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 21 of 22

1. Q401 CBSE 2019
   Two ​right ​triangles $ABC$ ​and $DBC$ ​are drawn ​on the same hypotenuse $BC$ ​and on the same side of $BC$. If $AC$ ​and $BD$ intersect at $P$, prove that $AP\times PC=BP\times DP$.
   Show answer

   Since $△APB\sim △DPC$ (AA), $\frac{AP}{DP}=\frac{PB}{PC}$, hence $AP\times PC=BP\times DP$. Proved.

   Step-by-step solution →
2. Q402 CBSE 2019
   Diagonals of a trapezium $PQRS$ intersect ​each other at ​the point $O$, $PQ\parallel RS$ ​and $PQ=3RS$. Find the ratio of ​the ​areas of ​triangles $POQ$ and $ROS$.
   Show answer

   $\text{ar}(△POQ):\text{ar}(△ROS)=9:1$.

   Step-by-step solution →
3. Q403 CBSE 2019
   In ​Figure 1, $BL$ and $CM$ ​are ​medians ​of ​a $△ABC$ right-angled at $A$. Prove ​that $4(B{L}^2+C{M}^2)=5B{C}^2$.

   Show answer

   Proved: $4(B{L}^2+C{M}^2)=5B{C}^2$, since $B{L}^2+C{M}^2=\frac{5}{4}(A{B}^2+A{C}^2)$ and $B{C}^2=A{B}^2+A{C}^2$.

   Step-by-step solution →
4. Q404 CBSE 2019
   In $△ABC$, $\angle B=90^{\circ }$ ​and $D$ ​is ​the ​mid-point of $BC$. Prove ​that $A{C}^2=A{D}^2+3C{D}^2$.
   Show answer

   Proved: $A{C}^2=A{D}^2+3C{D}^2$.

   Step-by-step solution →
5. Q405 CBSE 2019
   In Figure ​1, $E$ is a point on $CB$ produced ​of ​an isosceles $△ABC$, with ​side $AB=AC$. If $AD\perp BC$ ​and $EF\perp AC$, prove ​that $△ABD\sim △ECF$.

   Show answer

   $△ABD\sim △ECF$ (by the AA similarity criterion).

   Step-by-step solution →
6. Q406 CBSE 2019
   In Fig. ​3, $\angle ACB=90^{\circ }$ ​and $CD\perp AB$, ​prove ​that $C{D}^2=BD\times AD$.

   Show answer

   Proved: $C{D}^2=BD\times AD$ (since $△ADC\sim △CDB$ gives $\frac{AD}{CD}=\frac{CD}{BD}$).

   Step-by-step solution →
7. Q407 CBSE 2019
   If $P$ ​and $Q$ ​are ​the points on side $CA$ and $CB$ respectively ​of $△ABC$, right angled at $C$, prove ​that $(A{Q}^2+B{P}^2)=(A{B}^2+P{Q}^2)$.
   Show answer

   $A{Q}^2+B{P}^2=A{C}^2+B{C}^2+C{P}^2+C{Q}^2=A{B}^2+P{Q}^2$. Proved.

   Step-by-step solution →
8. Q408 CBSE 2019
   $D$, $E$ and $F$ ​are ​respectively ​the mid-points of ​sides $AB$, $BC$ and $CA$ of $△ABC$. Find the ​ratio of the areas of $△DEF$ and $△ABC$.
   Show answer

   $\frac{\text{ar}(△DEF)}{\text{ar}(△ABC)}=\frac{1}{4}$, i.e. $1:4$.

   Step-by-step solution →
9. Q409 CBSE 2019
   Prove that the ​area of an ​equilateral ​triangle described on one ​side of a square ​is equal to half the area of ​the equilateral triangle described on ​one ​of its diagonals.
   Show answer

   Area of the equilateral triangle on a side $=\frac{1}{2}\times$ area of the equilateral triangle on the diagonal. Hence proved.

   Step-by-step solution →
10. Q410 CBSE 2019
    In a triangle, if the ​square of ​one side ​is equal to the sum of the squares of the other two sides, then prove ​that the ​angle ​opposite to the first side is a right ​angle.
    Show answer

    $\angle ABC=90^{\circ }$ (the angle opposite $AC$ is a right angle).

