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 15 of 22

1. Q281 CBSE 2024
   CD and ​GH ​are respectively the bisectors ​of $\angle ACB$ ​and $\angle EGF$ ​such that D and H lie on sides AB and FE of $△ABC$ ​and $△FEG$ respectively. If $△ABC\sim △FEG$, ​show that :

   1. (i)$\frac{CD}{GH}=\frac{AC}{FG}$
   2. (ii)$△DCB\sim △HGE$

   1. (i)$\frac{CD}{GH}=\frac{AC}{FG}$[2]
   2. (ii)$△DCB\sim △HGE$[1]
   Show answer

   (i) Since $△ACD\sim △FGH$ (AA), $\frac{CD}{GH}=\frac{AC}{FG}$; (ii) $△DCB\sim △HGE$ by AA, as $\angle DCB=\angle HGE$ and $\angle B=\angle E$. Hence proved.

   Step-by-step solution →
2. Q282 CBSE 2024
   In ​the ​given ​figure $PA$, $QB$ and $RC$ ​are ​each ​perpendicular to $AC$. ​If $AP=x$, $BQ=y$ and $CR=z$, then prove ​that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

   Show answer

   From the two similarities, $\frac{y}{x}+\frac{y}{z}=\frac{AB+BC}{AC}=1$, so $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$. Hence proved.

   Step-by-step solution →
3. Q283 CBSE 2024
   If a line is drawn parallel to one side of a triangle intersecting the ​other ​sides in distinct points, prove that ​it divides ​the other sides ​in the ​same ratio. Use the above result to prove the following: In the ​given figure, if $EG\parallel CB$ and $FG\parallel DB$, ​then prove that $\frac{AE}{EC}=\frac{AF}{FD}$.

   Show answer

   If $ST\parallel QR$ in $△PQR$, comparing areas gives $\frac{PS}{SQ}=\frac{PT}{TR}$, which proves the theorem. Applying it in $△ABC$ ($EG\parallel CB$) and in $△ABD$ ($FG\parallel DB$) gives $\frac{AE}{EC}=\frac{AG}{GB}=\frac{AF}{FD}$. Hence $\frac{AE}{EC}=\frac{AF}{FD}$.

   Step-by-step solution →
4. Q284 CBSE 2024
   In the figure given below, ​a ​folding table is shown. The legs of the table are represented by line segments ​AB and CD intersecting at O. Join AC and BD. Considering table top is parallel to the ground, and $OB=x$, $OD=x+3$, $OC=3x+19$ and $OA=3x+4$, ​answer the following questions :

   1. (i)Prove that $△OAC$ is similar to $△OBD$.
   2. (ii)​Prove that $\frac{OA}{AC}=\frac{OB}{BD}$. (iii)(a) Observe the figure and find the value of $x$. Hence, find the ​length of ​OC. OR ​(iii)(b) Observe the figure and find $\frac{BD}{AC}$.

   1. (i)Prove that $△OAC$ is similar to $△OBD$.[1]
   2. (ii)Prove that $\frac{OA}{AC}=\frac{OB}{BD}$.[1]
   3. (iii)(a)Find the value of $x$. Hence, find the length of OC.[2]
   4. (iii)(b)Find $\frac{BD}{AC}$.[2]
   Show answer

   (i) $△OAC\sim △OBD$ by the AA criterion (alternate angles $\angle OAC=\angle OBD$ and vertically opposite angles $\angle AOC=\angle BOD$). (ii) Hence $\frac{OA}{AC}=\frac{OB}{BD}$. (iii)(a) $x=2$, so $OC=25$. (iii)(b) $\frac{BD}{AC}=\frac{1}{5}$.

   Step-by-step solution →
5. Q285 CBSE 2024
   Sukesh is having a ​garden ​at the back of his house with trees and flower plants. One ​day, due to heavy rain and storm, one of the trees broke such that the height of the ​unbroken part is 15 m and ​the broken part of the ​tree bends and touches the ground at 20 m away from ​the base ​of the tree. Based on the above information, answer the following questions.

   1. (i)Find the length of the broken part of the tree.[1]
   2. (ii)What was the height of the full tree?[1]
   3. (iii)(a)Find the perimeter of the triangle formed.[2]
   4. (iii)(b)OR Find the area of the triangle formed.[2]
   Show answer

   (i) The broken part is $25$ m long. (ii) The full tree was $40$ m tall. (iii)(a) The perimeter of the triangle is $60$ m. (iii)(b) The area of the triangle is $150{\text{m}}^2$.

   Step-by-step solution →
6. Q286 CBSE 2023
   If $△ABC\sim △PQR$ ​with $\angle A=32^{\circ }$ and $\angle R=65^{\circ }$, then the measure ​of $\angle B$ ​is:

   1. (a)$32^{\circ }$
   2. (b)$65^{\circ }$
   3. (c)$83^{\circ }$
   4. (d)$97^{\circ }$
   Show answer

   (c) $83^{\circ }$

   Step-by-step solution →
7. Q287 CBSE 2023
   In ​the ​given ​figure, $DE\parallel BC$. ​The value of $x$ is ​:

   1. (a)6
   2. (b)12.5
   3. (c)8
   4. (d)10
   Show answer

   (d) 10

   Step-by-step solution →
8. Q288 CBSE 2023
   The ​perimeters of two similar triangles are $42$ cm and $35$ cm respectively. If ​one ​side of the first triangle is $12$ ​cm, then the corresponding side ​of the second triangle ​is:

