Similar figures

A passport photo blown up into a poster, a toy car built to a fixed scale of the real vehicle, a small classroom map standing in for a whole city — in every case the shape stays exactly the same while the size changes. Geometry gives this everyday idea a precise name: similar figures. Two figures are similar when one is a pure enlargement or reduction of the other, with no stretching, squashing, or twisting allowed. This idea opens the CBSE Class 10 chapter Triangles, and it is the vocabulary every later theorem in that chapter — the Basic Proportionality Theorem, the similarity criteria for triangles, the area relation for similar triangles — quietly depends on. Before any of that can make sense, a student must be able to say, crisply and correctly, exactly what "same shape" means in mathematics, and how it differs from "same shape and same size," which is congruence.

The core insight

Two polygons are similar exactly when their corresponding angles are all equal and their corresponding sides are all in one common ratio, called the scale factor.

The core idea

Two figures are congruent if one can be placed exactly on top of the other by sliding, turning, or flipping it — same shape and same size, with scale factor $1$ . Two figures are similar if they have the same shape but not necessarily the same size — one is an enlargement or reduction of the other. Every congruent pair is automatically similar, but a similar pair need not be congruent. For two polygons with the same number of sides to be similar, both of the following must hold, vertex by matching vertex:

\angle A=\angle P,\ \angle B=\angle Q,\ \angle C=\angle R,\ \ldots \qquad\text{and}\qquad \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CD}{RS}=\cdots=k

where: $ABCDE\sim PQRST$ denotes the similarity with vertices matched in order, and $k$ is the scale factor (also called the ratio of similarity) — the single number that turns every side of one figure into the corresponding side of the other. Neither condition alone is enough. A square and a non-square rectangle have all angles equal ( $90^\circ$ ) but unequal side ratios, so they are not similar. A rhombus and a non-square rectangle can occasionally share a side ratio but never share all four angles, so they too fail. Some shapes have no freedom left once one measurement is fixed, so any two of them are automatically similar: all circles (fixed once the radius is known), all squares (fixed once the side is known), and all equilateral triangles (fixed once every angle is forced to $60^\circ$ ). Rectangles, rhombuses, and general triangles each carry an extra free parameter — a length-to-breadth ratio, an angle, a shape freedom — so two of them are similar only sometimes, when that extra parameter happens to match. This is exactly why triangles, having three independent angles, need their own dedicated similarity criteria, taken up next in the chapter. It helps to think of the scale factor $k$ as a single dial that stretches or shrinks a whole figure at once, never distorting it. Turning that dial to $k>1$ enlarges the figure — every side gets longer by the same multiple, and every angle stays exactly where it was. Turning it to $k<1$ shrinks the figure the same way, and $k=1$ leaves it unchanged, which is precisely the congruent case. Because angles never move as $k$ changes, similar figures always "look the same" at any zoom level — this is the mathematical reason a photograph enlarged to poster size does not look stretched or squashed, only bigger.

Figure

Two quadrilaterals of different size with matching angle marks and sides labelled to show the same ratio k between corresponding sides

Derivation AThe ratio of perimeters of similar polygons equals the scale factor

Given two similar polygons $ABCD\sim PQRS$ with scale factor $k$ , so that every side of $PQRS$ is $k$ times the matching side of $ABCD$ , show that the perimeter of $PQRS$ divided by the perimeter of $ABCD$ is also exactly $k$ , and not some other combination of the sides.

Step 1. Write the definition of similarity as four separate side equations. Since $PQRS\sim ABCD$ with scale factor $k$ , $PQ=k\cdot AB$ , $QR=k\cdot BC$ , $RS=k\cdot CD$ , and $SP=k\cdot DA$ . Each side of the new figure is the old side multiplied by the same fixed number $k$ — that is what "corresponding sides in the same ratio" means.

Step 2. Add the four sides of $PQRS$ to get its perimeter. Perimeter of $PQRS = PQ+QR+RS+SP = k\cdot AB + k\cdot BC + k\cdot CD + k\cdot DA$ .

Step 3. Factor $k$ out of every term, since it is common to all four. Perimeter of $PQRS = k\,(AB+BC+CD+DA) = k \times \text{Perimeter of } ABCD$ .

