Class 10 Maths · Chapter 6

Triangles

424 Class 10 Maths questions on Triangles from the NCERT textbook, NCERT Exemplar and CBSE board papers 2018–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 7 of 22

1. Q121 CBSE 2026
   In the given figure, $AB\parallel DC$. ​If $OB=3OD$ ​and $CD=1.8$ cm, ​then find ​the ​length $AB$.

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   $AB=5.4$ cm.

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2. Q122 CBSE 2026
   In ​the given ​figure, $QR\parallel CB$ and $RP\parallel AC$. ​If $BR=10$ cm, $QA=12$ cm, $BP=12$ ​cm and $PC=18$ ​cm, then ​find the ​lengths ​of $AR$ and $QC$.

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   $AR=15$ cm and $QC=8$ cm.

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3. Q123 CBSE 2026
   $S$ is any ​point on ​the side $QR$ of ​a $△PQR$ such ​that $\angle PSR=\angle QPR$. ​Prove that $\frac{QR}{RP}=\frac{RP}{RS}$.
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   By AA similarity $△RPQ\sim △RSP$, so $\frac{QR}{PR}=\frac{RP}{RS}$, i.e. $\frac{QR}{RP}=\frac{RP}{RS}$ (equivalently $R{P}^2=QR\times RS$). Hence proved.

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4. Q124 CBSE 2026
   T is ​a ​point on the line PS ​produced ​of ​a parallelogram ​PQRS ​and QT intersects RS at V. ​Prove that $△PQT\sim △RVQ$.
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   In $△PQT$ and $△RVQ$, $\angle QPT=\angle VRQ$ (opposite angles of parallelogram PQRS) and $\angle QTP=\angle VQR$ (alternate interior angles, $PT\parallel QR$ with transversal QT). Hence $△PQT\sim △RVQ$ by the AA similarity criterion.

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5. Q125 CBSE 2026
   In the given ​figure, $△ABC$ ​is a right angled triangle with $\angle A=90^{\circ }$. $AD$ is ​perpendicular to $BC$. Prove ​that: ​

   1. (i)$△DBA\sim △DAC$
   2. (ii)$D{A}^2=DB\times DC$
   3. (iii)Find the area of $△ABC$ when $DB=9$ ​cm and $DC=16$ cm.

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   (i) $△DBA\sim △DAC$ by the AA criterion. (ii) $D{A}^2=DB\times DC$. (iii) Area of $△ABC=150{\text{ cm}}^2$.

   Step-by-step solution →
6. Q126 CBSE 2026
   $CD$ and $GH$ ​are respectively ​the bisectors ​of $\angle ACB$ and $\angle EGF$ such ​that $D$ and $H$ ​lie ​on sides $AB$ and $FE$ ​of $△ABC$ and $△EFG$ ​respectively. If $△ABC\sim △FEG$, then show that

   1. (i)$\frac{CD}{GH}=\frac{AC}{FG}$ and
   2. (ii)$△DCB\sim △HGE$.
   Show answer

   Both results are proved. (i) $△ACD\sim △FGH$ (AA: $\angle A=\angle F$, $\angle ACD=\angle FGH$) gives $\frac{CD}{GH}=\frac{AC}{FG}$. (ii) $△DCB\sim △HGE$ (AA: $\angle DBC=\angle HEG$, $\angle DCB=\angle HGE$).

   Step-by-step solution →
7. Q127 CBSE 2020
   Prove ​that if ​a line is drawn ​parallel to one side ​of a triangle to intersect the ​other two sides ​in distinct points, then the ​other two sides are divided in the same ​ratio.
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   Joining $BE$, $CD$ and using half base times height, $\frac{\text{ar}(△ADE)}{\text{ar}(△BDE)}=\frac{AD}{DB}$ and $\frac{\text{ar}(△ADE)}{\text{ar}(△CDE)}=\frac{AE}{EC}$; since $\text{ar}(△BDE)=\text{ar}(△CDE)$ (same base $DE$, between the parallels $DE$ and $BC$), it follows that $\frac{AD}{DB}=\frac{AE}{EC}$.

