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 18 of 22

1. Q341 CBSE 2023
   If $AD$ ​and $PM$ ​are medians ​of ​triangles $ABC$ and $PQR$ respectively, where $△ABC\sim △PQR$, prove ​that $\frac{AB}{PQ}=\frac{AD}{PM}$.
   Show answer

   Since $△ABD\sim △PQM$, we get $\frac{AB}{PQ}=\frac{AD}{PM}$.

   Step-by-step solution →
2. Q342 CBSE 2023
   Prove that a ​line drawn parallel to ​one side of a triangle to intersect the ​other two sides in distinct ​points, divides the ​two ​sides in the ​same ​ratio.
   Show answer

   Proved: $\frac{AD}{DB}=\frac{AE}{EC}$ (Basic Proportionality Theorem).

   Step-by-step solution →
3. Q343 CBSE 2023
   Sides $AB$ and $BC$ ​and ​median $AD$ of a triangle $ABC$ are respectively ​proportional ​to sides $PQ$ ​and $QR$ and median $PM$ of $△PQR$. Show ​that $△ABC\sim △PQR$.

   Show answer

   Hence $△ABC\sim △PQR$.

   Step-by-step solution →
4. Q344 CBSE 2023
   Through the mid-point $M$ ​of ​the side $CD$ of a parallelogram $ABCD$, the line $BM$ ​is ​drawn intersecting $AC$ in $L$ and $AD$ (produced) ​in $E$. ​Prove that $EL=2BL$.

   Show answer

   Hence $EL=2BL$, as required.

   Step-by-step solution →
5. Q345 CBSE 2023
   In ​the ​given figure, $\frac{QR}{QS}=\frac{QT}{PR}$ and $\angle 1=\angle 2$. Prove ​that $△PQS\sim △TQR$.

   Show answer

   $△PQS\sim △TQR$ by the SAS similarity criterion.

   Step-by-step solution →
6. Q346 CBSE 2023
   Observe ​the figures given below carefully and ​answer the questions. Figure A shows ​two quadrilaterals $ABCD$ ​(with $AB=1.2$ cm, $BC=2.5$ cm, $CD=2.4$ cm, $DA=2$ cm ​and marked angles) and $PQRS$ (with $PQ=2.4$ cm, $QR=5$ cm, $RS=4.8$ ​cm, $SP=4$ cm and the same marked angles). Figure B shows two squares of sides $3.5$ cm and $3$ cm. Figure ​C shows two triangles $ABC$ and $PQR$ with sides $2$ cm, $3$ cm and $4.5$ cm each.

   1. (i)Name the figure(s) wherein two figures are similar.[1]
   2. (ii)Name the figure(s) wherein the figures are congruent.[1]
   3. (iii)(a)Prove that congruent triangles are also similar but not the converse.[2]
   4. (iii)(b)What more is least needed for two similar triangles to be congruent?[2]
   Show answer

   (i) Figures A, B and C all show similar pairs — A with scale factor $2$, B because any two squares are similar, C because congruent triangles are similar with ratio $1$; (ii) Only Figure C shows congruent figures (SSS); (iii)(a) Congruent triangles have equal angles and all side ratios equal to $1$, so they are similar; the converse fails — the squares in B, or equilateral triangles of sides $2$ cm and $4$ cm, are similar but not congruent; (iii)(b) Equality of any one pair of corresponding sides (scale factor $1$) is the least that is needed.

