Class 10 Maths · Chapter 6

Triangles

101 Class 10 Maths questions on Triangles from the NCERT textbook, NCERT Exemplar and CBSE board papers 2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 1 of 6

1. Q1 NCERT · Exercise 6.1
   Fill in the blanks using the correct word given in brackets:

   1. (i)​All circles ​are ​____. (congruent, similar)
   2. (ii)All ​squares are ____. (similar, congruent) ​
   3. (iii)All ____ triangles are similar. ​(isosceles, equilateral)
   4. (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)
   Show answer

   (i) Similar; (ii) Similar; (iii) Equilateral; (iv)(a) Equal; (iv)(b) Proportional.

   Step-by-step solution →
2. Q2 NCERT · Exercise 6.1
   Give two ​different examples ​of ​pair ​of: ​

   1. (i)similar ​figures
   2. (ii)non-similar figures
   Show answer

   (i) Similar: any two circles; any two squares. (ii) Non-similar: a square and a circle; a square and a rectangle of unequal sides.

   Step-by-step solution →
3. Q3 NCERT · Exercise 6.1
   State ​whether the following ​quadrilaterals ​are ​similar ​or not:

   Show answer

   No — the quadrilaterals are not similar. Their corresponding sides are in the same ratio $\left(\frac{1.5}{3}=\frac{1}{2}\right)$, but their corresponding angles are not equal: every angle of $ABCD$ is $90^{\circ }$ while the angles of the rhombus $PQRS$ are not.

   Step-by-step solution →
4. Q4 NCERT · Exercise 6.2
   In Fig. 6.17, ​(i) and (ii), DE || BC. Find EC in ​(i) and AD ​in (ii).

   1. (i)​Find $EC$, ​given $AD=1.5$ cm, $DB=3$ ​cm, $AE=1$ cm (Fig. 6.17 (i)).
   2. (ii)​Find $AD$, given $DB=7.2$ cm, $AE=1.8$ cm, $EC=5.4$ cm (Fig. 6.17 (ii)).

   Show answer

   (i) $EC=2$ cm. (ii) $AD=2.4$ cm.

   Step-by-step solution →
5. Q5 NCERT · Exercise 6.2
   E and F are points on the sides ​PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR:

   1. (i)PE ​= 3.9 cm, EQ = 3 cm, PF = 3.6 cm and ​FR = 2.4 cm
   2. (ii)PE = 4 cm, QE = 4.5 cm, ​PF ​= ​8 cm and RF = 9 cm
   3. (iii)PQ = ​1.28 cm, PR = 2.56 cm, PE = 0.18 cm ​and PF = 0.36 cm
   Show answer

   (i) $EF\nparallel QR$ (since $1.3\ne 1.5$); (ii) And (iii) $EF\parallel QR$.

   Step-by-step solution →
6. Q6 NCERT · Exercise 6.2
   In ​Fig. 6.18, ​if $LM\parallel CB$ and $LN\parallel CD$, prove ​that $\frac{AM}{AB}=\frac{AN}{AD}$.

   Show answer

   Proved: $\frac{AM}{AB}=\frac{AN}{AD}$, using the Basic Proportionality Theorem in $△ABC$ and $△ACD$.

   Step-by-step solution →
7. Q7 NCERT · Exercise 6.2
   In ​Fig. 6.19, $DE\parallel AC$ and $DF\parallel AE$. ​Prove ​that $\frac{BF}{FE}=\frac{BE}{EC}$.

   Show answer

   Proved. By the Basic Proportionality Theorem in $△ABE$ (where $DF\parallel AE$), $\frac{BF}{FE}=\frac{BD}{DA}$; by the same theorem in $△ABC$ (where $DE\parallel AC$), $\frac{BE}{EC}=\frac{BD}{DA}$. Hence $\frac{BF}{FE}=\frac{BE}{EC}$.

   Step-by-step solution →
8. Q8 NCERT · Exercise 6.2
   In ​Fig. ​6.20, $DE\parallel OQ$ ​and $DF\parallel OR$. Show ​that $EF\parallel QR$.

   Show answer

   Proved: $EF\parallel QR$, using BPT in $△POQ$ and $△POR$ and the converse of BPT in $△PQR$.

   Step-by-step solution →
9. Q9 NCERT · Exercise 6.2
   In Fig. 6.21, $A,B$ ​and $C$ ​are ​points ​on $OP$, $OQ$ and $OR$ respectively ​such that $AB\parallel PQ$ ​and $AC\parallel PR$. ​Show that $BC\parallel QR$.

   Show answer

   Since $\frac{OA}{AP}=\frac{OB}{BQ}$ and $\frac{OA}{AP}=\frac{OC}{CR}$, we get $\frac{OB}{BQ}=\frac{OC}{CR}$, so by the converse of the Basic Proportionality Theorem, $BC\parallel QR$.

   Step-by-step solution →
10. Q10 NCERT · Exercise 6.2
    Using Theorem 6.1, ​prove ​that a line drawn through the mid-point of one side ​of a triangle parallel to another ​side bisects the third side. (Recall that you have ​proved it in Class IX).
    Show answer

    Since $MN\parallel BC$, the Basic Proportionality Theorem (Theorem 6.1) gives $\frac{AM}{MB}=\frac{AN}{NC}$. As $M$ is the mid-point of $AB$, $AM=MB$, so the ratio is $1$ and therefore $AN=NC$. Hence $N$ is the mid-point of $AC$, i.e. the line bisects the third side.

