What kind of questions are asked in Exercise 8.4?

Most questions ask you to verify identities or simplify expressions using known trigonometric identities. You'll apply algebraic manipulations and substitutions.

How can I be sure my steps are correct when verifying an identity?

Always start from the more complex side (LHS or RHS) and simplify it using identities. Your goal is to transform it into the simpler side. Justify each step with the identity used.

Are these identities useful beyond Class 10?

Absolutely! These identities are foundational for higher classes (11th and 12th) and competitive exams like JEE, NEET, etc.

What if I make a mistake midway in verifying an identity?

If you get stuck or the two sides aren’t matching, retrace your steps and check for: Incorrect substitution Wrong sign (especially minus vs plus) Basic arithmetic/algebra mistakes

Exercise 6.3

NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.3

NCERT Solutions Class 10 Maths Chapter 6 Triangle Exercise 6.3 can help students enhance their skills to solve mathematical problems related to the similarity of triangles. The exercise will introduce you to the various criteria for the similarity of triangles, which is important to prove the similarity of any two triangles. Exercise 6.3 is crucial to thoroughly understand these criteria, which ultimately help solve more advanced problems and topics of geometry. So, let’s get a deeper insight into this important topic of Chapter 6 — Triangles.

1.0Download NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.3 : Free PDF

NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.3

2.0Introduction to Similarity of Triangles:

Before starting off with the criteria of similarity of triangles, let’s take a quick look at the similarity of a triangle itself. In geometry, similar figures are a pair of figures with equal shape but not necessarily equal size. Similarly, in a pair of triangles, if the lengths of corresponding sides are in equal proportion or all the angles of both triangles are equal, then the two triangles are said to be similar. In the exercise, this concept of similarity plays an important part in many geometric proofs and uses, such as when scaling shapes or exploring proportional relationships.

3.0NCERT Class 10 Maths Chapter 6 Triangles Exercise 6.3 Overview: Key Concepts

The exercise includes various questions, solved via the basic concepts and theorems related to the similarity of triangles. Let’s explore these key concepts and criteria of similarity to solve these questions with ease:

Angle-Angle-Angle (AAA) Criterion:

The AAA similarity criterion tells us that if two triangles have equal corresponding angles, then the triangles are similar. In other words, if all the angles of a triangle are equal to the corresponding angles of the second triangle, then the two triangles will be similar. Which ultimately means that the sides of these triangles will also be in equal proportion. To understand this, take a pair of triangles, say ABC and MNR. If these two triangles are similar, then ∠A = ∠M, ∠B = ∠N, and ∠C = ∠R.

Note that when solving these questions, if two angles of a pair of triangles are equal, then the third angle will automatically be equal to its corresponding angle.

Side-Side-Side (SSS) Criterion:

The SSS similarity criterion defines that if the corresponding sides of any two triangles are proportional, then the two triangles are similar. In other terms, if in a pair of triangles, say ABC and PQR:

$\frac{A B}{PQ} = \frac{BC}{QR} = \frac{A C}{PR}$

Then $Δ A BC \sim Δ PQR$ by the SSS criterion. Here " $\sim$ " is the symbol used to denote the similarity of a pair of triangles. This test is important when we have the sides of both triangles, and we need to verify proportionality to prove similarity.

Side-Angle-Side (SAS) Criterion:

In the exercise, another criterion for solving the questions is side-angle-side. In this criterion, when one angle of a triangle is congruent to one angle of another triangle, and the sides covering these angles are in the same ratio. In other words, when two triangles, say ABC and DEF, are similar, then:

$\frac{A B}{D E} = \frac{A C}{D F}$
∠A = ∠D

This is a criterion that we use when we only know a little information, like the length of one side and the angle between them.

To perform well in the exercise, start practising the above-mentioned concepts today from NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.3 and enhance your knowledge of the criteria for the similarity of triangles.

4.0NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.3 : Detailed Solutions

1. The states in which pairs of triangles in the figure 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 :

The states in which pairs of triangles in the figure 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 :

(ii)

(iii)

https://res.cloudinary.com/dpzpn3dkw/image/upload/w_800,f_auto,q_auto/v1743743771/si57ylq1zuddkcli0uul.png?_upload_ref=ic_img_tool&__ar__=1.07

(iv)

(v)

(vi)

Solutions:

(i) Yes. ∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°.

Therefore, △ABC ~ △PQR.

By AAA similarity criterion.

(ii) Yes.

AB/QR = 2/4 = 1/2, BC/RP = 2.5/5 = 1/2, CA/PQ = 3/6 = 1/2.

Therefore, △ABC ~ △QRP.

By SSS similarity criterion.

(iii) No.

MP/DE = 2/4 = 1/2, LP/DF = 3/6 = 1/2, LM/EF = 2.7/5 ≠ 1/2.

i.e., MP/DE = LP/DF ≠ LM/EF.

Thus, the two triangles are not similar.

(iv) Yes,

MN/QP = ML/QR = 1/2

and ∠NML = ∠PQR = 70°.

By SAS similarity criterion,

△NML ~ △PQR.

