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 11 of 22

1. Q201 CBSE 2025
   Point $E$ ​lies ​on the extended side $AD$ ​of parallelogram $ABCD$. $BE$ intersects $CD$ at $F$. ​Show that

   1. (i)$△DFE\sim △CFB$ ​
   2. (ii)$△AEB\sim △CBF$.
   Show answer

   (i) In $△DFE$ and $△CFB$: $\angle DEF=\angle CBF$ and $\angle EDF=\angle BCF$ (alternate angles, $AE\parallel BC$), so $△DFE\sim △CFB$ by AA. (ii) In $△AEB$ and $△CBF$: $\angle EAB=\angle BCF$ (opposite angles of the parallelogram) and $\angle AEB=\angle CBF$ (alternate angles), so $△AEB\sim △CBF$ by AA. Hence proved.

   Step-by-step solution →
2. Q202 CBSE 2025
   In ​the ​figure, $MNOP$ is a ​trapezium with $MN\parallel PO$ and $PO=2MN$. ​A ​line ​segment $FE$ drawn parallel to $MN$ intersects $MP$ ​at $F$ and $NO$ at $E$ such ​that $\frac{NE}{EO}=\frac{3}{4}$. Diagonal $PN$ intersects $FE$ at $X$. Prove that $7FE=10MN$.

   Show answer

   Hence proved: $FE=FX+XE=\frac{4}{7}MN+\frac{6}{7}MN=\frac{10}{7}MN$, so $7FE=10MN$.

   Step-by-step solution →
3. Q203 CBSE 2025
   State 'Basic Proportionality Theorem' and ​use it ​to prove the ​following : ​In a quadrilateral ​ABCD, diagonals AC and BD intersect each other at ​O such that $\frac{AO}{BO}=\frac{CO}{DO}$ ​as shown in the ​given figure. Prove that ABCD is a trapezium.

   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. Proof: draw $OE\parallel AB$ meeting $AD$ at $E$. In $△DAB$, BPT gives $\frac{DE}{EA}=\frac{DO}{OB}$; the given ratio gives $\frac{DO}{BO}=\frac{CO}{AO}$, so $\frac{DE}{EA}=\frac{CO}{OA}$. By the converse of BPT in $△DAC$, $EO\parallel DC$. Since $OE\parallel AB$ and $OE\parallel DC$, $AB\parallel DC$, so ABCD is a trapezium. Hence proved.

   Step-by-step solution →
4. Q204 CBSE 2025
   State the ​SAS ​criteria of ​similarity ​of ​two triangles. In the given ​figure, it is given ​that $OA\cdot OC=OB\cdot OD$. Use the SAS criteria ​to prove that $AD\parallel CB$.

   Show answer

   **SAS criterion:** if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar. **Proof:** from $OA\cdot OC=OB\cdot OD$, $\frac{OA}{OB}=\frac{OD}{OC}$, and $\angle AOD=\angle BOC$ (vertically opposite angles), so $△OAD\sim △OBC$ by SAS. Hence $\angle OAD=\angle OBC$, which are alternate angles on the transversal $AB$, so $AD\parallel CB$. Hence proved.

   Step-by-step solution →
5. Q205 CBSE 2025
   If a ​line is ​drawn parallel to one side of a triangle intersecting ​the ​other two sides in distinct points then it divides the two sides in the same ratio. Prove it. Also, state the ​converse of the ​above ​statement.
   Show answer

   Hence proved: if a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, then $\frac{AD}{DB}=\frac{AE}{EC}$. Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

   Step-by-step solution →
6. Q206 CBSE 2025
   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.
   Show answer

   In $△ABC$ with $DE\parallel BC$, $\frac{AD}{DB}=\frac{AE}{EC}$, so the other two sides are divided in the same ratio. Hence proved.

   Step-by-step solution →
7. Q207 CBSE 2025
   State ​and prove the ​"Basic ​Proportionality Theorem".
   Show answer

   **Statement:** 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. **Proof:** In $△ABC$ with $DE\parallel BC$, join $BE$, $CD$ and draw $EM\perp AB$, $DN\perp AC$. Then $\frac{\text{ar}(△ADE)}{\text{ar}(△BDE)}=\frac{AD}{DB}$ and $\frac{\text{ar}(△ADE)}{\text{ar}(△CDE)}=\frac{AE}{EC}$. Since $△BDE$ and $△CDE$ are on the same base $DE$ between the parallels $DE$ and $BC$, $\text{ar}(△BDE)=\text{ar}(△CDE)$. Therefore $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   Step-by-step solution →
8. Q208 CBSE 2025
   If ​a line ​drawn parallel ​to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the ​same ratio, then it is parallel ​to the third side. State and prove the converse of ​the above statement.
   Show answer

   The converse is the 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 $△ABC$, $DE\parallel BC$ gives $\frac{AD}{DB}=\frac{AE}{EC}$.

   Step-by-step solution →
9. Q209 CBSE 2025
   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. (Basic Proportionality Theorem)
   Show answer

   In $△ABC$ with $DE\parallel BC$, $D$ on $AB$ and $E$ on $AC$: $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   Step-by-step solution →
10. Q210 CBSE 2025
    The corresponding sides of $△ABC$ ​and $△PQR$ ​are in the ​ratio $3:5$. $AD\perp BC$ and $PS\perp QR$ as shown ​in the ​following ​figures:

    1. (i)Prove that $△ADC\sim △PSR$.
    2. (ii)If $AD=4$ cm, find the length ​of $PS$.
    3. (iii)Using (ii) find $\text{ar}(△ABC):\text{ar}(△PQR)$.

