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Which transformations preserve size and which only preserve shape.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you sort transformations by what they keep. A translation, a reflection and a rotation move a figure without changing it, so the image is congruent to the original. A dilation changes its size but not its shape, so the image is similar. You will use that to decide how two figures are related and to find a missing side of similar triangles with a scale factor.
You have moved shapes on a grid: sliding them, flipping them over a line and turning them about a point. You have also used scale drawings and maps, where every real length is shrunk by the same number, and you can solve a proportion such as $\frac{x}{4} = \frac{9}{6}$. This lesson puts those pieces together. It asks one question about any two figures: are they the same size and shape, or only the same shape?
| Term | What it means |
|---|---|
| Rigid motion | A translation (slide), reflection (flip) or rotation (turn). It moves a figure without stretching or shrinking it. |
| Congruent | Same size and same shape: one figure can be moved exactly onto the other by rigid motions. The symbol is $\cong$. |
| Dilation | A transformation that multiplies every distance from a center point by the same number, the scale factor. |
| Similar | Same shape, possibly a different size: one figure can be moved onto the other by rigid motions and a dilation. The symbol is $\sim$. |
| Scale factor | The number every length of the first figure is multiplied by to give the matching length of the second. |
| Corresponding (matching) parts | Sides or angles that sit in the same position in two figures, such as side $AB$ and side $DE$ when $A$ matches $D$ and $B$ matches $E$. |
| Image | The figure you get after a transformation. Its points are often named with a prime: $A'$ is the image of $A$. |
Two figures are congruent when a sequence of rigid motions (translations, reflections and rotations) moves one exactly onto the other. Rigid motions never change a length or an angle, so congruent figures have all matching sides equal and all matching angles equal.
Two figures are similar when a sequence of rigid motions and dilations moves one onto the other. A dilation multiplies every length by the same scale factor $k$ but keeps every angle. So similar figures have:
Every pair of congruent figures is also similar, with $k = 1$. The reverse is not true: a triangle with sides $3, 4, 5$ and one with sides $6, 8, 10$ are similar ($k = 2$) but not congruent.
For triangles there is a shortcut. If two angles of one triangle equal two angles of another, the third angles must be equal too, because each triangle's angles add to $180^\circ$. Then the triangles are similar. This is called the angle-angle (AA) rule.
Another way: picture
Picture a photo on a phone. Dragging it across the screen is a translation, and spinning it is a rotation: the photo is the same size, so it stays congruent to itself. Pinching to zoom is a dilation: every face in the photo gets bigger by the same amount, nobody gets stretched sideways, and every angle stays the same. The zoomed photo is similar to the original. If you could stretch only the width, faces would look wide, angles would change, and the result would not be similar at all.
Another way: steps
To decide how two figures are related: (1) match the corners by position or by angles; (2) compare the matching angles, which must all be equal for either word to apply; (3) divide each matching pair of sides, large by small; (4) if every ratio is $1$, the figures are congruent; if every ratio is the same number other than $1$, they are similar; if the ratios differ, they are neither.
A translation slides every point the same distance in the same direction. On a grid, moving $3$ right and $2$ down adds $3$ to each $x$ and takes $2$ from each $y$: the point $(1, 4)$ goes to $(4, 2)$.
A reflection flips a figure over a line. Over the $y$-axis, $(x, y)$ goes to $(-x, y)$. The image is a mirror image: it faces the other way, but every side and every angle is unchanged. Mirror images are congruent. Your left hand and right hand are a good model.
A rotation turns a figure about a point. A quarter turn counterclockwise about the origin sends $(x, y)$ to $(-y, x)$, so $(3, 1)$ goes to $(-1, 3)$.
A dilation with center at the origin and scale factor $k$ sends $(x, y)$ to $(kx, ky)$. With $k = 2$, the point $(3, 1)$ goes to $(6, 2)$. A factor greater than $1$ enlarges, a factor between $0$ and $1$ shrinks, and $k = 1$ changes nothing.
| Move | Lengths | Angles | Result |
|---|---|---|---|
| Translation | kept | kept | congruent |
| Reflection | kept | kept | congruent |
| Rotation | kept | kept | congruent |
| Dilation by $k$ | multiplied by $k$ | kept | similar |
Notice that no move on this list changes an angle. That is why equal angles are the first thing to check for both words.