    Step-by-step solution →
11. Q411 CBSE 2019
    In $△ABC$ (Figure ​3), $AD\perp BC$. Prove that $A{C}^2=A{B}^2+B{C}^2-2BC\times BD$.

    Show answer

    Proved: $A{C}^2=A{B}^2+B{C}^2-2BC\times BD$, using the Pythagoras theorem on triangles $ADC$ and $ADB$ with $DC=BC-BD$.

    Step-by-step solution →
12. Q412 CBSE 2019
    Prove ​that in ​a ​right triangle, ​the square of the hypotenuse is equal ​to the sum ​of the ​squares of the other two sides.
    Show answer

    $A{C}^2=A{B}^2+B{C}^2$, i.e. the square of the hypotenuse equals the sum of the squares of the other two sides. Hence proved.

    Step-by-step solution →
13. Q413 CBSE 2019
    Prove that in ​a right angle triangle, ​the square of the ​hypotenuse is ​equal to the ​sum of ​squares of the other two sides.
    Show answer

    Hence $A{C}^2=A{B}^2+B{C}^2$, i.e. the square of the hypotenuse equals the sum of the squares of the other two sides. Proved.

    Step-by-step solution →
14. Q414 CBSE 2018
    Given $△ABC\sim △PQR$, ​if $\frac{AB}{PQ}=\frac{1}{3}$, then find $\frac{\mathrm{ar}△ABC}{\mathrm{ar}△PQR}$.
    Show answer

    $\frac{\mathrm{ar}△ABC}{\mathrm{ar}△PQR}=\frac{1}{9}$

    Step-by-step solution →
15. Q415 CBSE 2018
    $DE$ is ​drawn parallel ​to the ​base $BC$ of a $△ABC$, meeting $AB$ ​at $D$ ​and $AC$ at $E$. If $\frac{AB}{BD}=4$ ​and $CE=2$ ​cm, find $AE$.
    Show answer

    $AE=6$ cm

    Step-by-step solution →
16. Q416 CBSE 2018
    Given $△ABC\sim △PQR$, ​if $\frac{AB}{PQ}=\frac{1}{3}$, ​then find $\frac{\text{ar}△ABC}{\text{ar}△PQR}$.
    Show answer

    $\frac{\text{ar}△ABC}{\text{ar}△PQR}=\frac{1}{9}$.

    Step-by-step solution →
17. Q417 CBSE 2018
    Prove that ​the area ​of an equilateral triangle described on one side of the square ​is equal to ​half the ​area of the equilateral triangle described on ​one of its ​diagonals.
    Show answer

    Since $\frac{A_1}{A_2}=\frac{1}{2}$, the equilateral triangle on a side has half the area of the equilateral triangle on the diagonal. Hence proved.

    Step-by-step solution →
18. Q418 CBSE 2018
    Prove that the area of an equilateral triangle described on ​one side of ​the square is equal ​to half ​the ​area ​of ​the equilateral triangle ​described on one of its diagonal.
    Show answer

    $\frac{A_{\text{side}}}{A_{\text{diagonal}}}=\frac{\frac{\sqrt{3}}{4}{a}^2}{\frac{\sqrt{3}}{4}(2a^2)}=\frac{1}{2}$, so the equilateral triangle on a side is half the equilateral triangle on a diagonal.

    Step-by-step solution →
19. Q419 CBSE 2018
    If the ​area ​of two ​similar triangles ​are equal, prove ​that ​they are congruent.
    Show answer

    The corresponding sides are equal ($AB=DE$, $BC=EF$, $CA=FD$), so $△ABC\cong △DEF$ by SSS.

    Step-by-step solution →
20. Q420 CBSE 2018
    $S$ and $T$ ​are ​points on ​the sides $PR$ and $QR$ ​of $△PQR$ such that $\angle P=\angle RTS$. ​Show ​that $△RPQ\sim △RTS$.
    Show answer

    $△RPQ\sim △RTS$ by the AA similarity criterion (common $\angle R$ and given $\angle P=\angle RTS$). Hence proved.

    Step-by-step solution →