   1. (a)$5$ cm
   2. (b)$7.5$ cm
   3. (c)$8$ cm
   4. (d)$10$ cm
   Show answer

   (d) $10$ cm

   Step-by-step solution →
9. Q289 CBSE 2023
   The sides of two similar ​triangles are in ​the ratio $4:7$. ​The ​ratio of ​their ​perimeters is

   1. (a)$4:7$
   2. (b)$12:21$
   3. (c)$16:49$
   4. (d)$7:4$
   Show answer

   (a) $4:7$

   Step-by-step solution →
10. Q290 CBSE 2023
    If $△ABC\sim △DEF$ and $\angle A=47^{\circ }$, $\angle E=83^{\circ }$, ​then $\angle C$ is ​equal ​to :

    1. (a)$47^{\circ }$
    2. (b)$50^{\circ }$
    3. (c)$83^{\circ }$
    4. (d)$130^{\circ }$
    Show answer

    (b) $50^{\circ }$

    Step-by-step solution →
11. Q291 CBSE 2023
    If $△PQR\sim △ABC$; $PQ=6$ ​cm, $AB=8$ ​cm ​and the perimeter ​of $△ABC$ is $36$ ​cm, then the ​perimeter ​of $△PQR$ is

    1. (a)20.25 cm
    2. (b)27 cm
    3. (c)48 cm
    4. (d)64 cm
    Show answer

    (b) 27 cm

    Step-by-step solution →
12. Q292 CBSE 2023
    In ​the ​given ​figure, $AB\parallel CD$. ​If $AB=5$ ​cm, $CD=2$ cm and $OB=3$ cm, ​then the length of $OC$ is

    1. (a)$\frac{15}{2}$ cm
    2. (b)$\frac{10}{3}$ cm
    3. (c)$\frac{6}{5}$ cm
    4. (d)$\frac{3}{5}$ cm
    Show answer

    (c) $\frac{6}{5}$ cm

    Step-by-step solution →
13. Q293 CBSE 2023
    The perimeters of ​two ​similar triangles are 28 cm ​and 35 ​cm respectively. If one side ​of ​the first triangle is 8 cm, then the ​corresponding side of the second triangle is :

    1. (a)10 cm
    2. (b)8 cm
    3. (c)16 cm
    4. (d)5 cm
    Show answer

    (a) 10 cm

    Step-by-step solution →
14. Q294 CBSE 2023
    If ​in two ​triangles $ABC$ and $PQR$, $\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}$, ​then ​:

    1. (a)$△PQR\sim △CAB$
    2. (b)$△PQR\sim △ABC$
    3. (c)$△CBA\sim △PQR$
    4. (d)$△BCA\sim △PQR$
    Show answer

    (a) $△PQR\sim △CAB$

    Step-by-step solution →
15. Q295 CBSE 2023
    In ​two triangles PQR ​and ​LMN, ​if $\frac{PQ}{MN}=\frac{QR}{LN}=\frac{PR}{LM}$, ​then ​:

    1. (a)$△LMN\sim △RPQ$
    2. (b)$△LMN\sim △PQR$
    3. (c)$△RQP\sim △LMN$
    4. (d)$△QRP\sim △LMN$
    Show answer

    (a) $△LMN\sim △RPQ$

    Step-by-step solution →
16. Q296 CBSE 2023
    $E$ ​and $F$ ​are ​points on the sides $PQ$ and $PR$ ​respectively of ​a $△PQR$. If $EF\parallel QR$ and $PE=4$ cm, $QE=3$ cm and $EF=4$ cm, then ​the ​length ​of $QR$ is:

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

    (c) $7$ cm

    Step-by-step solution →
17. Q297 CBSE 2023
    $△ABC\sim △DEF$ and their ​perimeters ​are 32 cm ​and ​24 cm ​respectively. If $AB=10$ ​cm, ​then $DE$ equals ​:

    1. (a)8 cm
    2. (b)$7.5$ cm
    3. (c)15 cm
    4. (d)$5\sqrt{3}$ cm
    Show answer

    (b) $7.5$ cm

    Step-by-step solution →
18. Q298 CBSE 2023
    If in ​triangles ABC and ​PQR, $\angle B=\angle Q$, $\angle R=\angle C$ and $AB=2PQ$, then ​the ​two ​triangles are ​:

    1. (a)congruent but not similar
    2. (b)similar but not congruent
    3. (c)neither congruent nor similar
    4. (d)congruent as well as similar
    Show answer

    (b) similar but not congruent

    Step-by-step solution →
19. Q299 CBSE 2023
    The ​line segments ​joining the mid-points of the adjacent sides ​of a ​quadrilateral form ​a :

    1. (a)rectangle
    2. (b)square
    3. (c)rhombus
    4. (d)parallelogram
    Show answer

    (d) parallelogram

    Step-by-step solution →
20. Q300 CBSE 2023
    In the ​given figure, $DE\parallel BC$ ​and ​all ​measurements are given ​in centimetres. The length ​of AE is ​:

    1. (a)2 cm
    2. (b)2.25 cm
    3. (c)2.5 cm
    4. (d)2.75 cm
    Show answer

    (b) 2.25 cm

    Step-by-step solution →