Class 10 Maths · Chapter 6

Triangles

136 Class 10 Maths questions on Triangles from the NCERT textbook, NCERT Exemplar and CBSE board papers 2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 5 of 7

1. Q81 Exemplar · Exercise 6.4
   In ​a quadrilateral ABCD, $\angle A+\angle D=90^{\circ }$. ​Prove that $A{C}^2+B{D}^2=A{D}^2+B{C}^2$. [Hint: ​Produce ​AB ​and DC to ​meet at ​E.]
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   Proved: producing AB and DC to meet at E gives $\angle E=90^{\circ }$; Pythagoras in the four right triangles shows both $A{C}^2+B{D}^2$ and $A{D}^2+B{C}^2$ equal $A{E}^2+B{E}^2+E{C}^2+E{D}^2$.

   Step-by-step solution →
2. Q82 Exemplar · Exercise 6.4
   In ​Fig. ​6.20, $l\parallel m$ and line ​segments $AB$, $CD$ ​and $EF$ are concurrent at point $P$. ​Prove ​that $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$.

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   Proved: $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$, each equal to $\frac{AP}{BP}$.

   Step-by-step solution →
3. Q83 Exemplar · Exercise 6.4
   In Fig. ​6.21, ​PA, ​QB, RC and SD are all perpendiculars ​to ​a line $l$, $AB=6$ ​cm, $BC=9$ cm, $CD=12$ cm and $SP=36$ cm. Find PQ, QR ​and RS.

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   $PQ=8$ cm, $QR=12$ cm, $RS=16$ cm.

   Step-by-step solution →
4. Q84 Exemplar · Exercise 6.4
   O ​is the point of ​intersection of the diagonals AC and BD of a trapezium ABCD ​with $AB\parallel DC$. Through O, a ​line segment PQ is drawn ​parallel to AB meeting AD in P and BC in ​Q. Prove that $PO=QO$.
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   Proved: $\frac{PO}{DC}=\frac{AO}{AC}=\frac{BO}{BD}=\frac{OQ}{DC}$, therefore $PO=QO$.

   Step-by-step solution →
5. Q85 Exemplar · Exercise 6.4
   In Fig. 6.22, line ​segment DF intersects the side AC of a ​triangle ​ABC ​at the point E such that E ​is the mid-point of CA ​and $\angle AEF=\angle AFE$. Prove that $\frac{BD}{CD}=\frac{BF}{CE}$. [Hint: Take point G on AB such ​that CG $\parallel$ DF.]

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   Proved: with $CG\parallel DF$, BPT gives $\frac{BD}{DC}=\frac{BF}{FG}$, and since $FG=AF=AE=EC$, we get $\frac{BD}{CD}=\frac{BF}{CE}$.

   Step-by-step solution →
6. Q86 Exemplar · Exercise 6.4
   Prove ​that the area of the semicircle drawn on the hypotenuse of a ​right angled ​triangle is equal ​to the sum of the areas of the ​semicircles drawn on the other ​two sides of ​the triangle.
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   Proved: $A_a+A_b=\frac{\pi }{8}(a^2+b^2)=\frac{\pi }{8}{c}^2=A_c$.

   Step-by-step solution →
7. Q87 Exemplar · Exercise 6.4
   Prove that the area of the ​equilateral triangle drawn on the hypotenuse of ​a ​right ​angled triangle ​is equal to the sum of the areas of the equilateral triangles drawn on the other ​two sides ​of the triangle.
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   $A_c=A_a+A_b$: the equilateral triangle on the hypotenuse equals the sum of those on the legs.

   Step-by-step solution →
8. Q88 CBSE 2026
   It is given ​that $△ABC\sim △EDF$. ​Which of ​the following is ​not true ​?

   1. (a)$\frac{\text{Perimeter of }△ABC}{\text{Perimeter of }△EDF}=\frac{AB}{ED}$
   2. (b)$\frac{AB}{ED}=\frac{AC}{EF}$
   3. (c)$\angle A=\angle D,\angle C=\angle F$
   4. (d)$\frac{AB+BC}{AC}=\frac{DE+DF}{EF}$
   Show answer

   (c) $\angle A=\angle D,\angle C=\angle F$

   Step-by-step solution →
9. Q89 CBSE 2026
   In $△DEF$, $AB\parallel EF$. The ​value ​of $x$ is ​:

   1. (a)$0,2$
   2. (b)$2$ only
   3. (c)$-2$
   4. (d)$1$
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   (b) $2$ only

   Step-by-step solution →
10. Q90 CBSE 2026
    Which of ​the following statements ​is ​not ​always ​true?

    1. (a)Two circles are similar.
    2. (b)Two isosceles right triangles are similar.
    3. (c)Two rectangles are similar.
    4. (d)Two equilateral triangles are similar.
    Show answer

    (c) Two rectangles are similar.

