Unit 1: Right Triangles, Distance & Rigid Motion | Geometry

Geometry • Unit 1

Right Triangles, Distance & Rigid Motion

One unit, two big ideas. First, the Pythagorean Theorem and what it lets you measure — including distance on the coordinate plane. Second, rigid motion: moving a figure by a vector or across a line without changing its size or shape.

Lesson Map Content Summary Vocabulary Sample Problems Equations & Rules Video Library Review Sheet Inquiry Project

Lesson Map

This is the order the unit is taught in class. Each row names the one skill that shows up on the assessment.

Lesson	Focus	The skill being checked
Lesson 1 — Day 1	The Pythagorean Theorem	Identify the hypotenuse, then decide whether the problem adds or subtracts.
Lesson 1 — Day 2	Distance on the coordinate plane	Find a distance either by building a right triangle on the grid or by using the distance formula.
Lesson 1 — Day 3	Applications and word problems	Turn a written situation into a labeled sketch before doing any algebra.
Lesson 2	The converse and Pythagorean triples	Decide whether three given side lengths make a right, acute, or obtuse triangle.
Lesson 3	Translations by vector	Apply a translation vector to every vertex, and recover the vector from a preimage and image.
Lesson 4 — Day 1	Reflections across the axes	Apply \((x,y)\rightarrow(x,-y)\) and \((x,y)\rightarrow(-x,y)\) from memory.
Lesson 4 — Day 2	Reflections across \(y=x\) and \(y=-x\)	Apply \((x,y)\rightarrow(y,x)\) and \((x,y)\rightarrow(-y,-x)\) from memory.
Inquiry	Stake It Out	Use the whole unit outdoors: build a right angle, survey a space, and lay out a transformed design.
Assessment	Unit 1 review sheet	Twelve questions in assessment order, with an answer key and a reteach guide.

Content Summary

The idea first, then the method, then a worked example in the format you are expected to reproduce.

Lesson 1 — Day 1

The Pythagorean Theorem

In a right triangle, the two shorter sides that form the right angle are the legs. The side opposite the right angle is the hypotenuse, and it is always the longest side. The theorem relates the three:

\[a^2+b^2=c^2\]

The small square at the corner marks the right angle. \(c\) sits opposite it.

Two things decide every problem:

Find the hypotenuse first. It is the side opposite the right angle, never one of the two sides that touch it.
Then decide add or subtract. If both legs are known you are looking for the hypotenuse, so you add the squares. If the hypotenuse and one leg are known, you subtract.

The most common error in this unit is adding when you should subtract. Before substituting, say out loud which side is the hypotenuse. If the unknown is the hypotenuse, add. If the unknown is a leg, subtract.

Worked example. A right triangle has legs of 6 and 8. Find the hypotenuse.

\[6^2+8^2=c^2\;\Rightarrow\;36+64=c^2\;\Rightarrow\;100=c^2\;\Rightarrow\;c=10\]

Worked example. A right triangle has a hypotenuse of 13 and one leg of 5. Find the other leg.

\[5^2+b^2=13^2\;\Rightarrow\;25+b^2=169\;\Rightarrow\;b^2=144\;\Rightarrow\;b=12\]

Answers are rounded to two decimal places unless the square root is exact. \(\sqrt{50}\) is reported as \(7.07\), not as \(7\).

Lesson 1 — Day 2

Distance on the Coordinate Plane

Two points on a grid are the ends of an invisible hypotenuse. Draw a horizontal leg and a vertical leg to meet at a right angle, and the distance between the points is exactly the Pythagorean Theorem again.

From \((-2,-1)\) to \((4,7)\): the horizontal leg is 6, the vertical leg is 8, so the distance is 10.

Method 1 — count the legs. Count the horizontal run and the vertical rise on the grid, then use \(a^2+b^2=c^2\). This works well when the points are plotted for you and lands on whole numbers often enough to be worth checking.

Method 2 — the distance formula. The formula is the same idea written once, so you do not have to draw:

\[d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

The subtraction order does not matter, because each difference is squared. \((-2-4)^2\) and \((4-(-2))^2\) are both 36.

