NCERT Class 10 Exercise 6.1 Solutions: Similar Figures

Solu­tions to Exer­cise 6.1#

Ques­tion 1#

Fill in the blanks using the cor­rect word given in brack­ets.

Ques­tion 1 (i)#

All cir­cles are ______. (con­gru­ent, sim­i­lar)

Solu­tion. A cir­cle is fixed com­pletely by its radius, and every cir­cle has exactly the same round shape. Enlarg­ing a cir­cle of radius $2$ cm by the fac­tor $5$ gives a cir­cle of radius $10$ cm, so any cir­cle is a scaled copy of any other. But cir­cles need not have the same size: radii of $2$ cm and $10$ cm are dif­fer­ent, so "con­gru­ent" can­not be true of all cir­cles.

Same shape, size may dif­fer: that is exactly what "sim­i­lar" means.

Answer: All cir­cles are sim­i­lar.

Ques­tion 1 (ii)#

All squares are ______. (sim­i­lar, con­gru­ent)

Solu­tion. Take squares of side $2$ cm and $5$ cm and test both con­di­tions.

• Angles: every angle of each square is $90^\circ$, so all cor­re­spond­ing angles are equal.
• Sides: all sides of a square are equal, so every pair of cor­re­spond­ing sides gives the same ratio $\displaystyle \frac{2}{5}$.

Both con­di­tions hold, and the same argu­ment works for any two squares. They are con­gru­ent only when their sides hap­pen to be equal, so "con­gru­ent" is not true of all squares.

Answer: All squares are sim­i­lar.

Ques­tion 1 (iii)#

All ______ tri­an­gles are sim­i­lar. (isosce­les, equi­lat­eral)

Solu­tion. Every angle of an equi­lat­eral tri­an­gle is $60^\circ$ (the three equal angles add up to $180^\circ$, and $3 \times 60^\circ = 180^\circ$). So two equi­lat­eral tri­an­gles always have equal cor­re­spond­ing angles. Their sides are also pro­por­tional: for sides $3$ cm and $9$ cm, every ratio is $\displaystyle \frac{3}{9} = \frac{1}{3}$. Both con­di­tions hold.

Isosce­les tri­an­gles fail. An isosce­les tri­an­gle with ver­tex angle $20^\circ$ has angles $20^\circ, 80^\circ, 80^\circ$, while one with ver­tex angle $120^\circ$ has angles $120^\circ, 30^\circ, 30^\circ$. These angle sets do not match, so the two tri­an­gles are not sim­i­lar.

Answer: All equi­lat­eral tri­an­gles are sim­i­lar.

Ques­tion 1 (iv)#

Two poly­gons of the same num­ber of sides are sim­i­lar, if (a) their cor­re­spond­ing angles are ______ and (b) their cor­re­spond­ing sides are ______. (equal, pro­por­tional)

Solu­tion. This is the def­i­n­i­tion of sim­i­lar poly­gons. The angles fix the shape, so they must match exactly. The sides fix the size, and for sim­i­lar­ity they may all be scaled, but by one com­mon fac­tor.

The words can­not be swapped. If the sides had to be equal as well, every pair of sim­i­lar fig­ures would be con­gru­ent, which con­tra­dicts part (ii). And "pro­por­tional angles" does not work either: dou­bling every angle of a quadri­lat­eral would give an angle sum of $720^\circ$ instead of $360^\circ$, which is impos­si­ble.

Answer: (a) equal, (b) pro­por­tional.

Ques­tion 2#

Give two dif­fer­ent exam­ples of pair of (i) sim­i­lar fig­ures, (ii) non-sim­i­lar fig­ures.

Ques­tion 2 (i)#

Solu­tion. We choose fig­ures and check both con­di­tions with actual num­bers.

Exam­ple A: two equi­lat­eral tri­an­gles with sides $4$ cm and $12$ cm.

• Angles: $60^\circ, 60^\circ, 60^\circ$ in both, so cor­re­spond­ing angles are equal.
• Sides: every pair gives $\displaystyle \frac{4}{12} = \frac{1}{3}$, so the sides are pro­por­tional.

Exam­ple B: two squares with sides $3$ cm and $7$ cm.

• Angles: $90^\circ$ at every cor­ner of both, so they are equal.
• Sides: every pair gives $\displaystyle \frac{3}{7}$, so the sides are pro­por­tional.