Take a rectangle $2$ inches by $4$ inches, so it is twice as long as it is wide. Adding $3$ to each side gives $5$ by $7$, which is only $1.4$ times as long as it is wide: a fatter shape. Multiplying each side by $3$ gives $6$ by $12$, still exactly twice as long as it is wide: the same shape.
So similar figures come from multiplying. The test for similarity is to divide matching sides and see whether every quotient is the same. For triangles with sides $4, 6, 8$ and $6, 9, 12$:
$$\frac{6}{4} = 1.5, \qquad \frac{9}{6} = 1.5, \qquad \frac{12}{8} = 1.5.$$
All three ratios agree, so the triangles are similar with scale factor $1.5$. If one ratio had been different, they would not be similar, however close the numbers looked.
The scale factor also has a direction. Going from the small triangle to the large one, $k = 1.5$. Going back from the large one to the small one, $k = \frac{4}{6} = \frac{2}{3}$. Both describe the same pair; just be clear which way you are going, and put the figure you are going to on top.
Most problems in this lesson ask for a missing side of two similar figures. The same five moves work every time.
Why each move is allowed. Step 1 works because rigid motions and dilations keep angles, so equal angles really do sit at matching corners. Steps 2 to 4 work because a dilation multiplies every length by the same $k$. The checks in step 5 work because multiplying by a positive number keeps the order of sizes.
To decide congruent or similar, use the same matching, then look at the factor: $k = 1$ means congruent; any other single factor means similar; no single factor means neither. Remember that a reflection is a rigid motion, so a flipped copy of the same size is congruent.
Around 600 BC, the Greek thinker Thales is said to have found the height of an Egyptian pyramid without climbing it, using shadows. You can do the same with a tree. On a sunny afternoon, stand a yardstick ($3$ feet) straight up. Suppose its shadow is $2$ feet long, and at the same moment the tree's shadow is $28$ feet long.
The sun is so far away that its rays arrive parallel, so they meet the yardstick and the tree at the same angle. Each object, its shadow and a ray make a right triangle, and the two triangles share two angles: similar by AA. The shadows match, so the scale factor is $28 \div 2 = 14$. The heights match too, so the tree is $3 \times 14 = 42$ feet tall. Check: the tree's shadow is $\frac{2}{3}$ of its height, just like the yardstick's ($\frac{28}{42} = \frac{2}{3}$). Do it twice, an hour apart: the shadows change, but the height you get should not.
A standard photo print is $4$ by $6$ inches. An $8$ by $12$ print is similar to it: both sides are doubled ($8 \div 4 = 2$ and $12 \div 6 = 2$), so the whole picture fits with nothing cut off. An $8$ by $10$ print is not similar: $8 \div 4 = 2$ but $10 \div 6 \approx 1.67$. That is why a store has to crop a $4$ by $6$ picture to print it at $8$ by $10$.
Model cars work the same way. A popular size for collectors' models is $1 : 18$, meaning every real length is $18$ times the model's. A real car about $189$ inches long becomes a model $189 \div 18 = 10.5$ inches long. Because the model is similar to the car, the angles of the windshield and the curve of the roof look exactly right, only smaller. Architects' models, globes and the maps on your phone all rely on the same idea: one scale factor for every length, and every angle kept.
Matching sides by where they are on the page. Two similar triangles may be turned or flipped. Match sides by the angles at their ends or by the letters, not by which one is on the bottom.
Adding instead of multiplying. A side going from $4$ to $6$ has been multiplied by $1.5$, not increased by $2$. Test your factor on a second pair: $8 + 2 = 10$, but the matching side is $12$.
Thinking a flip is not congruent. A reflection is a rigid motion. A mirror image has every side and angle the same, so it is congruent.