   Step-by-step solution →
8. Q128 CBSE 2026
   Prove that, if a line is drawn parallel to one side of a triangle ​to intersect the other two ​sides in ​distinct points, the other two ​sides ​are ​divided in the same ratio.
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   Hence $\frac{AD}{DB}=\frac{AE}{EC}$: a line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality / Thales Theorem).

   Step-by-step solution →
9. Q129 CBSE 2026
   In $△ABC$, $AD$ ​is ​a ​median. $X$ is ​a point on $AD$ such that $AX:XD=2:3$. $BX$ ​is extended so that it intersects $AC$ at $Y$. ​Prove that $BX=4XY$.

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   By the construction $DZ\parallel BY$, we get $BY=2DZ$ and $XY=\frac{2}{5}DZ$, so $BY=5XY$ and therefore $BX=BY-XY=4XY$. Hence proved.

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10. Q130 CBSE 2026
    In ​the ​adjoining ​figure, $△ABE\cong △ACD$. Prove ​that :

    1. (i)$△ADE\sim △ABC$
    2. (ii)$△BOD\sim △COE$

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    (i) By CPCT, $AD=AE$ and $AB=AC$, so $\frac{AD}{AB}=\frac{AE}{AC}$ with the common angle $\angle A$; hence $△ADE\sim △ABC$ by the SAS similarity criterion. (ii) $\angle BOD=\angle COE$ (vertically opposite angles) and $\angle DBO=\angle ECO$ (from the CPCT pair $\angle ABE=\angle ACD$); hence $△BOD\sim △COE$ by the AA similarity criterion.

    Step-by-step solution →
11. Q131 CBSE 2024
    If a line is drawn ​parallel to one ​side ​of a ​triangle to intersect the other two sides in ​distinct points, then ​prove ​that the other two sides ​are divided in the same ratio.
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    $\frac{AD}{DB}=\frac{AE}{EC}$, so the line $DE\parallel BC$ divides the sides $AB$ and $AC$ in the same ratio. Hence proved.

    Step-by-step solution →
12. Q132 CBSE 2026
    $D$ is the mid-point of ​side $BC$ of $△ABC$. $CE$ and $BF$ intersect at $O$, a point on $AD$. $AD$ is ​produced to $G$ such that $OD=DG$. ​Prove that ​:

    1. (i)$OBGC$ ​is a ​parallelogram.
    2. (ii)$EF\parallel BC$.
    3. (iii)$△AEF\sim △ABC$.

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    (i) $OBGC$ is a parallelogram, because its diagonals $OG$ and $BC$ bisect each other at $D$. (ii) $EF\parallel BC$, because $\frac{AE}{EB}=\frac{AO}{OG}=\frac{AF}{FC}$ by the Basic Proportionality Theorem, so its converse applies. (iii) $△AEF\sim △ABC$ by the AA criterion, since $\angle A$ is common and $EF\parallel BC$ gives $\angle AEF=\angle ABC$.

    Step-by-step solution →
13. Q133 CBSE 2026
    Sides $AB$, $AC$ and ​altitude $AD$ ​of $△ABC$ are proportional to ​sides $PQ$, $PR$ and altitude $PS$ ​of another triangle $PQR$. ​Prove ​that $△ABC\sim △PQR$.
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    The altitudes give right angles at $D$ and at $S$, and Pythagoras turns $\frac{AB}{PQ}=\frac{AD}{PS}=k$ into $BD=kQS$, so $△ABD\sim △PQS$ and likewise $△ACD\sim △PRS$. These give $\angle B=\angle Q$ and $\angle C=\angle R$, hence $△ABC\sim △PQR$ by the AA criterion.

    Step-by-step solution →
14. Q134 CBSE 2026
    Through the mid-point $Q$ ​of side $CD$ of ​a parallelogram $ABCD$, the ​line $AR$ is drawn which intersects $BD$ at $P$ ​and meets $BC$ ​produced at $R$. Prove that:

    1. (i)$AQ=QR$
    2. (ii)$AP=2PQ$ ​
    3. (iii)$PR=2AP$

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    All three results are proved. (i) $△ADQ\cong △RCQ$ (ASA) gives $AQ=QR$. (ii) $△DPQ\sim △BPA$ with ratio $\frac{1}{2}$ gives $AP=2PQ$. (iii) Writing $PQ=k$ gives $AP=2k$, $AQ=QR=3k$ and $PR=4k$, so $PR=2AP$.