Step 4. Divide both sides by the perimeter of $ABCD$ to isolate the ratio. $\dfrac{\text{Perimeter of } PQRS}{\text{Perimeter of } ABCD} = k$ . So the perimeters of two similar polygons are in exactly the same ratio as their corresponding sides — the scale factor governs the whole figure at once, not just individual sides.

Worked example

Quadrilateral $ABCD$ has $AB=4\text{ cm}$ , $BC=6\text{ cm}$ , $CD=5\text{ cm}$ , $DA=3\text{ cm}$ . Quadrilateral $PQRS\sim ABCD$ with scale factor $k=2.5$ . Find the sides of $PQRS$ , its perimeter, and confirm the perimeter ratio.

Step 1. Apply the scale factor to each side using $PQ=k\cdot AB$ and so on: $PQ=2.5\times4=10\text{ cm}$ , $QR=2.5\times6=15\text{ cm}$ , $RS=2.5\times5=12.5\text{ cm}$ , $SP=2.5\times3=7.5\text{ cm}$ .

Step 2. Add the sides of $ABCD$ : Perimeter of $ABCD=4+6+5+3=18\text{ cm}$ .

Step 3. Add the sides of $PQRS$ : Perimeter of $PQRS=10+15+12.5+7.5=45\text{ cm}$ .

Step 4. Check the ratio against the derivation above: $\dfrac{45}{18}=2.5=k$ . The perimeters land in exactly the scale factor's ratio, confirming Perimeter of $PQRS = 45\text{ cm}$ .

Competitive stretch

If instead only the perimeter of $PQRS$ were given as $45\text{ cm}$ and the perimeter of $ABCD$ as $18\text{ cm}$ , the derivation lets you find $k$ without knowing a single individual side: $k=45/18=2.5$ , and then every unknown side of $PQRS$ follows at once as $k$ times the matching side of $ABCD$ — a shortcut worth remembering when a question gives perimeters but hides the sides.

Common traps

Checking only one of the two conditions — equal angles and proportional sides are both required for general polygons; a match on angles alone (as in a rectangle) or on sides alone does not prove similarity.
Assuming any two rectangles, or any two rhombuses, are automatically similar; only shapes with no free shape-parameter (circles, squares, equilateral triangles) are similar by default.
Mismatching vertex correspondence — in $ABCD\sim PQRS$ , $A\leftrightarrow P$ and $B\leftrightarrow Q$ ; comparing $AB$ with $QR$ instead of $PQ$ produces a wrong ratio.
Treating "similar" and "congruent" as the same word — congruent figures are the special case of similar figures with $k=1$ , but similar figures are congruent only then.

Ask the tutor: Still unsure why a rhombus can dodge being similar to another rhombus even with all sides equal? Ask the AI tutor to sketch two rhombuses with different angles side by side and test the similarity conditions live.

Why do a square and a circle count as always similar but a rectangle does not?

A square is completely fixed once you know its side length — every square has four equal sides and four right angles, so there is no other way its shape could vary. A circle is the same, fixed once you know its radius. A rectangle, though, has a length and a breadth that can vary independently of each other, so two rectangles can have the same angles ( $90^\circ$ everywhere) yet completely different length-to-breadth ratios, like $2:1$ versus $5:1$ . That extra freedom is what makes rectangles only sometimes similar, never always.

Is every congruent pair also a similar pair?

Yes. Congruent figures are identical in both shape and size, which means their corresponding sides are in the ratio $1:1$ and their corresponding angles are equal — exactly the two conditions similarity requires, just with scale factor $k=1$ . So congruence is really a special, size-matching case sitting inside the bigger idea of similarity. The reverse is not true: two similar figures with $k\neq1$ have the same shape but different sizes, so they are not congruent.

If two polygons have proportional sides, do the angles have to be equal too?

Not automatically, and this is the classic trap. Proportional sides alone can still allow the polygon to "flex" into a different shape while keeping the same side ratios — think of a square's four equal sides being reshaped into a rhombus with the same side length but different angles. For triangles specifically this cannot happen (fixing all three side ratios pins down the angles too, which is why the SSS similarity criterion for triangles works), but for quadrilaterals and other polygons with four or more sides, proportional sides are not enough on their own — you must check the angles separately.

Similar figures

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads