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 2 of 6

1. Q21 Exemplar · Exercise 6.1
   It ​is given that $△ABC\sim △PQR$, with $\frac{BC}{QR}=\frac{1}{3}$. ​Then, $\frac{\mathrm{ar}(PRQ)}{\mathrm{ar}(BCA)}$ ​is ​equal ​to

   1. (a)$9$
   2. (b)$3$
   3. (c)$\frac{1}{3}$
   4. (d)$\frac{1}{9}$
   Show answer

   (a) $9$

   Step-by-step solution →
2. Q22 Exemplar · Exercise 6.1
   It is ​given ​that $△ABC\sim △DFE$, $\angle A=30^{\circ }$, $\angle C=50^{\circ }$, $AB=5$ cm, $AC=8$ ​cm ​and $DF=7.5$ cm. ​Then, ​the ​following ​is true:

   1. (a)$DE=12$ cm, $\angle F=50^{\circ }$
   2. (b)$DE=12$ cm, $\angle F=100^{\circ }$
   3. (c)$EF=12$ cm, $\angle D=100^{\circ }$
   4. (d)$EF=12$ cm, $\angle D=30^{\circ }$
   Show answer

   (b) $DE=12$ cm, $\angle F=100^{\circ }$

   Step-by-step solution →
3. Q23 Exemplar · Exercise 6.1
   If ​in triangles $ABC$ ​and $DEF$, $\frac{AB}{DE}=\frac{BC}{FD}$, ​then they will ​be similar, ​when

   1. (a)$\angle B=\angle E$
   2. (b)$\angle A=\angle D$
   3. (c)$\angle B=\angle D$
   4. (d)$\angle A=\angle F$
   Show answer

   (c) $\angle B=\angle D$

   Step-by-step solution →
4. Q24 Exemplar · Exercise 6.1
   If $△ABC\sim △QRP$, $\frac{\text{ar}(ABC)}{\text{ar}(PQR)}=\frac{9}{4}$, $AB=18$ ​cm ​and $BC=15$ cm, ​then ​PR is ​equal ​to

   1. (a)$10$ cm
   2. (b)$12$ cm
   3. (c)$\frac{20}{3}$ cm
   4. (d)$8$ cm
   Show answer

   (a) $10$ cm

   Step-by-step solution →
5. Q25 Exemplar · Exercise 6.1
   If ​S ​is ​a point on side PQ ​of a $△PQR$ such ​that $PS=QS=RS$, ​then

   1. (a)$PR\cdot QR=R{S}^2$
   2. (b)$Q{S}^2+R{S}^2=Q{R}^2$
   3. (c)$P{R}^2+Q{R}^2=P{Q}^2$
   4. (d)$P{S}^2+R{S}^2=P{R}^2$
   Show answer

   (c) $P{R}^2+Q{R}^2=P{Q}^2$

   Step-by-step solution →
6. Q26 Exemplar · Exercise 6.2
   Is ​the ​triangle with ​sides 25 cm, 5 cm ​and 24 ​cm a ​right triangle? ​Give reasons for your answer.
   Show answer

   No. Since $25^2=625\ne 5^2+24^2=601$, the triangle is not a right triangle.

   Step-by-step solution →
7. Q27 Exemplar · Exercise 6.2
   It is given that triangle DEF ~ triangle RPQ. ​Is it ​true to say ​that angle D = ​angle R and angle ​F = ​angle P? Why?
   Show answer

   Only partly true: angle D = angle R is correct, but angle F = angle P is wrong. From DEF ~ RPQ we get angle F = angle Q (and angle E = angle P).

   Step-by-step solution →
8. Q28 Exemplar · Exercise 6.2
   A and ​B are respectively the points on the sides PQ and PR ​of a triangle PQR ​such that $PQ=12.5$ ​cm, $PA=5$ cm, $BR=6$ cm and $PB=4$ cm. ​Is $AB\parallel QR$? ​Give reasons for your answer.
   Show answer

   Yes, AB is parallel to QR. Since PA/AQ = 5/7.5 = 2/3 and PB/BR = 4/6 = 2/3 are equal, by the converse of the Basic Proportionality Theorem AB || QR.

   Step-by-step solution →
9. Q29 Exemplar · Exercise 6.2
   In Fig. ​6.4, BD ​and ​CE ​intersect each ​other at the ​point ​P. Is $△PBC\sim △PDE$? Why?

   Show answer

   Yes, $△PBC\sim △PDE$ by the SAS similarity criterion: $\frac{PB}{PD}=\frac{PC}{PE}=\frac{1}{2}$ and $\angle BPC=\angle DPE$ (vertically opposite).