    Step-by-step solution →
11. Q11 NCERT · Exercise 6.2
    Using ​Theorem 6.2, prove that the line joining ​the mid-points of any two sides of a ​triangle is parallel ​to the third side. (Recall that you have done it in Class ​IX).
    Show answer

    Since $D$ and $E$ are mid-points, $\frac{AD}{DB}=\frac{AE}{EC}=1$; by Theorem 6.2 (the converse of the Basic Proportionality Theorem), the line $DE$ joining the mid-points is parallel to the third side, so $DE\parallel BC$.

    Step-by-step solution →
12. Q12 NCERT · Exercise 6.2
    ABCD is a trapezium in ​which $AB\parallel DC$ and ​its diagonals intersect each ​other at ​the point O. Show ​that $\frac{AO}{BO}=\frac{CO}{DO}$.
    Show answer

    Proved: $\frac{AO}{BO}=\frac{CO}{DO}$, using $△AOB\sim △COD$ (AA similarity).

    Step-by-step solution →
13. Q13 NCERT · Exercise 6.2
    The diagonals of a ​quadrilateral $ABCD$ intersect each ​other at ​the point $O$ ​such that $\frac{AO}{BO}=\frac{CO}{DO}$. ​Show that $ABCD$ ​is ​a trapezium.
    Show answer

    Proved: $AB\parallel DC$, so $ABCD$ is a trapezium (one pair of parallel sides).

    Step-by-step solution →
14. Q14 Exemplar · Exercise 6.1
    In ​Fig. ​6.2, $\angle BAC=90^{\circ }$ ​and $AD\perp BC$. Then,

    1. (a)$BD\cdot CD=B{C}^2$
    2. (b)$AB\cdot AC=B{C}^2$
    3. (c)$BD\cdot CD=A{D}^2$
    4. (d)$AB\cdot AC=A{D}^2$
    Show answer

    (c) $BD\cdot CD=A{D}^2$

    Step-by-step solution →
15. Q15 Exemplar · Exercise 6.1
    The lengths ​of the diagonals of a ​rhombus ​are 16 cm ​and 12 ​cm. Then, the length ​of the side of the rhombus ​is

    1. (a)$9$ cm
    2. (b)$10$ cm
    3. (c)$8$ cm
    4. (d)$20$ cm
    Show answer

    (b) $10$ cm

    Step-by-step solution →
16. Q16 Exemplar · Exercise 6.1
    If $△ABC\sim △EDF$ and $△ABC$ ​is ​not ​similar to $△DEF$, then which ​of the following is not true?

    1. (a)$BC\cdot EF=AC\cdot FD$
    2. (b)$AB\cdot EF=AC\cdot DE$
    3. (c)$BC\cdot DE=AB\cdot EF$
    4. (d)$BC\cdot DE=AB\cdot FD$
    Show answer

    (c) $BC\cdot DE=AB\cdot EF$

    Step-by-step solution →
17. Q17 Exemplar · Exercise 6.1
    If ​in ​two triangles ABC and PQR, $\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}$, ​then

    1. (a)$△PQR\sim △CAB$
    2. (b)$△PQR\sim △ABC$
    3. (c)$△CBA\sim △PQR$
    4. (d)$△BCA\sim △PQR$
    Show answer

    (a) $△PQR\sim △CAB$

    Step-by-step solution →
18. Q18 Exemplar · Exercise 6.1
    In Fig. 6.3, two line ​segments $AC$ ​and $BD$ ​intersect ​each other at the point $P$ such ​that $PA=6$ cm, $PB=3$ cm, $PC=2.5$ cm, $PD=5$ cm, $\angle APB=50^{\circ }$ and $\angle CDP=30^{\circ }$. Then, $\angle PBA$ is ​equal ​to

    1. (a)$50^{\circ }$
    2. (b)$30^{\circ }$
    3. (c)$60^{\circ }$
    4. (d)$100^{\circ }$
    Show answer

    (d) $100^{\circ }$

    Step-by-step solution →
19. Q19 Exemplar · Exercise 6.1
    If in ​two ​triangles DEF and PQR, $\angle D=\angle Q$ ​and $\angle R=\angle E$, then ​which of the ​following ​is not ​true?

    1. (a)$\frac{EF}{PR}=\frac{DF}{PQ}$
    2. (b)$\frac{DE}{PQ}=\frac{EF}{RP}$
    3. (c)$\frac{DE}{QR}=\frac{DF}{PQ}$
    4. (d)$\frac{EF}{RP}=\frac{DE}{QR}$
    Show answer

    (b) $\frac{DE}{PQ}=\frac{EF}{RP}$

    Step-by-step solution →
20. Q20 Exemplar · Exercise 6.1
    In triangles $ABC$ ​and $DEF$, $\angle B=\angle E$, $\angle F=\angle C$ and $AB=3DE$. ​Then, ​the two ​triangles ​are

    1. (a)congruent but not similar
    2. (b)similar but not congruent
    3. (c)neither congruent nor similar
    4. (d)congruent as well as similar
    Show answer

    (b) similar but not congruent

    Step-by-step solution →