Similar Figures in Maths: Meaning, Scale Factor and Examples

How was the height of Mount Ever­est worked out with­out any­one climb­ing it with a mea­sur­ing tape? How do we know the dis­tance to the Moon with­out a ruler long enough to reach it? Nei­ther was mea­sured directly. Both were cal­cu­lated using one sim­ple but pow­er­ful idea: sim­i­lar­ity, the idea of shapes that look exactly alike but come in dif­fer­ent sizes.

This les­son opens Chap­ter 6 (Tri­an­gles) of the NCERT Class 10 text­book. It explains what "sim­i­lar" means, how it dif­fers from "con­gru­ent", and the two-part test that decides whether two poly­gons are sim­i­lar. Maps, scale mod­els, photo enlarge­ments and the heights-and-dis­tances prob­lems you will meet later all rest on this idea.

Con­gru­ent and sim­i­lar fig­ures#

Con­gru­ent fig­ures#

In ear­lier classes you met con­gru­ent fig­ures: two fig­ures with exactly the same shape and exactly the same size. If you cut one out and place it on the other, they cover each other exactly, edge to edge.

Sim­i­lar fig­ures#

Now think about cir­cles. Take a small coin and a large din­ner plate. They are not con­gru­ent, because one is big­ger than the other. Yet they are clearly the same kind of shape: every cir­cle looks like every other cir­cle, only scaled up or down. The same is true of squares, and of equi­lat­eral tri­an­gles.

Fig­ures that have the same shape, but not nec­es­sar­ily the same size, are called sim­i­lar fig­ures. So:

• All cir­cles are sim­i­lar to each other.
• All squares are sim­i­lar to each other.
• All equi­lat­eral tri­an­gles are sim­i­lar to each other.

Every pair of con­gru­ent fig­ures is also sim­i­lar (same shape, and the same size as well). But sim­i­lar fig­ures need not be con­gru­ent. Sim­i­lar­ity is the broader idea; con­gru­ence is the spe­cial case in which the scale fac­tor is $1$.

Can a cir­cle be sim­i­lar to a square? Can a tri­an­gle be sim­i­lar to a square? No: their basic shapes dif­fer how­ever much you shrink or enlarge them. So not every pair of fig­ures is sim­i­lar, and for shapes less obvi­ous than cir­cles and squares (two quadri­lat­er­als that merely look alike, say) we need a pre­cise test.

What a pho­to­graph teaches us#

Same sub­ject, dif­fer­ent sizes#

Imag­ine three prints of the same pho­to­graph of the Taj Mahal: a small print, a medium one and a large poster. All three show the same mon­u­ment, only enlarged or reduced. They are sim­i­lar.

Now take two pho­tographs of exactly the same size, one of a per­son at age 10 and one of the same per­son at age 40. The sizes match, but the shapes do not: a ten-year-old does not look like a smaller forty-year-old. These two pho­tographs are not sim­i­lar.

So sim­i­lar­ity is about shape, not size. Two fig­ures can be the same size and not sim­i­lar (the two por­traits), and two fig­ures can be very dif­fer­ent in size and still be sim­i­lar (the three Taj Mahal prints).

Enlarg­ing a pho­to­graph#

Sup­pose a pho­tog­ra­pher starts with a neg­a­tive $35$ mm wide and enlarges it to $45$ mm, or to $55$ mm. Every line seg­ment in the small pic­ture grows in exactly the same pro­por­tion. Enlarged to $45$ mm, every seg­ment becomes $\displaystyle \frac{45}{35} = \frac{9}{7}$ times as long; enlarged to $55$ mm, every seg­ment becomes $\displaystyle \frac{55}{35} = \frac{11}{7}$ times as long. We say every seg­ment of the smaller pho­to­graph is enlarged in the ratio $35 : 45$ (or $35 : 55$), and every seg­ment of the larger one is reduced in the ratio $45 : 35$ (or $55 : 35$) to get back to the smaller one.

