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 10 of 22

1. Q181 CBSE 2025
   $PQRS$ ​is a trapezium in which $PQ\parallel SR$ and ​its diagonals intersect each ​other ​at ​the ​point $O$. Show that $\frac{PO}{QO}=\frac{RO}{SO}$.
   Show answer

   Hence $\frac{PO}{QO}=\frac{RO}{SO}$, since $△POQ\sim △ROS$ by the AA criterion.

   Step-by-step solution →
2. Q182 CBSE 2025
   In ​a trapezium $ABCD$, $AB\parallel DC$ ​and ​its diagonals ​intersect ​at $O$. Prove that $\frac{OA}{OC}=\frac{OB}{OD}$.
   Show answer

   In $△OAB$ and $△OCD$, $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ (alternate angles, $AB\parallel DC$), so $△OAB\sim △OCD$ (AA). Hence $\frac{OA}{OC}=\frac{OB}{OD}$. Hence proved.

   Step-by-step solution →
3. Q183 CBSE 2025
   A $1.5$ ​m ​tall boy is walking away from the base of ​a lamp post ​which is $12$ m high, at ​the speed ​of $2.5$ m/sec. Find the length ​of his ​shadow after $3$ seconds.
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   The length of the boy's shadow after $3$ seconds is $\frac{15}{14}$ m, that is about $1.07$ m.

   Step-by-step solution →
4. Q184 CBSE 2025
   $P$ and $Q$ are ​points on the sides $DE$ ​and $DF$ respectively ​of a $△DEF$ ​such that $DP=5$ cm, $DQ=6$ cm, $DE=15$ ​cm and $QF=18$ cm. ​Find ​whether $PQ\parallel EF$ or not.
   Show answer

   $PQ$ is **not** parallel to $EF$, because $\frac{DP}{PE}=\frac{5}{10}=\frac{1}{2}$ while $\frac{DQ}{QF}=\frac{6}{18}=\frac{1}{3}$, and $\frac{1}{2}\ne \frac{1}{3}$.

   Step-by-step solution →
5. Q185 CBSE 2025
   In ​the ​given figure, if $PQ\parallel RS$, ​then ​prove ​that $△POQ\sim △SOR$.

   Show answer

   $△POQ\sim △SOR$ by the AA similarity criterion, since $\angle OPQ=\angle OSR$ and $\angle OQP=\angle ORS$ (alternate interior angles, as $PQ\parallel RS$). Hence proved.

   Step-by-step solution →
6. Q186 CBSE 2025
   A vertical pole of height $10$ m ​casts a shadow of $15$ ​m on the ground and at the same time, ​a tower casts ​a shadow of $45$ m on the ​ground. ​Find the height of the ​tower.
   Show answer

   The height of the tower is $30$ m.

   Step-by-step solution →
7. Q187 CBSE 2025
   In ​the given ​figure, $△OSR\sim △OQP$, $\angle ROQ=125^{\circ }$ ​and $\angle ORS=70^{\circ }$. ​Find ​the measures ​of $\angle OSR$ ​and $\angle OQP$.

   Show answer

   $\angle OSR=55^{\circ }$ and $\angle OQP=55^{\circ }$.

   Step-by-step solution →
8. Q188 CBSE 2025
   In ​the given ​figure, $OA\cdot OB=OC\cdot OD$. Show ​that $\angle A=\angle C$ ​and $\angle B=\angle D$.

   Show answer

   Since $△AOD\sim △COB$ by SAS similarity, $\angle A=\angle C$ and $\angle B=\angle D$. Hence proved.

   Step-by-step solution →
9. Q189 CBSE 2025
   In parallelogram ​ABCD, side AD ​is produced ​to a point E and ​BE intersects CD at F. ​Prove ​that $△ABE\sim △CFB$.
   Show answer

   In $△ABE$ and $△CFB$, $\angle A=\angle C$ and $\angle AEB=\angle CBF$, so $△ABE\sim △CFB$ by the AA similarity criterion. Hence proved.

   Step-by-step solution →
10. Q190 CBSE 2025
    In ​the ​given ​figure, $\frac{QR}{QS}=\frac{QT}{PR}$ and $\angle 1=\angle 2$. Prove ​that $△PQS\sim △TQR$.

    Show answer

    Since $\angle 1=\angle 2$, $PQ=PR$, so $\frac{QP}{QT}=\frac{QS}{QR}$; with the common angle $\angle PQS=\angle TQR$, $△PQS\sim △TQR$ by the SAS similarity criterion. Hence proved.