Worked example. Find the distance between \((-2,-1)\) and \((4,7)\).

\[d=\sqrt{(-2-4)^2+(-1-7)^2}=\sqrt{36+64}=\sqrt{100}=10.00\]

Watch the negatives. Subtracting a negative is the step students lose most often. Write the substitution with parentheses around every coordinate before simplifying.

Lesson 1 — Day 3

Applications: Sketch First

Word problems in this unit are Pythagorean problems in disguise. The work is not the algebra; the work is figuring out which measurement is the hypotenuse. A sketch settles it.

Three situations cover almost every problem you will meet:

A leaning object. A ladder, a ramp, a guy wire, a zip line. The leaning object is the hypotenuse; the ground and the vertical height are the legs.
A diagonal across a rectangle. A television screen, a garden gate, a soccer field. The diagonal is the hypotenuse; the length and width are the legs.
Two legs of a journey at right angles. Walking 3 blocks north then 4 blocks east. The straight-line shortcut home is the hypotenuse.

The routine. Draw the triangle. Mark the right angle. Label the two known measurements with their units. Put a variable on the unknown. Only then substitute.

On the assessment, the sketch earns points on its own. A correct number with no diagram loses credit on the questions that ask for a picture.

Lesson 2

The Converse and Pythagorean Triples

The theorem tells you a missing side when you already know the triangle is right. The converse runs the other direction: if the three sides satisfy \(a^2+b^2=c^2\), then the triangle must be right.

\[\text{If } a^2+b^2=c^2 \text{, then the triangle is a right triangle.}\]

To test three lengths, identify the longest side first, square it, and compare that with the sum of the squares of the other two. You are comparing, not solving.

Comparison	Conclusion
\(a^2+b^2=c^2\)	Right triangle
\(a^2+b^2>c^2\)	Acute triangle
\(a^2+b^2<c^2\)	Obtuse triangle

Pythagorean triples are whole-number side lengths that satisfy the theorem exactly. Recognizing them saves time, and any multiple of a triple is also a triple.

3 – 4 – 5

Multiples: 6–8–10, 9–12–15, 30–40–50.

5 – 12 – 13

Multiples: 10–24–26, 15–36–39.

8 – 15 – 17

Multiples: 16–30–34.

7 – 24 – 25

Multiples: 14–48–50.

Worked example. Do 9, 40, and 41 form a right triangle?

\[9^2+40^2=81+1600=1681 \qquad 41^2=1681\]

The two sides are equal, so yes — this is a right triangle, and 9–40–41 is a triple.

Lesson 3

Translations by Vector

A translation slides every point of a figure the same distance in the same direction. Nothing turns, nothing flips, nothing changes size. A translation is described by a vector written \([h,\,k]\): move \(h\) units horizontally and \(k\) units vertically.

\[[h,\,k]:\quad (x,y)\rightarrow(x+h,\;y+k)\]

Positive \(h\) moves right and negative \(h\) moves left. Positive \(k\) moves up and negative \(k\) moves down.

Triangle \(PQR\) translated by \([6,-3]\): every vertex moves right 6 and down 3.

Worked example. Translate \(P(-5,1)\), \(Q(-2,4)\), \(R(-1,0)\) by \([6,-3]\).

Preimage	Rule applied	Image
P(−5, 1)	(−5 + 6, 1 − 3)	P′(1, −2)
Q(−2, 4)	(−2 + 6, 4 − 3)	Q′(4, 1)
R(−1, 0)	(−1 + 6, 0 − 3)	R′(5, −3)

Running it backwards. Given a point and its image, the vector is image minus preimage. If \(A(2,5)\rightarrow A'(-1,9)\), then the vector is \([-1-2,\;9-5]=[-3,\,4]\).

Notation: the image of point \(A\) is written \(A'\) and read “A prime.” Always name image points with primes, and give every corner point of the image, not just one.

Lesson 4 — Day 1

Reflections Across the Axes

A reflection flips a figure across a line, called the line of reflection. Every point of the image lands the same distance from that line as the original point, on the opposite side. Reflections reverse orientation: a figure labeled clockwise comes back labeled counterclockwise.

