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 20 of 22

1. Q381 CBSE 2020
   In a triangle, ​if 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

   By constructing a right triangle on the same legs and using Pythagoras with SSS congruence, $\angle B=90^{\circ }$, so the angle opposite the longest side is a right angle. Hence proved.

   Step-by-step solution →
2. Q382 CBSE 2020
   In ​a rectangle $ABCD$, $P$ ​is ​any interior ​point. Then ​prove that $P{A}^2+P{C}^2=P{B}^2+P{D}^2$.
   Show answer

   $P{A}^2+P{C}^2=p^2+q^2+a^2+b^2=P{B}^2+P{D}^2$, which proves the statement.

   Step-by-step solution →
3. Q383 CBSE 2020
   Prove that in a ​right triangle, ​the square ​of the ​hypotenuse ​is equal to sum of the squares of the ​remaining ​two sides.
   Show answer

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

   Step-by-step solution →
4. Q384 CBSE 2020
   In ​a right-angled ​triangle, prove that ​the square of the hypotenuse is ​equal ​to the sum of the ​squares ​of the remaining two sides.
   Show answer

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

   Step-by-step solution →
5. Q385 CBSE 2020
   In ​an equilateral triangle $ABC$, $D$ ​is a point ​on ​the ​side $BC$ such ​that $BD=\frac{1}{3}BC$. Prove that $9A{D}^2=7A{B}^2$.
   Show answer

   Since $A{D}^2=\frac{7}{9}{a}^2$ and $AB=a$, we get $9A{D}^2=7a^2=7A{B}^2$. Hence proved.

   Step-by-step solution →
6. Q386 CBSE 2020
   In a ​right-angled triangle, prove that ​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 →
7. Q387 CBSE 2020
   Prove that ​the ​sum of ​squares of the sides of a rhombus is ​equal ​to ​the sum of ​the squares of ​its diagonals.
   Show answer

   Proved: $A{B}^2+B{C}^2+C{D}^2+D{A}^2=A{C}^2+B{D}^2$ (sum of squares of sides $=$ sum of squares of diagonals).

   Step-by-step solution →
8. Q388 CBSE 2020
   Prove that ​in ​a ​triangle, ​if the square of one side is equal to the sum of the squares of the other two ​sides, then ​the angle opposite ​to the ​first side is a right angle.
   Show answer

   Since $△ABC\cong △DEF$ by SSS, $\angle B=\angle E=90^{\circ }$. Hence the angle opposite the longest side $AC$ is a right angle. (Converse of the Pythagoras theorem proved.)

   Step-by-step solution →
9. Q389 CBSE 2020
   In a right ​triangle, prove ​that the square of the hypotenuse ​is equal ​to sum of ​squares of ​the other ​two ​sides.
   Show answer

   By the altitude-on-hypotenuse similarity, $A{B}^2+B{C}^2=A{C}^2$; the square of the hypotenuse equals the sum of the squares of the other two sides.

   Step-by-step solution →
10. Q390 CBSE 2020
    Prove that the ​ratio of the ​areas ​of two ​similar triangles ​is equal to the ratio of the squares of their ​corresponding sides.
    Show answer

    Proved: $\frac{\text{ar}(△ABC)}{\text{ar}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$.

    Step-by-step solution →
11. Q391 CBSE 2020
    In ​Fig. ​6, DEFG is a ​square in a triangle ​ABC right angled at ​A. Prove that

    1. (i)$△AGF\sim △DBG$
    2. (ii)$△AGF\sim △EFC$

    Show answer

    Both results are proved: (i) $△AGF\sim △DBG$ and (ii) $△AGF\sim △EFC$, each by the AA similarity criterion.

    Step-by-step solution →
12. Q392 CBSE 2020
    In an ​obtuse $△ABC$ ​($\angle B$ ​is obtuse), $AD$ is perpendicular ​to $CB$ produced. Then prove ​that $A{C}^2=A{B}^2+B{C}^2+2BC\times BD$.
    Show answer

    Using Pythagoras in $△ADC$ and $△ADB$ with $DC=BC+BD$ gives $A{C}^2=A{B}^2+B{C}^2+2BC\times BD$; hence proved.

    Step-by-step solution →
13. Q393 CBSE 2019
    Let $△ABC\sim △DEF$ ​and ​their ​areas be respectively, $64$ cm${}^2$ ​and $121$ ​cm${}^2$. If $EF=15.4$ cm, find $BC$.
    Show answer

    $BC=11.2$ cm.

    Step-by-step solution →
14. Q394 CBSE 2019
    The area of ​two similar triangles ​are $25$ ​sq. cm ​and $121$ sq. cm. Find the ​ratio of their ​corresponding ​sides.
    Show answer

    Ratio of corresponding sides $=5:11$.

    Step-by-step solution →
15. Q395 CBSE 2019
    In ​Fig. ​1, $DE\parallel BC$, $AD=1$ ​cm and $BD=2$ cm. ​What is the ratio ​of the $\text{ar}(△ABC)$ ​to the $\text{ar}(△ADE)$?

    Show answer

    $\text{ar}(△ABC):\text{ar}(△ADE)=9:1$

    Step-by-step solution →
16. Q396 CBSE 2019
    $DE$ is ​drawn parallel ​to base $BC$ of $△ABC$, ​meeting $AB$ ​in $D$ ​and $AC$ at $E$. If $\frac{AB}{BD}=4$ and $CE=2$ ​cm, ​find ​the length of $AE$.
    Show answer

    $AE=6$ cm.

    Step-by-step solution →
17. Q397 CBSE 2019
    In ​Figure ​1, $ABC$ ​is an isosceles triangle right ​angled at $C$ ​with $AC=4$ ​cm. Find the length ​of $AB$.

    Show answer

    $AB=4\sqrt{2}\approx 5.66$ cm

    Step-by-step solution →
18. Q398 CBSE 2019
    In ​Figure ​2, $DE\parallel BC$. ​Find ​the length of side $AD$, given that $AE=1.8$ cm, $BD=7.2$ ​cm and $CE=5.4$ ​cm.

    Show answer

    $AD=2.4$ cm.

    Step-by-step solution →
19. Q399 CBSE 2019
    The ​perpendicular ​from $A$ ​on side $BC$ ​of a $△ABC$ meets $BC$ ​at $D$ such ​that $DB=3CD$. Prove ​that $2A{B}^2=2A{C}^2+B{C}^2$.
    Show answer

    Hence $2A{B}^2=2A{C}^2+B{C}^2$, as required.

    Step-by-step solution →
20. Q400 CBSE 2019
    $AD$ and $PM$ ​are medians ​of triangles $ABC$ ​and $PQR$ respectively ​where $△ABC\sim △PQR$. ​Prove ​that $\frac{AB}{PQ}=\frac{AD}{PM}$.
    Show answer

    By proving $△ABD\sim △PQM$ (SAS), corresponding sides give $\frac{AB}{PQ}=\frac{AD}{PM}$. Hence proved.

    Step-by-step solution →