Answer: (a) two equi­lat­eral tri­an­gles of sides $4$ cm and $12$ cm; (b) two squares of sides $3$ cm and $7$ cm. (Any two cir­cles would also do.)

Ques­tion 2 (ii)#

Solu­tion.

Exam­ple A: a square and a rec­tan­gle that is not a square, say a square of side $4$ cm and a rec­tan­gle $6$ cm by $4$ cm.

• Angles: all $90^\circ$, so this con­di­tion holds.
• Sides: match­ing the $4$ cm sides gives $\displaystyle \frac{4}{4} = 1$, but match­ing a $4$ cm side of the square with the $6$ cm side of the rec­tan­gle gives $\displaystyle \frac{4}{6} = \frac{2}{3}$. Since $\displaystyle 1 \ne \frac{2}{3}$, the sides are not pro­por­tional.

The side con­di­tion fails, so the fig­ures are not sim­i­lar, even though the angles match.

Exam­ple B: an equi­lat­eral tri­an­gle and a right-angled isosce­les tri­an­gle.

• Angles: $60^\circ, 60^\circ, 60^\circ$ against $90^\circ, 45^\circ, 45^\circ$. No match­ing of ver­tices makes these equal.

The angle con­di­tion already fails, so there is no need to check the sides.

Answer: (a) a square of side $4$ cm and a $6$ cm by $4$ cm rec­tan­gle; (b) an equi­lat­eral tri­an­gle and a right-angled isosce­les tri­an­gle.

Why this works: a sin­gle failed con­di­tion is enough to rule out sim­i­lar­ity.

Ques­tion 3#

State whether the fol­low­ing quadri­lat­er­als are sim­i­lar or not (Fig. 6.8 of the text­book: a square $PQRS$ of side $1.5$ cm and a rhom­bus $ABCD$ of side $3$ cm whose angles are not right angles).

Solu­tion.

• Sides: every side of the square is $1.5$ cm and every side of the rhom­bus is $3$ cm, so every pair of cor­re­spond­ing sides gives $\displaystyle \frac{1.5}{3} = \frac{1}{2}$. The sides are pro­por­tional.
• Angles: every angle of the square is $90^\circ$, but the angles of the rhom­bus are not $90^\circ$. So the cor­re­spond­ing angles are not equal.

Con­di­tion (i) fails, so the quadri­lat­er­als are not sim­i­lar. This is the square and rhom­bus case from the les­son: pro­por­tional sides alone are not enough.

Answer: The quadri­lat­er­als are not sim­i­lar, because their cor­re­spond­ing angles are not equal.

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.

Cor­re­spond­ing angles

Angles at ver­tices that are matched with each other when two fig­ures are com­pared.

Pro­por­tional sides

Cor­re­spond­ing sides that all give the same ratio.

Scale fac­tor

That com­mon ratio; it equals $1$ exactly when sim­i­lar fig­ures are con­gru­ent.

Equi­lat­eral tri­an­gle

A tri­an­gle with all three sides equal, and there­fore all three angles equal to $60^\circ$.

Rhom­bus

A quadri­lat­eral with all four sides equal; its angles need not be right angles.

Com­mon ques­tions#

Why are all cir­cles sim­i­lar but not all ellipses?#

A cir­cle's shape is fixed and only its radius changes, so any cir­cle is an enlarge­ment of any other. An ellipse can be more or less stretched, so two ellipses can have dif­fer­ent shapes.

Are all right-angled tri­an­gles sim­i­lar?#

No. A tri­an­gle with angles $90^\circ, 45^\circ, 45^\circ$ and one with angles $90^\circ, 60^\circ, 30^\circ$ are both right-angled, but their other angles dif­fer.

Are con­gru­ent fig­ures sim­i­lar?#

Yes. Their angles are equal and every side ratio is $1$, so both con­di­tions hold.

Is it enough to show one con­di­tion fails?#

Yes, to prove fig­ures are not sim­i­lar. To prove they are sim­i­lar, both con­di­tions must be shown (for tri­an­gles, the cri­te­ria in the next lessons shorten this).

Can two fig­ures with the same area fail to be sim­i­lar?#

Yes. A $4$ cm by $4$ cm square and a $2$ cm by $8$ cm rec­tan­gle both have area $16 \text{ cm}^2$, but they are not sim­i­lar.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.