Across the x-axis

\((x,y)\rightarrow(x,-y)\)

The x-coordinate stays. The y-coordinate changes sign.

Across the y-axis

\((x,y)\rightarrow(-x,y)\)

The y-coordinate stays. The x-coordinate changes sign.

Across the x-axis

\(A(2,1)\rightarrow A'(2,-1)\), \(B(6,2)\rightarrow B'(6,-2)\), \(C(4,6)\rightarrow C'(4,-6)\).

Across the y-axis

\(A(2,1)\rightarrow A'(-2,1)\), \(B(6,2)\rightarrow B'(-6,2)\), \(C(4,6)\rightarrow C'(-4,6)\).

A way to keep them straight: you cross the x-axis by moving vertically, so it is the y-coordinate that flips. You cross the y-axis by moving horizontally, so it is the x-coordinate that flips. The axis you cross is the coordinate that survives.

Lesson 4 — Day 2

Reflections Across \(y=x\) and \(y=-x\)

These two lines are the diagonals of the plane. Reflecting across either one swaps the roles of \(x\) and \(y\).

Across \(y=x\)

\((x,y)\rightarrow(y,x)\)

Switch the coordinates. Signs are untouched.

Across \(y=-x\)

\((x,y)\rightarrow(-y,-x)\)

Switch the coordinates and change both signs.

Across \(y=x\)

\(D(1,4)\rightarrow D'(4,1)\), \(E(5,6)\rightarrow E'(6,5)\), \(F(3,1)\rightarrow F'(1,3)\).

Across \(y=-x\)

\(G(2,3)\rightarrow G'(-3,-2)\), \(H(6,4)\rightarrow H'(-4,-6)\), \(J(4,8)\rightarrow J'(-8,-4)\).

Check your work with the line itself. Any point already sitting on the line of reflection does not move. If your image point moved a point that was on the line, the rule was applied incorrectly.

Tying it together

What Makes a Motion “Rigid”

Translations and reflections are both rigid motions (also called isometries). A rigid motion preserves:

Side lengths. Every segment in the image is the same length as the matching segment in the preimage.
Angle measures. Corresponding angles are equal.
Area and perimeter. Nothing is stretched or shrunk.

What can change is position (both motions) and orientation (reflections only). This is why the distance work from Lesson 1 and the transformation work from Lessons 3 and 4 belong in the same unit: you can use the distance formula to prove a transformation was rigid by showing a side of the image measures the same as the matching side of the preimage.

Vocabulary

These are the words used in the questions. Misreading one of them is the same as not knowing the math.

Leg

Either of the two sides of a right triangle that form the right angle.

Hypotenuse

The side of a right triangle opposite the right angle. Always the longest side.

Pythagorean Theorem

In a right triangle, \(a^2+b^2=c^2\), where \(c\) is the hypotenuse.

Converse of the Pythagorean Theorem

If three side lengths satisfy \(a^2+b^2=c^2\), then the triangle they form is a right triangle.

Pythagorean triple

A set of three whole numbers that satisfies the theorem exactly, such as 3–4–5 or 5–12–13.

Distance formula

\(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\); the Pythagorean Theorem applied to two points on the coordinate plane.

Transformation

A rule that moves or changes a figure on the plane.

Preimage

The original figure, before the transformation is applied.

Image

The resulting figure, after the transformation. Its points are named with primes: \(A'\), \(B'\), \(C'\).

Rigid motion (isometry)

A transformation that preserves all side lengths and angle measures. Translations and reflections are rigid motions.

Translation

A rigid motion that slides every point the same distance in the same direction.

Vector

The instruction for a translation, written \([h,\,k]\): \(h\) units horizontally, \(k\) units vertically.

Reflection

A rigid motion that flips a figure across a line, producing a mirror image.

Line of reflection

The mirror line. Every point of the image is the same distance from it as the matching preimage point, on the opposite side.

Orientation

The order corner points run in, clockwise or counterclockwise. Translations keep it; reflections reverse it.

Sample Problems

Sixteen problems in homework and assessment format, grouped by lesson. Work each one on paper first — the solutions are one click away, which makes them very easy to read too early.