    Step-by-step solution →
15. Q135 CBSE 2026
    $AD$ ​and $PS$ are ​respectively ​the medians of $△ABC$ ​and $△PQR$. If $△ABC\sim △PQR$, ​then prove ​that

    1. (i)$△ADC\sim △PSR$
    2. (ii)$\frac{AD}{PS}=\frac{BC}{QR}$.

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    (i) $△ADC\sim △PSR$ (proved by SAS similarity using $\angle C=\angle R$ and $\frac{AC}{PR}=\frac{DC}{SR}$). (ii) Hence $\frac{AD}{PS}=\frac{BC}{QR}$. Proved.

    Step-by-step solution →
16. Q136 CBSE 2026
    In the ​given figure, $PQRS$ is ​a ​trapezium with $PQ\parallel SR$. ​Diagonals $PR$ and $QS$ ​intersect each ​other at a point $T$. Prove that $\frac{PT}{TR}=\frac{QT}{TS}$. Further, if $TS=TR$, ​prove that $△PTS\sim △QTR$.

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    Since $PQ\parallel SR$, the alternate angles give $△PTQ\sim △RTS$, so $\frac{PT}{TR}=\frac{QT}{TS}$. If further $TS=TR$, this forces $PT=QT$, and together with the vertically opposite angles $\angle PTS=\angle QTR$ it gives $△PTS\sim △QTR$ by the SAS criterion. Hence proved in either alternative.

    Step-by-step solution →
17. Q137 CBSE 2026
    State and ​prove ​Basic ​Proportionality ​Theorem.
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    Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Proof as above establishes $\frac{AD}{DB}=\frac{AE}{EC}$ using areas of triangles.

    Step-by-step solution →
18. Q138 CBSE 2026
    In ​the given figure, $CM$ and $RN$ ​are respectively the ​medians of $△ABC$ ​and $△PQR$. If $△ABC\sim △PQR$, then ​prove that ​:

    1. (i)$△AMC\sim △PNR$
    2. (ii)$△CMB\sim △RNQ$

    Show answer

    (i) $△AMC\sim △PNR$ is proved by SAS similarity using $\frac{AM}{PN}=\frac{CA}{RP}$ and $\angle A=\angle P$; (ii) $△CMB\sim △RNQ$ is proved by SAS similarity using $\frac{MB}{NQ}=\frac{BC}{QR}$ and $\angle B=\angle Q$.

    Step-by-step solution →
19. Q139 CBSE 2026
    As ​shown ​in the given figure, a girl of height $90$ cm is walking away from ​the ​base of a lamp post at a speed ​of $1.2$ ​m/s. If the lamp is $3.6$ m ​above the ground, find the ​length of her shadow after $4$ seconds.

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    The length of her shadow after $4$ seconds is $1.6$ m.

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20. Q140 CBSE 2026
    Carom board is a very popular game. The board is a square of side length $65$ cm. It has circular pockets in each corner. Ansh strikes a disc, ​kept at position $P$ ​with a striker. ​The disc hits ​the boundary of the board at $R$ and goes ​straight to pocket at corner $C$. It is ​given that $PS=9$ cm, $PQ=35$ cm, $BR=x$, $\angle PRQ=\alpha$ and $\angle CRB=\theta$. Based on the above information, answer the following questions :

    1. (i)Using law of reflection i.e. $\angle PRT=\angle CRT$, prove that $\theta =\alpha$.[1]
    2. (ii)Prove that $△PQR\sim △CBR$, given that $PQ$ is perpendicular to $AB$.[1]
    3. (iii) (a)Find the value of $x$ using similarity of triangles.[2]
    4. (iii) (b)(OR) If $\frac{\text{Area }△PQR}{\text{Area }△CBR}=\frac{P{Q}^2}{C{B}^2}$, then find the value of $x$.[2]
    Show answer

    (i) $\theta =\alpha$, proved using the law of reflection. (ii) $△PQR\sim △CBR$ by the AA criterion. (iii) $x=36.4$ cm (both methods (a) and (b) give the same value).

    Step-by-step solution →