(v) No,

AB/FD ≠ AC/FE.

Thus, the two triangles are not similar.

(vi) In triangle DEF, ∠D + ∠E + ∠F = 180°.

70° + 80° + ∠F = 180°.

∠F = 30°.

In triangle PQR,

∠P + 80° + 30° = 180°.

∠P = 70°.

∠E = ∠Q = 80°.

∠D = ∠P = 70°.

∠F = ∠R = 30°.

By AAA similarity criterion,

△DEF ~ △PQR.

2. In figure, △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:

From the given figure:

∠DOC + 125° = 180° (Linear pair)

⇒ ∠DOC = 180° - 125° = 55°

In triangle ODC:

∠DCO + ∠CDO + ∠DOC = 180° (Sum of angles in a triangle)

⇒ ∠DCO + 70° + 55° = 180°

⇒ ∠DCO + 125° = 180°

⇒ ∠DCO = 180° - 125° = 55°

Given that ΔODC ~ ΔOBA (Triangle ODC is similar to triangle OBA)

Therefore, corresponding angles are equal:

∠OCD = ∠OAB

⇒ ∠OAB = ∠OCD = ∠DCO = 55°

i.e., ∠OAB = 55°

Hence, we have:

∠DOC = 55°

∠DCO = 55°

∠OAB = 55°

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

Solution:

Given that AB || DC in trapezium ABCD.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

=> ∠1 = ∠3, ∠2 = ∠4 (Alternate interior angles)

Also ∠DOC = ∠BOA (Vertically opposite angles)

=> ΔOCD ~ ΔOAB => OC/OA = OD/OB (Ratios of the corresponding sides of the similar triangle)

=> OA/OC = OB/OD (Taking reciprocals)

4. In figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Solution:

In the given figure, it is given that ∠1 = ∠2.

This implies that PQ = PR (Sides opposite to equal angles of triangle PQR).

We are also given that:

QR/QS = QT/PR

Substituting PQ for PR (since PQ = PR), we get:

QR/QS = QT/PQ

Taking the reciprocals of both sides, we have:

QS/QR = PQ/QT ...(1)

Now, consider triangles PQS and TQR. We have:

∠PQS = ∠TQR (Each equals ∠1)

and

QS/QR = PQ/QT (From equation (1))

Therefore, by the SAS (Side-Angle-Side) similarity criterion, we can conclude that:

ΔPQS ~ ΔTQR.

5. S and T are points on sides PR and QR of triangle PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Solution:

In the given figure, we have triangles RPQ and RTS.

It is given that:

∠RPQ = ∠RTS

∠PRQ = ∠SRT (Each equals ∠R)

S and T are points on sides PR and QR of triangle PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Then by AA similarity criterion, we have ΔRPQ ∼ ΔRTS.

6. In figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.

Solution:

In the given figure,

ΔABE ≅ ΔACD (Given)

⇒ AB = AC and AE = AD (CPCT - Corresponding Parts of Congruent Triangles)

⇒ AB/AC = 1 and AD/AE = 1

⇒ AB/AC = AD/AE (Each = 1)

Now, in ΔADE and ΔABC, we have

AD/AE = AB/AC (proved)

i.e., AD/AB = AE/AC

and also ∠DAE = ∠BAC (Each = ∠A)

⇒ ΔADE ~ ΔABC (By SAS similarity criterion - Side-Angle-Side)

7. In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

(i) Triangle AEP ~ Triangle CDP

(ii) Triangle ABD ~ Triangle CBE

(iii) Triangle AEP ~ Triangle ADB

(iv) Triangle PDC ~ Triangle BEC

Solution:

(i) In Triangle AEP and Triangle CDP,

∠APE = ∠CPD (vertically opposite angles)

∠AEP = ∠CDP = 90°

∴ By AA similarity

Triangle AEP ~ Triangle CDP

(ii) In Triangle ABD and Triangle CBE,

∠ABD = ∠CBE (common)

∠ADB = ∠CEB = 90°

∴ By AA similarity

Triangle ABD ~ Triangle CBE

(iii) In Triangle AEP and Triangle ADB,

∠PAE = ∠DAB (common)

∠AEP = ∠ADB = 90°

∴ By AA similarity

Triangle AEP ~ Triangle ADB

(iv) In Triangle PDC and Triangle BEC,

∠PCD = ∠BCE (common)

∠PDC = ∠BEC = 90°

∴ By AA similarity

Triangle PDC ~ Triangle BEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Triangle ABE ~ Triangle CFB.

Solution:

In triangle ABE and triangle CFB,

∠EAB = ∠BCF (opposite angles of parallelogram)

∠AEB = ∠CBF (Alternate interior angles, as AE || BC)

Therefore, by AA similarity,

triangle ABE ~ triangle CFB.

9. In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) triangle ABC ~ triangle AMP

(ii) CA/PA = BC/MP

Solution:

(i) In triangle ABC and triangle AMP,

angle CAB = angle PAM (common)

angle ABC = angle AMP = 90 degrees

Therefore, by AA similarity,

triangle ABC ~ triangle AMP

(ii) As triangle ABC ~ triangle AMP (proved above),

Therefore, CA/PA = BC/MP

10. 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 triangle ABC and triangle EFG respectively.