   Step-by-step solution →
7. Q347 CBSE 2020
   Sides ​of two ​similar triangles are ​in the ​ratio $4:9$. Areas of these triangles are in ​the ​ratio

   1. (a)$4:9$
   2. (b)$2:3$
   3. (c)$81:16$
   4. (d)$16:81$
   Show answer

   (d) $16:81$

   Step-by-step solution →
8. Q348 CBSE 2020
   In Figure-1, $ABC$ is ​an ​isosceles ​triangle, ​right-angled ​at $C$. Therefore

   1. (a)$A{B}^2=2A{C}^2$
   2. (b)$B{C}^2=2A{B}^2$
   3. (c)$A{C}^2=2A{B}^2$
   4. (d)$A{B}^2=4A{C}^2$
   Show answer

   (a) $A{B}^2=2A{C}^2$

   Step-by-step solution →
9. Q349 CBSE 2020
   If $△ABC\sim △DEF$ ​such ​that $AB=1.2$ ​cm ​and $DE=1.4$ cm, the ​ratio of the ​areas of $△ABC$ and $△DEF$ ​is

   1. (a)$49:36$
   2. (b)$6:7$
   3. (c)$7:6$
   4. (d)$36:49$
   Show answer

   (d) $36:49$

   Step-by-step solution →
10. Q350 CBSE 2020
    In fig. ​1, $MN\parallel BC$ and $AM:MB=1:2$, then $\frac{\text{ar}(△AMN)}{\text{ar}(△ABC)}=$ ​________ ​.

    Show answer

    $\frac{\mathrm{ar}(△AMN)}{\mathrm{ar}(△ABC)}=\frac{1}{9}$

    Step-by-step solution →
11. Q351 CBSE 2020
    Given $△ABC\sim △PQR$, if $\frac{AB}{PQ}=\frac{1}{3}$, ​then $\frac{ar(△ABC)}{ar(△PQR)}=$ ​______.
    Show answer

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

    Step-by-step solution →
12. Q352 CBSE 2020
    Let $△ABC\sim △DEF$ ​and their ​areas ​be respectively $81{\text{ cm}}^2$ and $144{\text{ cm}}^2$. If $EF=24$ ​cm, ​then length of side $BC$ is ​________ cm.
    Show answer

    $BC=18$ cm

    Step-by-step solution →
13. Q353 CBSE 2020
    $ABC$ ​is ​an equilateral triangle of side $2a$, ​then length ​of ​one ​of its ​altitude is ______.
    Show answer

    Altitude $=\sqrt{3}a$ (i.e. $a\sqrt{3}$)

    Step-by-step solution →
14. Q354 CBSE 2020
    If the corresponding sides of two similar ​triangles are ​in ​the ratio of $3:4$, then the ratio ​of the areas ​of these triangles is ____.
    Show answer

    The ratio of the areas is $9:16$.

    Step-by-step solution →
15. Q355 CBSE 2020
    In $△ABC$, $AB=6\sqrt{3}$ ​cm, $AC=12$ ​cm ​and $BC=6$ ​cm, then $\angle B=$ __________.
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    $\angle B=90^{\circ }$.

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16. Q356 CBSE 2020
    Two ​triangles ​are similar ​if ​their corresponding sides ​are ​__________.
    Show answer

    Proportional (i.e. in the same ratio)

    Step-by-step solution →
17. Q357 CBSE 2020
    The perimeters of two similar triangles are $25$ cm and $15$ cm ​respectively. If one ​side of the ​first triangle is $9$ cm, ​then ​the corresponding ​side of ​second triangle is ​__________.
    Show answer

    $5.4$ cm

    Step-by-step solution →
18. Q358 CBSE 2020
    If two ​triangles ​ABC and DEF ​are similar ​and $\angle A=67^{\circ }$, $\angle E=63^{\circ }$, ​then the measure ​of $\angle C$ is ________ .
    Show answer

    $\angle C=50^{\circ }$

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19. Q359 CBSE 2020
    If ratio ​of the corresponding sides ​of two similar ​triangles ​is $2:3$, ​then ​ratio of ​their perimeters is ________.
    Show answer

    The ratio of perimeters is $2:3$.

    Step-by-step solution →
20. Q360 CBSE 2020
    The corresponding sides of two similar ​triangles are in the ratio $3:4$, then the ​ratios ​of ​the area of triangles ​is ​________.
    Show answer

    The ratio of the areas is $9:16$.

    Step-by-step solution →