you took as the hypotenuse each time. In the lower corner, draw a square, add its diagonal, derive the root-two ratio, and use it on a ninety-foot diamond. Finally, in the margin, write the two size checks that catch most errors.
Every answer on your page should satisfy both size checks: the hypotenuse largest, and each side less than the sum of the other two. Those two conditions catch nearly every substitution error without any rechecking of arithmetic.
Recap
Five things, and the first decides the other four.
| If the question says | Your first move is |
|---|---|
| Find the hypotenuse | Add the squares of the legs |
| Find a leg | Subtract from the hypotenuse's square |
| The sides are expressions | Substitute and expect a quadratic |
| Is this a right triangle? | Use the converse with the longest side as c |
| Find a diagonal | Draw it and treat it as a hypotenuse |
Lesson 12.7 applies the theorem to the coordinate plane. The distance between two points is the hypotenuse of a right triangle whose legs are the horizontal and vertical gaps between them.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.6 The Pythagorean Theorem and Its Converse §12.6, pp. 724-729 — everything on these slides traces back here
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