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 14 of 22

1. Q261 CBSE 2024
   In ​the given ​figure, $OA\cdot OB=OC\cdot OD$. Prove ​that $△AOD\sim △COB$.

   Show answer

   Since $\frac{OA}{OC}=\frac{OD}{OB}$ and $\angle AOD=\angle COB$ (vertically opposite angles), $△AOD\sim △COB$ by the SAS similarity criterion. Hence proved.

   Step-by-step solution →
2. Q262 CBSE 2024
   In ​the ​given ​figure, $\angle D=\angle E$ ​and $\frac{AD}{DB}=\frac{AE}{EC}$. ​Prove that $△ABC$ ​is isosceles.

   Show answer

   Since $DE\parallel BC$, $\angle ABC=\angle ADE=\angle AED=\angle ACB$, so $AB=AC$ and $△ABC$ is isosceles. Hence proved.

   Step-by-step solution →
3. Q263 CBSE 2024
   In ​the ​given figure, $△AHK\sim △ABC$. ​If $AK=8$ cm, $BC=3.2$ cm and $HK=6.4$ ​cm, then ​find the length ​of $AC$.

   Show answer

   $AC=4$ cm.

   Step-by-step solution →
4. Q264 CBSE 2023
   State ​and prove ​Basic ​Proportionality ​theorem.
   Show answer

   **Basic Proportionality Theorem:** 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. In $△ABC$ with $DE\parallel BC$, $\frac{\text{ar}(△ADE)}{\text{ar}(△BDE)}=\frac{AD}{DB}$ and $\frac{\text{ar}(△ADE)}{\text{ar}(△DEC)}=\frac{AE}{EC}$, and $\text{ar}(△BDE)=\text{ar}(△DEC)$ (same base $DE$, same parallels), so $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   Step-by-step solution →
5. Q265 CBSE 2024
   In ​the ​given ​figure, $\angle ABC=\angle ACB$ ​and $\frac{BC}{BE}=\frac{BD}{AC}$. ​Show that $△ABE\sim △DBC$ and $AE\parallel DC$.

   Show answer

   $△ABE\sim △DBC$ by SAS similarity (common $\angle B$ and $\frac{BA}{BD}=\frac{BE}{BC}$, using $AB=AC$), with $A\leftrightarrow D$, $B\leftrightarrow B$, $E\leftrightarrow C$. Hence $\angle BAE=\angle BDC$, and these corresponding angles give $AE\parallel DC$.

   Step-by-step solution →
6. Q266 CBSE 2024
   In ​the given figure, MNOP ​is ​a parallelogram and $AB\parallel MP$. ​Prove that $QC\parallel PO$.

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   $\frac{BP}{PQ}=\frac{BO}{OC}$ in $△BQC$, so by the converse of the Basic Proportionality Theorem $QC\parallel PO$. Hence proved.

   Step-by-step solution →
7. Q267 CBSE 2024
   Sides $AB$ and $BC$ and median $AD$ of ​a $△ABC$ ​are ​respectively proportional to ​sides $PQ$ ​and $QR$ ​and median $PM$ of $△PQR$. Show ​that $△ABC\sim △PQR$.

   Show answer

   $△ABD\sim △PQM$ by SSS, so $\angle B=\angle Q$; with $\frac{AB}{PQ}=\frac{BC}{QR}$, SAS similarity gives $△ABC\sim △PQR$. Hence proved.

   Step-by-step solution →
8. Q268 CBSE 2024
   ABCD is ​a trapezium with $AB\parallel DC$. ​AC ​and BD intersect ​at ​E. If $△AED\sim △BEC$, ​then ​prove that $AD=BC$.
   Show answer

   $AD=BC$. Hence proved.

   Step-by-step solution →
9. Q269 CBSE 2024
   $E$ ​is ​a ​point ​on the side $AD$ produced ​of a parallelogram $ABCD$ and $BE$ ​intersects $CD$ ​at $F$. ​Show that $△ABE\sim △CFB$.

   Show answer

   $△ABE\sim △CFB$ by the AA similarity criterion, since $\angle BAE=\angle FCB$ (opposite angles of the parallelogram) and $\angle AEB=\angle CBF$ (alternate angles, $AE\parallel BC$). Hence proved.

   Step-by-step solution →
10. Q270 CBSE 2024
    Sides $AB$, $BC$ and ​the ​median $AD$ of $△ABC$ are respectively proportional to sides $PQ$, $QR$ ​and the median $PM$ ​of ​another $△PQR$. Prove ​that $△ABC\sim △PQR$.

    Show answer

    $△ABC\sim △PQR$: from $\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$ we get $△ABD\sim △PQM$ (SSS), so $\angle B=\angle Q$; with $\frac{AB}{PQ}=\frac{BC}{QR}$ this gives $△ABC\sim △PQR$ (SAS). Hence proved.

    Step-by-step solution →
11. Q271 CBSE 2024
    Sides $AB$ and $AC$ ​and ​median $AD$ to $△ABC$ ​are ​respectively ​proportional ​to ​sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $△ABC\sim △PQR$.
    Show answer

    $△ABC\sim △PQR$. Hence proved: doubling the medians gives parallelograms $ABEC$ and $PQNR$; SSS similarity of $△ABE,△PQN$ and of $△ACE,△PRN$ gives $\angle BAC=\angle QPR$, and SAS similarity with $\frac{AB}{PQ}=\frac{AC}{PR}$ completes the proof.

