Triangles

24 questions · step-by-step solutions

Get free step-by-step NCERT solutions for Class 10 Maths Triangles (Chapter 6). All 24 questions across 4 exercises are solved with clear reasoning, following the CBSE 2026–27 syllabus. Work through each solution to understand the method, not just the final answer.

Exercise 6.1

Question 1

Fill in the blanks using the correct word given in brackets :

(i) All circles are __________ . (congruent, similar)

(ii) All squares are __________ . (similar, congruent)

(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________ . (equal, proportional)

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Question 2

Give two different examples of pair of

(i) similar figures. (ii) non-similar figures.

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Question 3

State whether the following quadrilaterals are similar or not:

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Exercise 6.2

Question 1

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

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Question 2

E and F are points on the sides PQ and PR respectively of a Δ\Delta PQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

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Question 3

In Fig. 6.18, if LM || CB and LN || CD, prove that

AMAB=ANAD\frac{\text{AM}}{\text{AB}} = \frac{\text{AN}}{\text{AD}}.

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Question 4

In Fig. 6.19, DE || AC and DF || AE. Prove that

BFFE=BEEC\frac{\text{BF}}{\text{FE}} = \frac{\text{BE}}{\text{EC}}.

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Exercise 6.3

Question 1
  1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
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Question 2
  1. In Fig. 6.35, ΔODCΔOBA\Delta \text{ODC} \sim \Delta \text{OBA}, BOC=125\angle \text{BOC} = 125^\circ and CDO=70\angle \text{CDO} = 70^\circ. Find DOC\angle \text{DOC}, DCO\angle \text{DCO} and OAB\angle \text{OAB}.
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Question 3
  1. Diagonals AC and BD of a trapezium ABCD with ABDC\text{AB} \parallel \text{DC} intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD\frac{\text{OA}}{\text{OC}} = \frac{\text{OB}}{\text{OD}}.
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Question 4
  1. In Fig. 6.36, QRQS=QTPR\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}} and 1=2\angle 1 = \angle 2. Show that ΔPQSΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
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Question 5
  1. S and T are points on sides PR and QR of ΔPQR\Delta \text{PQR} such that P=RTS\angle \text{P} = \angle \text{RTS}. Show that ΔRPQΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.
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Question 6
  1. In Fig. 6.37, if ΔABEΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADEΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
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Question 7
  1. In Fig. 6.38, altitudes AD and CE of ΔABC\Delta \text{ABC} intersect each other at the point P. Show that:

(i) ΔAEPΔCDP\Delta \text{AEP} \sim \Delta \text{CDP} (ii) ΔABDΔCBE\Delta \text{ABD} \sim \Delta \text{CBE} (iii) ΔAEPΔADB\Delta \text{AEP} \sim \Delta \text{ADB} (iv) ΔPDCΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

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Question 8
  1. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABEΔCFB\Delta \text{ABE} \sim \Delta \text{CFB}.
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Question 9
  1. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) ΔABCΔAMP\Delta \text{ABC} \sim \Delta \text{AMP} (ii) CAPA=BCMP\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}

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Question 10
  1. CD and GH are respectively the bisectors of ACB\angle \text{ACB} and EGF\angle \text{EGF} such that D and H lie on sides AB and FE of ΔABC\Delta \text{ABC} and ΔEFG\Delta \text{EFG} respectively. If ΔABCΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\frac{\text{CD}}{\text{GH}} = \frac{\text{AC}}{\text{FG}} (ii) ΔDCBΔHGE\Delta \text{DCB} \sim \Delta \text{HGE} (iii) ΔDCAΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

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Question 11

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD \perp BC and EF \perp AC, prove that Δ\Delta ABD \sim Δ\Delta ECF.

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Question 12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ\Delta PQR (see Fig. 6.41). Show that Δ\Delta ABC \sim Δ\Delta PQR.

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Question 13

D is a point on the side BC of a triangle ABC such that \angle ADC = \angle BAC. Show that CA2=CB.CD\text{CA}^2 = \text{CB.CD}.

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Question 14

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ\Delta ABC \sim Δ\Delta PQR.

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Question 15

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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Question 16

If AD and PM are medians of triangles ABC and PQR, respectively where Δ\Delta ABC \sim Δ\Delta PQR, prove that

ABPQ=ADPM\frac{\text{AB}}{\text{PQ}} = \frac{\text{AD}}{\text{PM}}

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Activity

Question 4

Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).

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Frequently asked questions

Common questions about Class 10 Maths Triangles solutions.

How many questions are there in Class 10 Maths Triangles?

Triangles (Chapter 6) in Class 10 Maths has 24 questions across 4 exercises. Every question is solved step by step on this page.

Are these Triangles solutions based on the latest NCERT syllabus?

Yes. These solutions follow the current CBSE 2026–27 syllabus and the latest NCERT textbook for Class 10 Maths. If the exercises change, the solutions here are updated to match.

How should I use these Triangles solutions?

Try each question yourself first, then read the step-by-step solution to see where your approach diverged. Focus on understanding the method behind each step, not just the final answer.

Triangles Class 10 NCERT Solutions