   Step-by-step solution →
10. Q30 Exemplar · Exercise 6.2
    In triangles $PQR$ and $MST$, $\angle P=55^{\circ }$, $\angle Q=25^{\circ }$, $\angle M=100^{\circ }$ ​and $\angle S=25^{\circ }$. Is $△QPR\sim △TSM$? ​Why?
    Show answer

    No, $△QPR\sim △TSM$ as written is not correct: $\angle Q=25^{\circ }\ne \angle T=55^{\circ }$. The two triangles do have equal angle sets $\{25^{\circ },55^{\circ },100^{\circ }\}$, so they are similar, but in the correct correspondence $△QPR\sim △STM$.

    Step-by-step solution →
11. Q31 Exemplar · Exercise 6.2
    Is the ​following ​statement true? ​Why? "Two quadrilaterals are similar, if ​their ​corresponding angles ​are ​equal".
    Show answer

    No, the statement is false. A square and a non-square rectangle have equal corresponding angles but unequal side ratios, so they are not similar.

    Step-by-step solution →
12. Q32 Exemplar · Exercise 6.2
    Two sides and the perimeter of one triangle are respectively ​three times ​the ​corresponding sides and the perimeter of ​the other triangle. ​Are the two triangles ​similar? Why?
    Show answer

    Yes, the triangles are similar by the SSS criterion, since the third sides are also in ratio 3:1.

    Step-by-step solution →
13. Q33 Exemplar · Exercise 6.2
    If in two right triangles, ​one of ​the acute ​angles of one triangle is equal to ​an ​acute angle of the ​other triangle, can you say ​that the two triangles will be similar? Why?
    Show answer

    Yes, the two right triangles are similar. The right angles are equal, one pair of acute angles is equal (given), so by the angle sum property the third pair is also equal — the triangles are equiangular and hence similar by the AAA (AA) similarity criterion.

    Step-by-step solution →
14. Q34 Exemplar · Exercise 6.2
    The ratio of the corresponding altitudes of two similar triangles is $\frac{3}{5}$. Is it ​correct ​to say ​that ratio ​of ​their areas is $\frac{6}{5}$? Why?
    Show answer

    No, it is not correct. The ratio of areas is ${\left(\frac{3}{5}\right)}^2=\frac{9}{25}$, not $\frac{6}{5}$.

    Step-by-step solution →
15. Q35 Exemplar · Exercise 6.2
    D ​is a point ​on side QR ​of $△PQR$ such that $PD\perp QR$. Will it be ​correct to say that $△PQD\sim △RPD$? ​Why?
    Show answer

    No. $△PQD\sim △RPD$ only when $\angle QPR=90^{\circ }$, and that is not given.

    Step-by-step solution →
16. Q36 Exemplar · Exercise 6.2
    In ​Fig. 6.5, if $\angle D=\angle C$, ​then ​is it ​true that $△ADE\sim △ACB$? ​Why?

    Show answer

    Yes, $△ADE\sim △ACB$ by the AA similarity criterion, since $\angle A$ is common and $\angle D=\angle C$.

    Step-by-step solution →
17. Q37 Exemplar · Exercise 6.2
    Is it ​true to ​say that if in two triangles, an angle of one triangle is equal to an angle of another triangle ​and two ​sides ​of one ​triangle are proportional to the two sides of the other ​triangle, then the triangles are similar? Give reasons for your answer.
    Show answer

    No, it is not always true. The triangles are similar only if the equal angle is the angle **included** between the two proportional sides (SAS similarity criterion). If the equal angle is not the included angle, similarity is not guaranteed.

    Step-by-step solution →
18. Q38 Exemplar · Exercise 6.3
    In a triangle PQR, ​PR^2 ​- ​PQ^2 = QR^2 and M is ​a ​point on side PR such that QM is ​perpendicular to PR. Prove that ​QM^2 = PM x ​MR.
    Show answer

    $Q{M}^2=PM\times MR$, proved: the given relation makes angle Q a right angle, so QM is the altitude on the hypotenuse and the similar triangles PMQ and QMR give the geometric-mean relation.

    Step-by-step solution →
19. Q39 Exemplar · Exercise 6.3
    Find the ​value ​of $x$ ​for ​which $DE\parallel AB$ ​in Fig. 6.8.

    Show answer

    $x=2$

    Step-by-step solution →
20. Q40 Exemplar · Exercise 6.3
    In ​Fig. ​6.9, if $\angle 1=\angle 2$ ​and $△NSQ\cong △MTR$, ​then prove ​that $△PTS\sim △PRQ$.

    Show answer

    Proved: $△PTS\sim △PRQ$. Chain of reasoning: $SQ=TR$ (CPCT from the given congruence) and $PS=PT$ (sides opposite the equal angles $\angle 1$, $\angle 2$), so $\frac{PS}{SQ}=\frac{PT}{TR}$, giving $ST\parallel QR$ by the converse of the Basic Proportionality Theorem; the resulting corresponding angles together with the common angle $P$ make the two triangles equiangular (AAA).

    Step-by-step solution →