    Step-by-step solution →
11. Q91 CBSE 2026
    Devansh proved that $△ABC\sim △PQR$ using SAS ​similarity criteria. ​If ​he ​found $\angle C=\angle R$, ​then which ​of the following was proved true?

    1. (a)$\frac{AC}{AB}=\frac{PR}{PQ}$
    2. (b)$\frac{BC}{AC}=\frac{PR}{QR}$
    3. (c)$\frac{AC}{BC}=\frac{PR}{PQ}$
    4. (d)$\frac{AC}{BC}=\frac{PR}{QR}$
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    (d) $\frac{AC}{BC}=\frac{PR}{QR}$

    Step-by-step solution →
12. Q92 CBSE 2026
    In ​the ​given figure, $DE\parallel BC$. ​If $\frac{AD}{DB}=\frac{1}{3}$ ​and $AC=6$ cm, ​then length $AE$ ​is

    1. (a)$1.5$ cm
    2. (b)$1$ cm
    3. (c)$2$ cm
    4. (d)$3$ cm
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    (a) $1.5$ cm

    Step-by-step solution →
13. Q93 CBSE 2026
    $ABCD$ ​is a trapezium ​in which $AB\parallel DC$ and $E,F$ are points on $AD$ and $BC$ respectively ​such ​that $EF\parallel DC$. ​If $ED=36$ cm, $BF=70$ cm ​and $FC=30$ cm, then ​the length of $AD$ is :

    1. (a)$124$ cm
    2. (b)$120$ cm
    3. (c)$110$ cm
    4. (d)$114$ cm
    Show answer

    (b) $120$ cm

    Step-by-step solution →
14. Q94 CBSE 2026
    If $△ABC$ and $△DEF$ ​are ​similar such that $2AB=DE$ ​and $BC=8$ cm, ​then $EF$ is equal to ​:

    1. (a)$4$ cm
    2. (b)$8$ cm
    3. (c)$12$ cm
    4. (d)$16$ cm
    Show answer

    (d) $16$ cm

    Step-by-step solution →
15. Q95 CBSE 2026
    In ​the ​given figure, if $ME\parallel SU$, then ​which of ​the ​following ​statements ​is correct ​?

    1. (a)$△MOE\sim △SOU$
    2. (b)$△MOE\sim △SUO$
    3. (c)$△OEM\sim △USO$
    4. (d)$△OEM\sim △OSU$
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    (d) $△OEM\sim △OSU$

    Step-by-step solution →
16. Q96 CBSE 2026
    Observe the ​given ​triangles : Which of the ​following ​statements is ​true ​?

    1. (a)$△LMN\sim △ABC$
    2. (b)$△LMN\sim △ACB$
    3. (c)$△NML\sim △ABC$
    4. (d)$△NML\sim △ACB$
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    (c) $△NML\sim △ABC$

    Step-by-step solution →
17. Q97 CBSE 2026
    In ​triangles $ABC$ ​and $PQR$, $\angle A=\angle Q$ ​and $\angle B=\angle R$, ​then $AB:AC$ is ​equal ​to ​:

    1. (a)$PQ:PR$
    2. (b)$PQ:QR$
    3. (c)$QR:QP$
    4. (d)$PR:QR$
    Show answer

    (c) $QR:QP$

    Step-by-step solution →
18. Q98 CBSE 2026
    If $△DEF\sim △PQR$ such ​that $3DE=PQ$ ​and $EF=6$ ​cm, ​then ​the length of $QR$ ​is :

    1. (a)$12\text{cm}$
    2. (b)$3\text{cm}$
    3. (c)$2\text{cm}$
    4. (d)$18\text{cm}$
    Show answer

    (d) $18\text{cm}$

    Step-by-step solution →
19. Q99 CBSE 2026
    If $△ABC\sim △ADE$ ​in ​the adjoining ​figure, then which of the following ​is ​true ?

    1. (a)$\frac{AB}{BE}=\frac{AC}{CD}$
    2. (b)$\frac{AB}{AD}=\frac{AC}{AE}$
    3. (c)$\frac{AB}{BC}=\frac{AE}{DE}$
    4. (d)$\frac{AC}{AD}=\frac{AB}{AE}$
    Show answer

    (b) $\frac{AB}{AD}=\frac{AC}{AE}$

    Step-by-step solution →
20. Q100 CBSE 2026
    In ​the ​given ​figure, ​if $△ACP\sim △DCP$, then :

    1. (a)$\angle DCP=60^{\circ }$
    2. (b)$\angle DCP=30^{\circ }$
    3. (c)$\angle DCP=90^{\circ }$
    4. (d)$\angle DPC=30^{\circ }$
    Show answer

    (b) $\angle DCP=30^{\circ }$

    Step-by-step solution →