Lesson 1 — The Pythagorean Theorem

Find the hypotenuse.

A right triangle has legs measuring 9 and 12. Find the length of the hypotenuse.

Show solution

Both legs are known, so the squares are added.

\[9^2+12^2=c^2\;\Rightarrow\;81+144=c^2\;\Rightarrow\;225=c^2\]

\[c=\sqrt{225}=15.00\]

This is a 9–12–15 triangle, a multiple of 3–4–5.

Find the missing leg.

A right triangle has a hypotenuse of 26 and one leg of 10. Find the other leg.

Show solution

The hypotenuse is known, so this is a subtraction problem.

\[10^2+b^2=26^2\;\Rightarrow\;100+b^2=676\;\Rightarrow\;b^2=576\]

\[b=\sqrt{576}=24.00\]

10–24–26 is a multiple of the 5–12–13 triple.

An answer that is not a whole number.

A right triangle has legs of 5 and 9. Find the hypotenuse, rounded to two decimal places.

Show solution

\[5^2+9^2=c^2\;\Rightarrow\;25+81=c^2\;\Rightarrow\;106=c^2\]

\[c=\sqrt{106}\approx 10.30\]

Not every triangle is a triple. Leave the radical if an exact answer is requested; otherwise round.

Subtracting, with rounding.

A right triangle has a hypotenuse of 20 and one leg of 14. Find the other leg, rounded to two decimal places.

Show solution

\[14^2+b^2=20^2\;\Rightarrow\;196+b^2=400\;\Rightarrow\;b^2=204\]

\[b=\sqrt{204}\approx 14.28\]

Sanity check: a leg must be shorter than the hypotenuse. \(14.28 < 20\), so the answer is plausible. If your leg came out longer than the hypotenuse, you added when you should have subtracted.

Lesson 2 — The Converse and Triples

Is it a right triangle?

A triangle has sides of 8, 15, and 17. Is it a right triangle? Justify with the converse.

Show solution

The longest side is 17, so it plays the role of \(c\).

\[8^2+15^2=64+225=289 \qquad 17^2=289\]

The two results are equal, so by the converse of the Pythagorean Theorem the triangle is a right triangle. 8–15–17 is a Pythagorean triple.

Right, acute, or obtuse?

Classify the triangle with sides 6, 8, and 11.

Show solution

\[6^2+8^2=36+64=100 \qquad 11^2=121\]

Here \(100 < 121\), so the square of the longest side is larger than the sum of the other two squares. The triangle is obtuse, not right.

Notice how close this is to the 6–8–10 right triangle. Stretching the longest side from 10 to 11 opens the largest angle past \(90^\circ\).

A close call.

Classify the triangle with sides 10, 10, and 14.

Show solution

\[10^2+10^2=100+100=200 \qquad 14^2=196\]

Here \(200 > 196\), so the triangle is acute — but only barely. A longest side of \(\sqrt{200}\approx 14.14\) would have made it exactly right.

This is why the converse is a comparison, not a guess. The picture would look right-angled; the arithmetic is what decides.

Lesson 1 Day 2 — Distance

Distance from coordinates.

Find the distance between \((1,2)\) and \((7,10)\).

Show solution

\[d=\sqrt{(1-7)^2+(2-10)^2}=\sqrt{(-6)^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}\]

\[d=10.00\text{ units}\]

Distance with negative coordinates.

Find the distance between \((-3,5)\) and \((4,-1)\), rounded to two decimal places.

Show solution

Write every coordinate inside parentheses before simplifying.

\[d=\sqrt{(-3-4)^2+(5-(-1))^2}=\sqrt{(-7)^2+(6)^2}=\sqrt{49+36}=\sqrt{85}\]

\[d\approx 9.22\text{ units}\]

Distance from a graph.

Find the distance between the two plotted points.

Show solution

Read the coordinates as \((-5,-3)\) and \((3,3)\), then build the right triangle: the horizontal leg is 8 units and the vertical leg is 6 units.

\[8^2+6^2=d^2\;\Rightarrow\;64+36=d^2\;\Rightarrow\;100=d^2\]

\[d=10.00\text{ units}\]

Another 6–8–10. Counting the legs was faster than the formula here.