What does not change dur­ing an enlarge­ment? The angles. A line that is tilted in the small pho­to­graph is tilted by exactly the same amount in the big one. Lengths change in one fixed pro­por­tion; angles stay exactly the same. That sin­gle obser­va­tion con­tains the whole idea of sim­i­lar­ity.

The def­i­n­i­tion of sim­i­lar poly­gons#

The two con­di­tions#

Two poly­gons with the same num­ber of sides are sim­i­lar if (i) their cor­re­spond­ing angles are equal, and (ii) their cor­re­spond­ing sides are in the same ratio (that is, pro­por­tional).

For quadri­lat­er­als $ABCD$ and $PQRS$ with $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$, $D \leftrightarrow S$, this means

$$\angle A = \angle P,\quad \angle B = \angle Q,\quad \angle C = \angle R,\quad \angle D = \angle S$$

$$\displaystyle \frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD} = \frac{SP}{DA} = k.$$

We then write $ABCD \sim PQRS$. The sym­bol $\sim$ is read "is sim­i­lar to", and the order of the let­ters shows which ver­tices cor­re­spond.

The scale fac­tor#

The com­mon ratio $k$ of cor­re­spond­ing sides is called the scale fac­tor (map-mak­ers call it the rep­re­sen­ta­tive frac­tion). In the pho­to­graph exam­ple it is $\displaystyle \frac{45}{35}$ or $\displaystyle \frac{55}{35}$. If $k \gt 1$ the sec­ond fig­ure is an enlarge­ment, if $k \lt 1$ it is a reduc­tion, and if $k = 1$ the fig­ures are con­gru­ent.

Because every side is mul­ti­plied by $k$, the perime­ter is mul­ti­plied by $k$ too. So the ratio of the perime­ters of two sim­i­lar poly­gons equals the scale fac­tor.

Why both con­di­tions are needed#

Is it enough to check only the angles, or only the sides? For gen­eral poly­gons, no. Two famil­iar quadri­lat­er­als show why.

Angles equal, sides not pro­por­tional#

Take a square of side $3$ cm and a rec­tan­gle $5$ cm by $3$ cm. Every angle of both shapes is $90^\circ$, so all cor­re­spond­ing angles are equal. But the sides are not in one ratio: match­ing the $3$ cm sides gives $\displaystyle \frac{3}{3} = 1$, while match­ing a $3$ cm side of the square with the $5$ cm side of the rec­tan­gle gives $\displaystyle \frac{3}{5} = 0.6$. So a square is not sim­i­lar to a rec­tan­gle that is not a square, even though every angle matches.

Sides pro­por­tional, angles not equal#

Now take the same square and a rhom­bus of side $3$ cm whose angles are $70^\circ$ and $110^\circ$. All sides of both fig­ures are $3$ cm, so every ratio of cor­re­spond­ing sides is $1$. But the angles are $90^\circ$ in one fig­ure and $70^\circ$ or $110^\circ$ in the other. So a square is not sim­i­lar to a rhom­bus that is not a square, even though every side ratio matches.

This is why the def­i­n­i­tion insists on both con­di­tions together. For tri­an­gles, as the next lessons show, one con­di­tion auto­mat­i­cally brings the other with it; that spe­cial prop­erty is what makes tri­an­gles so use­ful.

Method: test­ing two poly­gons for sim­i­lar­ity#

1. Check that the poly­gons have the same num­ber of sides.
2. Decide the cor­re­spon­dence of ver­tices (which ver­tex matches which). Usu­ally the equal angles, or the order of the let­ters, tell you.
3. Check that every pair of cor­re­spond­ing angles is equal.
4. Work out the ratio of every pair of cor­re­spond­ing sides, writ­ing each ratio in the same order (sec­ond fig­ure over first, say).
5. If all the angles match and all the ratios are equal, the poly­gons are sim­i­lar and that com­mon ratio is the scale fac­tor. If either check fails, they are not sim­i­lar.