    Step-by-step solution →
12. Q272 CBSE 2023
    If $AD$ ​and $PM$ are medians ​of ​triangles $ABC$ and $PQR$, ​respectively where $△ABC\sim △PQR$, prove ​that $\frac{AB}{PQ}=\frac{AD}{PM}$.
    Show answer

    $\frac{AB}{PQ}=\frac{AD}{PM}$ (proved via SAS similarity of ABD and PQM).

    Step-by-step solution →
13. Q273 CBSE 2024
    In ​the ​given ​figure, $△FEC\cong △GDB$ and $\angle 1=\angle 2$. Prove ​that $△ADE\sim △ABC$.

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    Since $\angle 1=\angle 2$, $AD=AE$; by CPCT from $△FEC\cong △GDB$, $EC=DB$. So $\frac{AD}{DB}=\frac{AE}{EC}$ and $DE\parallel BC$ (converse of BPT), giving $\angle ADE=\angle ABC$. With $\angle A$ common, $△ADE\sim △ABC$ (AA). Hence proved.

    Step-by-step solution →
14. Q274 CBSE 2024
    If ​a ​line is drawn parallel to one ​side of a triangle to intersect the other ​two sides in distinct points, then show ​that the other two sides are divided by ​this line ​in the ​same ratio.
    Show answer

    Proved: $\frac{AD}{DB}=\frac{AE}{EC}$. A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem).

    Step-by-step solution →
15. Q275 CBSE 2024
    If ​BD ​and ​QM are ​medians of triangles ​ABC and PQR, respectively, where $△ABC\sim △PQR$, prove that $\frac{AB}{PQ}=\frac{BD}{QM}$.
    Show answer

    $△ABD\sim △PQM$ by SAS, since $\frac{AB}{PQ}=\frac{AD}{PM}$ and $\angle A=\angle P$; therefore $\frac{AB}{PQ}=\frac{BD}{QM}$. Hence proved.

    Step-by-step solution →
16. Q276 CBSE 2024
    Sides ​AB, BC and ​median AD ​of $△ABC$ are respectively ​proportional to sides PQ, QR and median ​PM of $△PQR$. ​Show that $△ABC\sim △PQR$.
    Show answer

    $△ABD\sim △PQM$ (SSS) gives $\angle B=\angle Q$; with $\frac{AB}{PQ}=\frac{BC}{QR}$, $△ABC\sim △PQR$ (SAS). Hence proved.

    Step-by-step solution →
17. Q277 CBSE 2024
    In the given figure, altitudes $CE$ and $AD$ ​of $△ABC$ intersect each ​other ​at the ​point $P$. ​Show ​that

    1. (i)$△AEP\sim △CDP$, ​
    2. (ii)$△ABD\sim △CBE$, ​
    3. (iii)$△AEP\sim △ADB$.

    Show answer

    (i) $△AEP\sim △CDP$ by AA, since $\angle AEP=\angle CDP=90^{\circ }$ and $\angle APE=\angle CPD$ (vertically opposite); (ii) $△ABD\sim △CBE$ by AA, since $\angle ADB=\angle CEB=90^{\circ }$ and $\angle B$ is common; (iii) $△AEP\sim △ADB$ by AA, since $\angle AEP=\angle ADB=90^{\circ }$ and $\angle BAD$ is common. Hence proved.

    Step-by-step solution →
18. Q278 CBSE 2024
    Sides $AB$ ​and $AC$ and median $AD$ ​of ​a $△ABC$ are respectively proportional ​to ​sides $PQ$ ​and $PR$ and ​median $PM$ of ​another $△PQR$. Show that $△ABC\sim △PQR$.
    Show answer

    Produce $AD$ to $E$ and $PM$ to $N$ with $DE=AD$, $MN=PM$. Then $ABEC$ and $PQNR$ are parallelograms, so $BE=AC$, $QN=PR$, and $\frac{AB}{PQ}=\frac{BE}{QN}=\frac{AE}{PN}$, giving $△ABE\sim △PQN$ (SSS). Hence $\angle BAE=\angle QPN$ and, through alternate angles, $\angle EAC=\angle NPR$, so $\angle BAC=\angle QPR$. With $\frac{AB}{PQ}=\frac{AC}{PR}$, $△ABC\sim △PQR$ (SAS). Hence proved.

    Step-by-step solution →
19. Q279 CBSE 2024
    $ABCD$ ​is a trapezium with $AB\parallel DC$ and the ​diagonals intersect each ​other ​at ​the point $O$. ​Show ​that $\frac{AO}{BO}=\frac{CO}{DO}$.
    Show answer

    Since $△AOB\sim △COD$ (AA), $\frac{AO}{CO}=\frac{BO}{DO}$, so $AO\times DO=BO\times CO$ and $\frac{AO}{BO}=\frac{CO}{DO}$. Hence proved.

    Step-by-step solution →
20. Q280 CBSE 2024
    Prove that if a ​line is drawn ​parallel to ​one side of a ​triangle ​to intersect ​the other two ​sides ​at distinct points, then the other two sides are divided in the same ratio.
    Show answer

    $\frac{AD}{DB}=\frac{AE}{EC}$, so the line parallel to $BC$ divides $AB$ and $AC$ in the same ratio. Hence proved.

    Step-by-step solution →