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 19 of 22

1. Q361 CBSE 2020
   In ​a $△ABC$, $DE\parallel BC$. ​If $AD=1.75$ ​cm, $AB=7$ ​cm and $AE=2$ cm, ​then ​find $AC$.
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   $AC=8$ cm

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2. Q362 CBSE 2020
   In ​Fig. ​3, in $△ABC$, $DE\parallel BC$ ​such that $AD=2.4$ cm, $AB=3.2$ cm ​and $AC=8$ cm, then ​what is ​the ​length ​of $AE$?

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   $AE=6$ cm.

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3. Q363 CBSE 2020
   In Fig. 1, $D$ and $E$ ​are points ​on sides $AB$ and $AC$ ​respectively ​of a $△ABC$ ​such that $DE\parallel BC$. If $AD=3.6$ cm, $AB=10$ ​cm and $AE=4.5$ cm, find $EC$ and $AC$.

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   $EC=8$ cm and $AC=12.5$ cm.

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4. Q364 CBSE 2020
   A ladder 25 m long just reaches the top ​of a building 24 m high from the ground. What is the ​distance of the foot ​of ladder from ​the ​base ​of ​the building?
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   The foot of the ladder is $7$ m from the base of the building.

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5. Q365 CBSE 2020
   $ABCD$ is a trapezium ​in ​which $AB\parallel DC$ ​and ​its ​diagonals intersect each other ​at $O$. If $AB=3CD$, find the ratio of the areas of triangles $AOB$ ​and $COD$.

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   $\mathrm{ar}(△AOB):\mathrm{ar}(△COD)=9:1$

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6. Q366 CBSE 2020
   In ​the given ​Figure-5, $DE\parallel AC$ ​and $DF\parallel AE$. ​Prove ​that $\frac{BF}{FE}=\frac{BE}{EC}$.

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   Proved: $\frac{BF}{FE}=\frac{BD}{DA}=\frac{BE}{EC}$, hence $\frac{BF}{FE}=\frac{BE}{EC}$.

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7. Q367 CBSE 2020
   If the areas ​of two ​similar triangles ​are ​equal, then prove ​that ​the triangles are congruent.

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   Proved: similar triangles of equal area have all corresponding sides equal, hence are congruent by SSS.

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8. Q368 CBSE 2020
   Prove ​that the sum of the squares of the sides ​of ​a ​rhombus is equal to the sum ​of the ​squares of its diagonals.

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   Proved: $A{B}^2+B{C}^2+C{D}^2+D{A}^2=A{C}^2+B{D}^2$ for rhombus $ABCD$.

   Step-by-step solution →
9. Q369 CBSE 2020
   In Fig. 5, $ABC$ and $DBC$ ​are ​two triangles ​on ​the same ​base $BC$. If $AD$ ​intersects $BC$ at $O$, ​show ​that $\frac{\text{ar}(△ABC)}{\text{ar}(△DBC)}=\frac{AO}{DO}$.

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   $\frac{\text{ar}(△ABC)}{\text{ar}(△DBC)}=\frac{AO}{DO}$, as required.

   Step-by-step solution →
10. Q370 CBSE 2020
    The perimeters of ​two similar triangles are 30 ​cm ​and 20 cm respectively. ​If one ​side of the first triangle is ​9 cm long, find the length of the corresponding side of the second ​triangle.
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    The corresponding side of the second triangle is $6$ cm.

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11. Q371 CBSE 2020
    In ​Fig. ​2, $DE\parallel AC$ ​and $DC\parallel AP$. ​Prove ​that $\frac{BE}{EC}=\frac{BC}{CP}$.

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    Since both ratios equal $\frac{BD}{DA}$, we conclude $\frac{BE}{EC}=\frac{BC}{CP}$. Hence proved.

    Step-by-step solution →
12. Q372 CBSE 2020
    In ​Figure-4, $△ABC$ and $△XYZ$ are shown. If $AB=3.8$ ​cm, $AC=3\sqrt{3}$ cm, $BC=6$ cm, $XY=6\sqrt{3}$ ​cm, $XZ=7.6$ cm, $YZ=12$ ​cm and $\angle A=65^{\circ }$, $\angle B=70^{\circ }$, then find ​the value ​of $\angle Y$.

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    $\angle Y=45^{\circ }$.

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13. Q373 CBSE 2020
    In Figure-5, $△PQR$ is ​right-angled at $P$. $M$ ​is ​a point ​on $QR$ such ​that $PM$ is perpendicular to $QR$. Show ​that $P{Q}^2=QM\times QR$.

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    Since $△QMP\sim △QPR$, $\frac{QM}{QP}=\frac{QP}{QR}$, so $P{Q}^2=QM\times QR$. Hence proved.

    Step-by-step solution →
14. Q374 CBSE 2020
    If the areas ​of ​two similar ​triangles are ​equal, ​show that ​they are ​congruent.
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    Since $\frac{\mathrm{ar}(△ABC)}{\mathrm{ar}(△DEF)}={\left(\frac{AB}{DE}\right)}^2=1$, every pair of corresponding sides is equal, so $△ABC\cong △DEF$ by SSS. Hence the triangles are congruent.

    Step-by-step solution →
15. Q375 CBSE 2020
    In fig. ​6, if $AD\perp BC$, ​then ​prove that $A{B}^2+C{D}^2=B{D}^2+A{C}^2$.

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    $A{B}^2+C{D}^2=B{D}^2+A{C}^2$ is proved.

    Step-by-step solution →
16. Q376 CBSE 2018
    Prove ​that the ratio of the areas ​of ​two similar triangles is equal ​to the square of the ratio of ​their corresponding sides.
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    $\frac{\text{ar}(△ABC)}{\text{ar}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$. Hence proved.

    Step-by-step solution →
17. Q377 CBSE 2020
    In ​Fig. ​5, $\angle D=\angle E$ and $\frac{AD}{DB}=\frac{AE}{EC}$, ​prove ​that $△BAC$ ​is ​an isosceles triangle.

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    Since $\frac{AD}{DB}=\frac{AE}{EC}$ gives $DE\parallel BC$, the corresponding angles make $\angle ABC=\angle ACB$ (using $\angle ADE=\angle AED$); hence $AB=AC$ and $△BAC$ is isosceles. Q.E.D.

    Step-by-step solution →
18. Q378 CBSE 2020
    In ​Fig. ​7, if $△ABC\sim △DEF$ and their sides of ​lengths (in cm) are marked along ​them, ​then ​find the lengths of ​sides of each triangle.

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    $x=5$; sides of $△ABC$: $9$, $12$, $15$ cm and of $△DEF$: $18$, $24$, $30$ cm.

    Step-by-step solution →
19. Q379 CBSE 2020
    In ​a ​right triangle, prove ​that the square of the hypotenuse is equal to the sum of squares ​of the ​other two sides.
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    Hence $A{C}^2=A{B}^2+B{C}^2$: the square of the hypotenuse equals the sum of the squares of the other two sides.

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20. Q380 CBSE 2020
    Prove ​that, in ​a right ​triangle, the square of the ​hypotenuse is equal ​to the ​sum of the squares of ​the ​other two sides.
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    Proved: $A{C}^2=A{B}^2+B{C}^2$ (Pythagoras theorem)

    Step-by-step solution →