Doubling the perimeter but also the area. A dilation by $k$ multiplies every length, including the perimeter, by $k$. Area grows faster: by $k^{2}$. A square with side $2$ has area $4$; with side $4$ its area is $16$, four times as much.
Putting the wrong figure on top. The scale factor from $A$ to $B$ is $B$'s length over $A$'s. Upside down, you get the factor going the other way.
Triangle $ABC$ has $A(1, 1)$, $B(4, 1)$ and $C(1, 5)$. Translate it $2$ right and $3$ up by adding $2$ to each $x$ and $3$ to each $y$.
$A'(3, 4), \quad B'(6, 4), \quad C'(3, 8)$
A translation adds the same amounts to every point.
Reflect the image over the $y$-axis by changing the sign of each $x$.
$A''(-3, 4), \quad B''(-6, 4), \quad C''(-3, 8)$
A reflection over the $y$-axis sends $(x, y)$ to $(-x, y)$.
Measure the two straight sides of the original triangle along the grid.
$AB = 4 - 1 = 3, \qquad AC = 5 - 1 = 4$
$AB$ is horizontal and $AC$ is vertical, so each length is a difference of one coordinate.
Measure the matching sides of the final image.
$A''B'' = -3 - (-6) = 3, \qquad A''C'' = 8 - 4 = 4$
The image's sides are also horizontal and vertical, so the same subtraction works.
Compare, and name the relationship.
$A''B'' = AB, \quad A''C'' = AC, \quad \angle A'' = \angle A = 90^\circ \quad\Rightarrow\quad \triangle A''B''C'' \cong \triangle ABC$
Translations and reflections are rigid motions, so every length and angle is kept. The flipped image faces the other way but is still congruent.
Dilate triangle $ABC$ with $A(1, 1)$, $B(4, 1)$, $C(1, 5)$ by a scale factor of $2$ with center the origin: multiply each coordinate by $2$.
$A'(2, 2), \quad B'(8, 2), \quad C'(2, 10)$
A dilation centered at the origin sends $(x, y)$ to $(kx, ky)$.
Measure the horizontal side of the image.
$A'B' = 8 - 2 = 6$
Both points have $y = 2$, so only the $x$-coordinates differ.
Measure the vertical side of the image.
$A'C' = 10 - 2 = 8$
Both points have $x = 2$, so only the $y$-coordinates differ.
Divide each image side by its partner in the original ($AB = 3$, $AC = 4$).
$\dfrac{A'B'}{AB} = \dfrac{6}{3} = 2, \qquad \dfrac{A'C'}{AC} = \dfrac{8}{4} = 2$
The ratios agree and equal the scale factor, as a dilation promises.
Check an angle: both triangles still have a horizontal side meeting a vertical side at $A$.
$\angle A' = \angle A = 90^\circ$
A dilation keeps every angle, so the shape is unchanged.
Name the relationship.
$\triangle A'B'C' \sim \triangle ABC, \quad k = 2, \quad \triangle A'B'C' \not\cong \triangle ABC$
Same angles and one common ratio make the triangles similar. The ratio is $2$, not $1$, so they are not congruent.
In $\triangle ABC$, $\angle A = 40^\circ$ and $\angle B = 75^\circ$. Find $\angle C$.
$\angle C = 180^\circ - 40^\circ - 75^\circ = 65^\circ$
The angles of a triangle add to $180^\circ$.
In $\triangle DEF$, $\angle D = 40^\circ$ and $\angle F = 65^\circ$. Find $\angle E$.
$\angle E = 180^\circ - 40^\circ - 65^\circ = 75^\circ$
The same rule; now every angle of each triangle is known.
Match the corners by equal angles.
$A \leftrightarrow D \;(40^\circ), \quad B \leftrightarrow E \;(75^\circ), \quad C \leftrightarrow F \;(65^\circ)$
Two pairs of equal angles is enough for similarity (AA), and it tells us which sides match.
We know $AB = 10$ and its partner $DE = 15$. Find the scale factor from $ABC$ to $DEF$.
$k = \dfrac{DE}{AB} = \dfrac{15}{10} = 1.5$
$DE$ runs between the $40^\circ$ and $75^\circ$ corners, just like $AB$, so they match.