If triangle ABC ~ triangle FEG, show that:

(i) CD/GH = AC/FG

(ii) triangle DCB ~ triangle HGE

(iii) triangle DCA ~ triangle HGF

Solution:

10. 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 triangle ABC and triangle EFG respectively. If triangle ABC ~ triangle FEG, show that: (i) CD/GH = AC/FG (ii) triangle DCB ~ triangle HGE (iii) triangle DCA ~ triangle HGF

△ABC ∼ △FEG

⇒ ∠ACB = ∠EGF

=> ∠DBC = ∠HEG

From (1) and (2), we have:

=> ΔDCB ~ ΔHGE

Similarly, we have:

ΔDCA ~ ΔHGF

Now, ΔDCA ~ ΔHGF

=> DC/HG = CA/GF => CD/GH = AC/FG

11. In the given figure, E is a point on side CB produced by an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Solution:

In the given figure, we are given that triangle ABC is isosceles and AB = AC.

Therefore, ∠B = ∠C ...(1)

Now, consider triangles ABD and ECF.

∠ABD = ∠ECF (from equation 1)

∠ADB = ∠EFC (each = 90°)

Therefore, triangle ABD is similar to triangle ECF (by AA similarity).

12. Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. We need to show that triangle ABC is similar to triangle PQR.

Solution:

https://res.cloudinary.com/dpzpn3dkw/image/upload/w_800,f_auto,q_auto/v1743743770/vjhlcniw83bch2vzbmnx.png?_upload_ref=ic_img_tool&__ar__=2.18

As, AB/PQ = BC/QR = AD/PM

So, AB/PQ = BD/QM = AD/PM

{∵ AB/PQ = (1/2)BC / (1/2)QR = BD/QM}

∴ By SSS similarity,

△ABD ∼ △PQM.

As, △ABD ∼ △PQM.

∴ ∠ABD = ∠PQM

Now, In △ABC and △PQR

AB/PQ = BC/QR

∠ABC = ∠PQR (Proved above)

∴ By SAS similarity

△ABC ∼ △PQR.

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

Solution:

For △ABC and △DAC, We have:

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

∠BAC = ∠ADC (Given)

And ∠ACB = ∠DCA (Each = ∠C)

⇒ ΔABC ~ ΔDAC (AA similarity)

⇒ AC/DC = CB/CA

⇒ CA/CD = CB/CA

⇒ CA × CA = CB × CD

⇒ CA² = CB × CD

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 ΔABC ~ ΔPQR.

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 ΔABC ~ ΔPQR.

Solution:

Given: Triangle ABC and triangle PQR. AD and PM are their medians respectively.

AB/PQ = AC/PR = AD/PM

To prove: Triangle ABC ~ triangle PQR.

Construction: Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN, and RN.

Proof: Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M respectively.

=> BE = AC and QN = PR.

=> BE/QN = AC/PR => BE/QN = AB/PQ

i.e., AB/PQ = BE/QN

From (1), AB/PQ = AD/PM = (2AD)/(2PM) = AE/PN

i.e., AB/PQ = AE/PN

From (2) and (3), we have

AB/PQ = BE/QN = AE/PN

=> ΔABE ~ ΔPQN => ∠1 = ∠2

Similarly, we can prove

=> ΔACE ~ ΔPRN => ∠3 = ∠4

Adding (4) and (5), we have

=> ∠1 + ∠3 = ∠2 + ∠4

=> ∠A = ∠P

=> ΔABC ~ ΔPQR (SAS similarity criterion)

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.

Solution:

△ABC ∼ △PQR

∴ AB/PQ = BC/QR

6/x = 4/28

⇒ x = 42 m

16. If AD and PM are medians of triangles ABC and PQR, respectively, where △ABC ∼ △PQR, prove that:

Solution:

△ABC ∼ △PQR (Given)

⇒ AB/PQ = BC/QR = AC/PR; ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

Now, BD = CD = (1/2)BC and QM = RM = (1/2)QR

(D is the midpoint of BC and M is the midpoint of QR)

From (1), AB/PQ = BC/QR => AB/PQ = 2BD/2QM (By (2))

=> AB/PQ = BD/QM

Thus, we have AB/PQ = BD/QM and ∠ABD = ∠PQM (∠B = ∠Q)

=> ΔABD ~ ΔPQM (By SAS similarity criterion)

=> AB/PQ = AD/PM

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.1

Similar triangles form the basis of concepts in coordinate geometry, mensuration, and trigonometry.
Enhances logical thinking and improves the ability to build structured geometric proofs.
NCERT solutions offer clear, detailed explanations that help students understand the why and how behind each step.
Students practice drawing triangles and marking corresponding parts, which improves their visual learning and accuracy.
Similar triangles are applied in real-life areas like map reading, construction, architecture, and scale drawing

NCERT Class 10 Maths Ch 6 Triangles Other Exercises:

Exercise 6.3

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

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