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)

View Solution
Question 2

Give two different examples of pair of

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

View Solution
Question 3

State whether the following quadrilaterals are similar or not:

View Solution

Exercise 6.2

Question 1

In Fig. 6.17, (i) and (ii), DE∥BCDE \parallel BC. Find ECEC in (i) and ADAD in (ii).

View Solution
Question 2

EE and FF are points on the sides PQPQ and PRPR respectively of a ΔPQR\Delta PQR. For each of the following cases, state whether EF∥QREF \parallel QR :

(i) PE=3.9 cmPE = 3.9\text{ cm}, EQ=3 cmEQ = 3\text{ cm}, PF=3.6 cmPF = 3.6\text{ cm}, and FR=2.4 cmFR = 2.4\text{ cm}

(ii) PE=4 cmPE = 4\text{ cm}, QE=4.5 cmQE = 4.5\text{ cm}, PF=8 cmPF = 8\text{ cm}, and RF=9 cmRF = 9\text{ cm}

(iii) PQ=1.28 cmPQ = 1.28\text{ cm}, PR=2.56 cmPR = 2.56\text{ cm}, PE=0.18 cmPE = 0.18\text{ cm}, and PF=0.36 cmPF = 0.36\text{ cm}

View Solution
Question 3

In Fig. 6.18, if LM∥CB\text{LM} \parallel \text{CB} and LN∥CD\text{LN} \parallel \text{CD}, prove that

AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

View Solution
Question 4

In Fig. 6.19, DE∥ACDE \parallel AC and DF∥AEDF \parallel AE. Prove that

BFFE=BEEC\dfrac{\text{BF}}{\text{FE}} = \dfrac{\text{BE}}{\text{EC}}

View Solution

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 :
View Solution
Question 2
  1. In Fig. 6.35, ΔODC∼ΔOBA\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}.
View Solution
Question 3
  1. Diagonals AC\text{AC} and BD\text{BD} of a trapezium ABCD\text{ABCD} with AB∥DC\text{AB} \parallel \text{DC} intersect each other at the point O\text{O}. Using a similarity criterion for two triangles, show that OAOC=OBOD\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}.
View Solution
Question 4
  1. In Fig. 6.36, QRQS=QTPR\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}} and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
View Solution
Question 5

S and T are points on sides PRPR and QRQR of ΔPQR\Delta \text{PQR} such that ∠P=∠RTS\angle \text{P} = \angle \text{RTS}. Show that ΔRPQ∼ΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.

View Solution
Question 6
  1. In Fig. 6.37, if ΔABE≅ΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADE∼ΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
View Solution
Question 7
  1. In Fig. 6.38, altitudes ADAD and CECE of ΔABC\Delta \text{ABC} intersect each other at the point PP. Show that:

(i) ΔAEP∼ΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

(ii) ΔABD∼ΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

(iii) ΔAEP∼ΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

(iv) ΔPDC∼ΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

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

(i) ΔABC∼ΔAMP\Delta \text{ABC} \sim \Delta \text{AMP}

(ii) CAPA=BCMP\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}

View Solution
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 ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

(ii) ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

View Solution
Question 11

In Fig. 6.40, EE is a point on side CBCB produced of an isosceles triangle ABCABC with AB=ACAB = AC. If AD⊥BCAD \perp BC and EF⊥ACEF \perp AC, prove that ΔABD∼ΔECF\Delta ABD \sim \Delta ECF.

View Solution
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 ΔPQR\Delta PQR (see Fig. 6.41). Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

View Solution
Question 13

D is a point on the side BCBC of a triangle ABCABC such that ∠ADC=∠BAC\angle \text{ADC} = \angle \text{BAC}. Show that CA2=CB⋅CD\text{CA}^2 = \text{CB} \cdot \text{CD}.

View Solution
Question 14

Sides ABAB and ACAC and median ADAD of a triangle ABCABC are respectively proportional to sides PQPQ and PRPR and median PMPM of another triangle PQRPQR. Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

View Solution
Question 15

A vertical pole of length 6 m6\text{ m} casts a shadow 4 m4\text{ m} long on the ground and at the same time a tower casts a shadow 28 m28\text{ m} long. Find the height of the tower.

View Solution
Question 16

If ADAD and PMPM are medians of triangles ABCABC and PQRPQR, respectively where ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR, prove that

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

View Solution

Activity

Question 4

Draw two line segments BCBC and EFEF of two different lengths, say 3 cm3\text{ cm} and 5 cm5\text{ cm} respectively. Then, at the points BB and CC respectively, construct angles ∠PBC\angle PBC and ∠QCB\angle QCB of some measures, say, 60∘60^\circ and 40∘40^\circ. Also, at the points EE and FF, construct angles ∠REF\angle REF and ∠SFE\angle SFE of 60∘60^\circ and 40∘40^\circ respectively (see Fig. 6.23).

View Solution

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