Lesson 1 Day 3 — Applications

The baseball diamond.

A baseball diamond is a square with 90 feet between consecutive bases. How far does a catcher at home plate have to throw to reach second base? Round to two decimal places.

Show solution

Home to first and first to second meet at a right angle, and the throw from home to second is the hypotenuse.

\[90^2+90^2=d^2\;\Rightarrow\;8100+8100=d^2\;\Rightarrow\;16200=d^2\]

\[d=\sqrt{16200}\approx 127.28\text{ feet}\]

The ramp.

A wheelchair ramp rises 3 feet over a horizontal run of 14 feet. How long is the ramp surface itself? Round to two decimal places.

Show solution

The rise and the run are the legs; the ramp surface is the hypotenuse.

\[3^2+14^2=c^2\;\Rightarrow\;9+196=c^2\;\Rightarrow\;205=c^2\]

\[c=\sqrt{205}\approx 14.32\text{ feet}\]

The ramp is only about 4 inches longer than its own run, which is what a gentle slope looks like numerically.

The television.

A television is advertised as 55 inches, which refers to its diagonal. The screen is 27 inches tall. How wide is it? Round to two decimal places.

Show solution

The diagonal is the hypotenuse; the height and width are the legs.

\[27^2+w^2=55^2\;\Rightarrow\;729+w^2=3025\;\Rightarrow\;w^2=2296\]

\[w=\sqrt{2296}\approx 47.92\text{ inches}\]

This is the standard complaint about screen sizes: a “55-inch” television is under 48 inches wide, because the advertised number is the diagonal.

Lesson 3 — Translations

Translate by a vector.

Translate triangle \(DEF\) by the vector \([-4,\,3]\). Sketch the image and state the coordinates of its corner points.

Show solution

The rule is \((x,y)\rightarrow(x-4,\;y+3)\): left 4 units, up 3 units.

Preimage	Rule applied	Image
D(1, −2)	(1 − 4, −2 + 3)	D′(−3, 1)
E(5, −2)	(5 − 4, −2 + 3)	E′(1, 1)
F(4, −6)	(4 − 4, −6 + 3)	F′(0, −3)

Recover the vector.

A translation maps \(P(-7,2)\) to \(P'(1,-3)\). Write the translation vector, then use it to find the image of \(Q(4,6)\).

Show solution

The vector is image minus preimage, taken coordinate by coordinate.

\[[\,1-(-7),\;\;-3-2\,]=[8,\,-5]\]

Now apply \((x,y)\rightarrow(x+8,\;y-5)\) to \(Q\):

\[Q(4,6)\rightarrow Q'(12,\,1)\]

Check the vector by substituting the original point back in: \((-7+8,\;2-5)=(1,-3)\), which is \(P'\). If it does not return the image, the subtraction was done backwards.

Lesson 4 — Reflections

Reflect across \(y=x\), then compare with \(y=-x\).

(a) Reflect quadrilateral \(KLMN\) across the line \(y=x\) and state the coordinates of the image.

(b) Without drawing it, state where \(K\) would land if the same figure were reflected across \(y=-x\) instead.

Show solution

(a) Rule: \((x,y)\rightarrow(y,x)\). Switch the coordinates and leave the signs alone.

Preimage	Image across \(y=x\)
K(2, −3)	K′(−3, 2)
L(6, −3)	L′(−3, 6)
M(6, −7)	M′(−7, 6)
N(2, −6)	N′(−6, 2)

(b) Rule: \((x,y)\rightarrow(-y,-x)\). Switch the coordinates and change both signs.

\[K(2,-3)\rightarrow K'(3,\,-2)\]

The two diagonal reflections send \(K\) to \((-3,2)\) and \((3,-2)\) — opposite corners of the plane. Mixing up the two rules is the single most common reflection error on the assessment.

Finished all sixteen? Move to the Unit 1 review sheet, which asks the same skills in assessment order and under time.

Equations & Rules

Everything the unit needs on one page. Print it and keep it in the notebook — the transformation rules are not provided on the assessment.