    Step-by-step solution →
11. Q191 CBSE 2025
    Prove that ​the line ​joining ​the mid-points of ​any ​two sides of ​a triangle is ​parallel to the third side. (Mid-point Theorem)
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    In $△ABC$ let $D$ and $E$ be the mid-points of $AB$ and $AC$. Then $\frac{AD}{DB}=1=\frac{AE}{EC}$, so by the converse of the Basic Proportionality Theorem, $DE\parallel BC$. Hence proved.

    Step-by-step solution →
12. Q192 CBSE 2025
    Prove ​that ​diagonals of ​a ​trapezium ​divide ​each other ​proportionally.
    Show answer

    The diagonals $AC$ and $BD$ meet at $O$ with $\frac{AO}{OC}=\frac{BO}{OD}$, so the diagonals of a trapezium divide each other proportionally. Hence proved.

    Step-by-step solution →
13. Q193 CBSE 2025
    $S$ ​is a ​point ​on ​the ​side $QR$ of ​a $△PQR$ such that $\angle PSR=\angle QPR$. Prove ​that $P{R}^2=QR\times SR$.
    Show answer

    Since $△RSP\sim △RPQ$ (AA: $\angle RSP=\angle RPQ$ given, $\angle R$ common), $\frac{RS}{RP}=\frac{RP}{RQ}$, so $P{R}^2=QR\times SR$. Hence proved.

    Step-by-step solution →
14. Q194 CBSE 2025
    $AD$ ​and $PS$ ​are ​medians of ​triangles $ABC$ and $PQR$ respectively ​such ​that $△ABD\sim △PQS$. Prove that $△ABC\sim △PQR$.
    Show answer

    $△ABC\sim △PQR$ by the SAS similarity criterion, because $\frac{AB}{PQ}=\frac{BD}{QS}=\frac{BC}{QR}$ and the included angles $\angle B$ and $\angle Q$ are equal.

    Step-by-step solution →
15. Q195 CBSE 2025
    $P$ is ​a ​point on ​the ​side $BC$ ​of $△ABC$ such ​that $\angle APC=\angle BAC$. Prove ​that $A{C}^2=BC\cdot CP$.
    Show answer

    Since $△APC\sim △BAC$ by AA (the given equal angles and the common angle $C$), $\frac{CP}{AC}=\frac{AC}{BC}$, so $A{C}^2=BC\cdot CP$. Hence proved.

    Step-by-step solution →
16. Q196 CBSE 2025
    If $△ABC\sim △PQR$ in which $AB=6$ cm, $BC=4$ ​cm, $AC=8$ cm and $PR=6$ ​cm, then find ​the ​length of $(PQ+QR)$.
    Show answer

    Since $PQ=4.5$ cm and $QR=3$ cm, the length of $(PQ+QR)$ is $7.5$ cm.

    Step-by-step solution →
17. Q197 CBSE 2025
    In ​the ​given ​figure, D is a point on the side BC ​of $△ABC$ such ​that $\angle ADC=\angle BAC$. Show ​that $C{A}^2=CD\cdot CB$.

    Show answer

    Since $△ACD\sim △BCA$ by AA similarity, $\frac{CA}{CB}=\frac{CD}{CA}$, so $C{A}^2=CD\times CB$. Hence proved.

    Step-by-step solution →
18. Q198 CBSE 2025
    In the ​given ​figure, $\frac{QR}{QS}=\frac{QT}{PR}$ ​and $\angle 1=\angle 2$. ​Show that $△PQS\sim △TQR$.

    Show answer

    $△PQS\sim △TQR$ by the SAS similarity criterion, since $\frac{PQ}{TQ}=\frac{QS}{QR}$ and the angle at $Q$ is common. Hence proved.

    Step-by-step solution →
19. Q199 CBSE 2025
    In the ​given ​figure $\angle ADE=\angle ACB$ ​and $\frac{AD}{DB}=\frac{AE}{EC}$. ​Prove ​that $△ABC$ is ​an ​isosceles triangle.

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    Since $\frac{AD}{DB}=\frac{AE}{EC}$, $DE\parallel BC$, so $\angle ABC=\angle ADE=\angle ACB$. The sides opposite these equal angles give $AB=AC$, so $△ABC$ is isosceles. Hence proved.

    Step-by-step solution →
20. Q200 CBSE 2025
    From ​the ​given ​figure, ​find ​the value ​of $\sin \alpha$.

    Show answer

    $\sin \alpha =\frac{1}{2}$

    Step-by-step solution →