We know $BC = 6$. Find its partner $EF$.
$EF = 1.5 \times 6 = 9$
Going from the small triangle to the large one, multiply by $k$.
We know $DF = 12$. Find its partner $AC$ by going back to the small triangle.
$AC = 12 \div 1.5 = 8$
Going from the large triangle to the small one undoes the multiplication, so divide by $k$.
Check every ratio.
$\dfrac{15}{10} = \dfrac{9}{6} = \dfrac{12}{8} = 1.5$
All three pairs give the same factor, so the matching and the arithmetic are right.
Match the sides: width with width, height with height.
$4 \leftrightarrow 10, \qquad 6 \leftrightarrow h$
An enlargement that keeps the picture's shape is a dilation, so matching sides share one factor.
Find the scale factor from the widths.
$k = \dfrac{10}{4} = 2.5$
The new width over the old width.
Multiply the old height by the factor.
Check with the other ratio.
A triangle with a perimeter of $10$ cm is dilated by a scale factor of $5$. What happens to its perimeter?
Triangle $PQR$ is similar to triangle $STU$, with $P$, $Q$, $R$ matching $S$, $T$, $U$. $PQ = 4$, $QR = 8$, $PR = 10$, and $ST = 12$. Complete the worked solution to find $SU$.
Match the sides by their letters.
$PQ \leftrightarrow ST, \quad QR \leftrightarrow TU, \quad PR \leftrightarrow SU$
Matching letters sit at matching angles, so the sides between them match.
Divide the known large side by its small partner to get the scale factor.
$k = \dfrac{ST}{PQ} = \dfrac{12}{4} =$ f
In similar figures every length is multiplied by one number.
Multiply the small partner of $SU$ by the scale factor.
$SU = k \times PR = k \times 10 =$ x
$SU$ matches $PR$, so it is $PR$ scaled up.
Check that the answer is longer than $PR$.
$SU > 10$
The factor is more than $1$, so every side of $STU$ is longer than its partner.
In triangle $ABC$, $\angle A = 65^\circ$ and $\angle B = 71^\circ$. In triangle $DEF$, $\angle D = 65^\circ$ and $\angle E = 71^\circ$. The sides of the two triangles are different lengths. Are the triangles similar?
Triangle $ABC$ is similar to triangle $DEF$, with $A$ matching $D$, $B$ matching $E$ and $C$ matching $F$. In the small triangle $AB = 3$, $BC = 6$ and $AC = 8$. In the large one $DE = 6$ and $EF = 12$. How long is $DF$?
$DF =$ answer
A triangle is dilated to make a smaller copy. A side that was $12$ cm long is $4$ cm long in the image. What is the scale factor of the dilation? Give it as a fraction.
Answer:
A model locomotive is built at a scale of $1 : 36$, so the model is similar to the real engine. The model is $8$ inches long. How long is the real locomotive, in feet?
Answer:
A stick $4$ feet tall stands straight up and casts a shadow $2$ feet long. At the same moment a flagpole nearby casts a shadow $8$ feet long. How tall is the flagpole, in feet?
The flagpole is answer feet tall.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Triangle $ABC$ is similar to triangle $DEF$, with $A$ matching $D$, $B$ matching $E$ and $C$ matching $F$. In the small triangle $AB = 3$, $BC = 7$ and $AC = 9$. In the large one $DE = 9$ and $EF = 21$. How long is $DF$?
$DF =$ answer
You can say whether two figures are congruent or similar and which transformations get from one to the other, and you can find a missing side with a scale factor. Without looking: which transformation changes size, and what stays the same when it does?
14. Your turn: a $4$ inch by $6$ inch photo is enlarged to $10$ inches wide. How tall is it?, step 3
$h = 6 \times 2.5 = 15 \text{ inches}$
The height is multiplied by the same factor as the width.
14. Your turn: a $4$ inch by $6$ inch photo is enlarged to $10$ inches wide. How tall is it?, step 4
$\dfrac{15}{6} = 2.5$
Both ratios agree, so the enlargement is similar to the original.