Worked exam­ples#

Exam­ple 1: two rec­tan­gles#

Are a $4$ cm by $6$ cm rec­tan­gle and a $6$ cm by $9$ cm rec­tan­gle sim­i­lar? What about a $4$ cm by $6$ cm rec­tan­gle and a $5$ cm by $8$ cm rec­tan­gle?

Solu­tion. All angles of a rec­tan­gle are $90^\circ$, so the angle con­di­tion holds in both cases. For the first pair, match short side with short side and long with long:

$$\displaystyle \frac{6}{4} = 1.5, \qquad \frac{9}{6} = 1.5.$$

The ratios are equal, so the rec­tan­gles are sim­i­lar with scale fac­tor $1.5$. For the sec­ond pair, $\displaystyle \frac{5}{4} = 1.25$ but $\displaystyle \frac{8}{6} \approx 1.33$. The ratios dif­fer, so these rec­tan­gles are not sim­i­lar.

Answer: the first pair is sim­i­lar ($k = 1.5$); the sec­ond pair is not.

Exam­ple 2: read­ing a scale fac­tor#

A pho­to­graph neg­a­tive $35$ mm wide is enlarged to a print $55$ mm wide. A tree in the neg­a­tive is $14$ mm tall. How tall is it in the print?

Solu­tion. Every length is mul­ti­plied by the same fac­tor $\displaystyle k = \frac{55}{35} = \frac{11}{7}$. So the tree's height in the print is

$$\displaystyle 14 \times \frac{11}{7} = 22 \text{ mm}.$$

Answer: $22$ mm.

Exam­ple 3: find­ing miss­ing sides and angles#

In the fig­ure above, $ABCD \sim PQRS$ with $AB = 6$ cm, $BC = 5$ cm, $CD = 3$ cm, $DA = 4$ cm, $\angle A = \angle D = 90^\circ$ and $\angle B \approx 53.1^\circ$. If $PQ = 9$ cm, find the other sides and all the angles of $PQRS$.

Solu­tion. The scale fac­tor is

$$\displaystyle k = \frac{PQ}{AB} = \frac{9}{6} = 1.5.$$

So $QR = 1.5 \times 5 = 7.5$ cm, $RS = 1.5 \times 3 = 4.5$ cm and $SP = 1.5 \times 4 = 6$ cm. Cor­re­spond­ing angles are equal, so $\angle P = \angle S = 90^\circ$ and $\angle Q = \angle B \approx 53.1^\circ$. The angles of a quadri­lat­eral add up to $360^\circ$, so

$$\angle R = 360^\circ - 90^\circ - 90^\circ - 53.1^\circ \approx 126.9^\circ.$$

Answer: $QR = 7.5$ cm, $RS = 4.5$ cm, $SP = 6$ cm; $\angle P = \angle S = 90^\circ$, $\angle Q \approx 53.1^\circ$, $\angle R \approx 126.9^\circ$.

Exam­ple 4: perime­ters of sim­i­lar poly­gons#

Find the perime­ters of $ABCD$ and $PQRS$ in Exam­ple 3 and com­pare their ratio with the scale fac­tor.

Solu­tion.

$$\displaystyle \begin{aligned} \text{Perimeter of } ABCD &= 6 + 5 + 3 + 4 = 18 \text{ cm} \\ \text{Perimeter of } PQRS &= 9 + 7.5 + 4.5 + 6 = 27 \text{ cm} \\ \frac{27}{18} &= 1.5 = k \end{aligned}$$

Answer: $18$ cm and $27$ cm; their ratio $1.5$ equals the scale fac­tor, as expected.

Exam­ple 5: a map#

A map is drawn to the scale $1 : 50000$. Two vil­lages are $3.2$ cm apart on the map. How far apart are they on the ground?

Solu­tion. The map is a reduc­tion of the ground with every length divided by $50000$, so the ground dis­tance is

$$3.2 \times 50000 = 160000 \text{ cm} = 1600 \text{ m} = 1.6 \text{ km}.$$

Answer: $1.6$ km.