    Show answer

    (i) $△ADC\sim △PSR$ by AA similarity ($\angle ADC=\angle PSR=90^{\circ }$ and $\angle C=\angle R$); (ii) $PS=\frac{20}{3}$ cm; (iii) $\text{ar}(△ABC):\text{ar}(△PQR)=9:25$.

    Step-by-step solution →
11. Q211 CBSE 2025
    The diagonal $BD$ of ​a parallelogram $ABCD$ ​intersects the line segment $AE$ ​at the point $F$, where $E$ is ​any point on the side $BC$. ​Prove that $DF\times EF=FB\times FA$.
    Show answer

    Since $△ADF\sim △EBF$ (AA similarity), $\frac{DF}{FB}=\frac{FA}{EF}$, and cross-multiplying gives $DF\times EF=FB\times FA$. Hence proved.

    Step-by-step solution →
12. Q212 CBSE 2025
    In ​the ​adjoining ​figure, $△CAB$ ​is a ​right triangle, right angled at $A$ and $AD\perp BC$. Prove that $△ADB\sim △CDA$. Further, if $BC=10$ cm and $CD=2$ cm, find ​the ​length ​of $AD$.

    Show answer

    $△ADB\sim △CDA$ is proved by the AA similarity criterion, and with $BC=10$ cm and $CD=2$ cm, the length of $AD$ is $4$ cm.

    Step-by-step solution →
13. Q213 CBSE 2025
    State basic ​proportionality theorem. ​Use it to ​prove the following: If three ​parallel lines $l,m,n$ are ​intersected ​by transversals $q$ and $s$ as shown ​in the adjoining ​figure, then $\frac{AB}{BC}=\frac{DE}{EF}$.

    Show answer

    BPT says a line parallel to one side of a triangle, meeting the other two sides at distinct points, divides them in the same ratio. Joining $AF$ (meeting $m$ at $G$) and applying it in $△ACF$ and $△AFD$ gives $\frac{AB}{BC}=\frac{AG}{GF}=\frac{DE}{EF}$, which proves the result.

    Step-by-step solution →
14. Q214 CBSE 2025
    In ​the ​given figure, $CM$ and $RN$ are respectively, the medians of $△ABC$ ​and $△PQR$. If $△ABC\sim △PQR$, ​prove that:

    1. (i)$△AMC\sim △PNR$;
    2. (ii)$\angle BCM=\angle QRN$; ​
    3. (iii)$△BMC\sim △QNR$.

    Show answer

    (i) $△AMC\sim △PNR$ by SAS similarity, since $\frac{AM}{PN}=\frac{AC}{PR}$ and $\angle A=\angle P$; (ii) $\angle BCM=\angle ACB-\angle ACM=\angle PRQ-\angle PRN=\angle QRN$; (iii) $△BMC\sim △QNR$ by SAS similarity, since $\frac{BM}{QN}=\frac{BC}{QR}$ and $\angle B=\angle Q$. Hence proved.

    Step-by-step solution →
15. Q215 CBSE 2025
    It is given ​that sides $AB$ ​and $AC$ and median $AD$ of $△ABC$ are respectively proportional to ​sides $PQ$ and $PR$ ​and median $PM$ ​of another $△PQR$. Show ​that $△ABC\sim △PQR$.
    Show answer

    $△ABC\sim △PQR$ by SAS similarity, since $\frac{AB}{PQ}=\frac{AC}{PR}$ and $\angle BAC=\angle QPR$ (obtained from $△AEC\sim △PNR$ by SSS). Hence proved.

    Step-by-step solution →
16. Q216 CBSE 2025
    Prove that a ​line drawn parallel to one side of ​a triangle to intersect the other two sides ​in distinct points divides the ​other two ​sides in the same ratio. Hence, in ​the ​figure given below, prove that $\frac{AM}{MB}=\frac{AN}{ND}$ where $LM\parallel CB$ and $LN\parallel CD$.

    Show answer

    The Basic Proportionality Theorem is proved by comparing areas of triangles; applying it in $△ABC$ and in $△ADC$ gives $\frac{AM}{MB}=\frac{AL}{LC}=\frac{AN}{ND}$, hence $\frac{AM}{MB}=\frac{AN}{ND}$.

    Step-by-step solution →
17. Q217 CBSE 2025
    If ​AD and ​PM ​are medians of triangles ABC ​and ​PQR ​respectively, where $△ABC\sim △PQR$, then ​prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.
    Show answer

    $△ABD\sim △PQM$ by SAS, so $\frac{AB}{PQ}=\frac{AD}{PM}$. Hence proved.

    Step-by-step solution →
18. Q218 CBSE 2025
    In ​the given figure, $PA$, $QB$ ​and $RC$ are perpendicular to $AC$. ​If $PA=x$ ​units, $QB=y$ ​units and $RC=z$ ​units, prove ​that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

    Show answer

    Using $△ABQ\sim △ACR$ and $△CBQ\sim △CAP$, $\frac{y}{z}+\frac{y}{x}=\frac{AB+BC}{AC}=1$, so $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$. Hence proved.

    Step-by-step solution →
19. Q219 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.
    Show answer

    $\frac{AD}{DB}=\frac{AE}{EC}$: a line drawn parallel to one side of a triangle divides the other two sides in the same ratio. Hence proved.

    Step-by-step solution →
20. Q220 CBSE 2025
    The ​perimeter of a ​right triangle ​is ​60 ​cm and its ​hypotenuse is 25 cm. Find the lengths ​of other two sides of the triangle.
    Show answer

    The other two sides of the triangle are $15$ cm and $20$ cm.

    Step-by-step solution →