Right triangles and distance

Pythagorean Theorem

\[a^2+b^2=c^2\]

\(c\) is the hypotenuse, opposite the right angle. Add for the hypotenuse, subtract for a leg.

Converse

\[a^2+b^2=c^2\Rightarrow\text{right}\]

Compare, do not solve. Larger left side means acute; larger right side means obtuse.

Distance formula

\[d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

Order of subtraction does not matter, because each difference is squared.

Pythagorean triples worth memorizing

Triple	Check	Common multiples
3 – 4 – 5	\(9+16=25\)	6–8–10, 9–12–15, 12–16–20, 30–40–50
5 – 12 – 13	\(25+144=169\)	10–24–26, 15–36–39
8 – 15 – 17	\(64+225=289\)	16–30–34
7 – 24 – 25	\(49+576=625\)	14–48–50
9 – 40 – 41	\(81+1600=1681\)	18–80–82

Transformation rules

Transformation	Rule	What happens
Translation by \([h,\,k]\)	\((x,y)\rightarrow(x+h,\;y+k)\)	Slides every point \(h\) horizontally and \(k\) vertically. Orientation is kept.
Reflection across the x-axis	\((x,y)\rightarrow(x,-y)\)	The y-coordinate changes sign.
Reflection across the y-axis	\((x,y)\rightarrow(-x,y)\)	The x-coordinate changes sign.
Reflection across \(y=x\)	\((x,y)\rightarrow(y,x)\)	The coordinates switch places.
Reflection across \(y=-x\)	\((x,y)\rightarrow(-y,-x)\)	The coordinates switch places and both change sign.
Finding a translation vector	\([x'-x,\;y'-y]\)	Image minus preimage, coordinate by coordinate.

Rigid motion: what is preserved

Side lengths
Angle measures
Perimeter and area
Parallel and perpendicular relationships

Rigid motion: what can change

Position on the plane (translations and reflections)
Orientation, clockwise versus counterclockwise (reflections only)

Before you turn in the assessment

Every Pythagorean answer rounded to two decimal places unless it is exact.
Every leg shorter than its hypotenuse.
A labeled sketch on every word problem, with the right angle marked.
Every image point named with a prime, and all corner points listed.
Translation vectors checked by substituting the original point back in.
Reflection rules double-checked: switch for \(y=x\), switch and negate for \(y=-x\).

Video Library

Khan Academy walkthroughs matched to the lessons in this unit. Use them after a first attempt, not instead of one. Videos do not appear on the printed copy.

Lesson 1 — The Pythagorean Theorem

Lesson 1 — the hypotenuse Pythagorean theorem — worked examples Khan Academy Open the video →

Lesson 1 — solving for a leg Pythagorean theorem 2 — solving for a leg Khan Academy Open the video →

Lesson 1 Day 2 — Distance

Lesson 1 Day 2 — count, then solve Example finding distance with the Pythagorean theorem Khan Academy Open the video →

Lesson 1 Day 2 — the formula Distance formula Khan Academy Open the video →

Lesson 1 Day 3 — Applications

Lesson 1 Day 3 — applications Multi-step word problem with the Pythagorean theorem Khan Academy Open the video →

Lesson 3 — Translations

Lesson 3 — draw the image Drawing the image of a translation Khan Academy Open the video →

Lesson 3 — find the vector Determining a translation between points Khan Academy Open the video →

Lesson 4 — Reflections

Lesson 4 Day 1 — over the axes Example reflecting a quadrilateral over the x-axis Khan Academy Open the video →

Lesson 4 Day 2 — over \(y=x\) Reflecting over \(y=x\) Khan Academy Open the video →

Lesson 4 Day 2 — over \(y=-x\) Reflecting over \(y=-x\) Khan Academy Open the video →

Where to Go Next

The other two Unit 1 pages use the same skills in different formats.

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Unit 1 Review Sheet

Twelve assessment-style questions with graphs, worked solutions, an answer key, and a reteach guide that points back to the lesson for anything missed.

Open the review sheet →

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Inquiry Project: Stake It Out

Build a rope right-angle tool from a Pythagorean triple, survey a real outdoor space, and stake out a design generated by vectors and reflections.

Open the project →