Com­mon mis­takes#

• Check­ing only the angles. A square and a non-square rec­tan­gle have equal angles but are not sim­i­lar.
• Check­ing only the sides. A square and a non-square rhom­bus have pro­por­tional sides but are not sim­i­lar.
• Mix­ing the order in the ratios, for exam­ple writ­ing $\displaystyle \frac{PQ}{AB}$ for one pair and $\displaystyle \frac{BC}{QR}$ for the next. Keep the same fig­ure on top every time.
• Match­ing the wrong sides: the short side of one rec­tan­gle must be com­pared with the short side of the other.
• Think­ing "sim­i­lar" means "con­gru­ent". Sim­i­lar fig­ures may dif­fer in size; con­gru­ent fig­ures may not.
• Writ­ing the ver­tices in the wrong order after $\sim$. The order records the cor­re­spon­dence.

Try these#

1. Are a $3$ cm by $5$ cm rec­tan­gle and a $6$ cm by $10$ cm rec­tan­gle sim­i­lar? Answer: Yes, scale fac­tor $2$.
2. Are a $2$ cm by $3$ cm rec­tan­gle and a $3$ cm by $4$ cm rec­tan­gle sim­i­lar? Answer: No, since $\displaystyle \frac{3}{2} \ne \frac{4}{3}$.
3. A quadri­lat­eral has sides $2$, $3$, $4$ and $5$ cm. It is enlarged with scale fac­tor $2.5$. Find the new sides. Answer: $5$, $7.5$, $10$ and $12.5$ cm.
4. A $10$ cm by $15$ cm pho­to­graph is enlarged so that its shorter side becomes $25$ cm. Find the longer side. Answer: $37.5$ cm.
5. Is a rhom­bus of side $4$ cm with angles $60^\circ$ and $120^\circ$ sim­i­lar to a square of side $6$ cm? Answer: No; the sides are pro­por­tional but the angles are not equal.
6. Two squares have sides $5$ cm and $8$ cm. Find the scale fac­tor from the smaller to the larger. Answer: $1.6$.

Key terms#

Con­gru­ent fig­ures

Fig­ures with the same shape and the same size.

Sim­i­lar fig­ures

Fig­ures with the same shape but not nec­es­sar­ily the same size.

Sim­i­lar poly­gons

Poly­gons with the same num­ber of sides whose cor­re­spond­ing angles are equal and whose cor­re­spond­ing sides are pro­por­tional.

Cor­re­spond­ing angles and sides

The angles and sides that match under the cho­sen pair­ing of ver­tices.

Pro­por­tional

In the same ratio: every pair of cor­re­spond­ing sides gives the same quo­tient.

Scale fac­tor

The com­mon ratio of cor­re­spond­ing sides; also called the rep­re­sen­ta­tive frac­tion.

Enlarge­ment and reduc­tion

A sim­i­lar copy with scale fac­tor greater than $1$, or less than $1$, respec­tively.

Com­mon ques­tions#

Are con­gru­ent fig­ures sim­i­lar?#

Yes. Con­gru­ent fig­ures have equal cor­re­spond­ing angles and a scale fac­tor of $1$, so they sat­isfy both con­di­tions of sim­i­lar­ity.

Are all rec­tan­gles sim­i­lar?#

No. All their angles are equal, but their length-to-breadth ratios can dif­fer, so their sides need not be pro­por­tional.

Are all rhom­buses sim­i­lar?#

No. All their sides are pro­por­tional, but their angles can dif­fer, so the angle con­di­tion can fail.

Does the order of let­ters mat­ter when writ­ing $ABCD \sim PQRS$?#

Yes. It states that $A$ cor­re­sponds to $P$, $B$ to $Q$, and so on. Writ­ing the let­ters in another order makes a dif­fer­ent, pos­si­bly false, claim.

Why do tri­an­gles get spe­cial treat­ment later in the chap­ter?#

For tri­an­gles, equal angles force pro­por­tional sides and pro­por­tional sides force equal angles, so only one of the two con­di­tions needs to be checked. The Basic Pro­por­tion­al­ity The­o­rem and the AA, SSS and SAS cri­te­ria build on this.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.