NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.3

Class 10 Maths Chapter 6 Triangles Ex 6.3:- In class 10 Maths Chapter 6 Exercise 6.3, the focus is on the principles of similarity in triangles. This exercise aims to explore and apply the criteria that determine the similarity between triangles. Similar triangles are characterised by having identical shapes, which implies that their corresponding angles are congruent and their corresponding sides are proportional.

The Class 10 Chapter 6 Maths exercise 6.3 examines key similarity criteria, including the AA (Angle-Angle) Criterion, the SSS (Side-Side-Side) Criterion, and the SAS (Side-Angle-Side) Criterion. By solving the problems in this exercise, students will enhance their ability to identify and prove triangle similarity, thereby reinforcing their understanding of geometric relationships and properties. Check out the NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.3 from the below article.

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 requires a deep understanding of the concepts of Applications of Derivatives. These detailed solutions for chapter 6 class 10 maths exercise 6.3 to help you understand the key principles and solve problems effectively. Explore the complete NCERT Solutions for Maths Class 10 Chapter 6 Exercise 6.3 below, where each answer is explained in alignment with CBSE guidelines.

Before starting the Class 10 Maths Chapter 6 Exercise 6.3, it is highly recommended to first go through the chapter and examples. This will help you develop a solid foundation and enable you to apply the concepts with greater clarity and ease.

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NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.3

Check out the Class 10 Maths Chapter 6 Triangles Ex 6.3 NCERT Solutions from the below table:-

1. State 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:

Answer:

(i) In  ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In  ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴  ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° - 70° - 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° - 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

2. In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer:

DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° − 125°

= 55°

In ΔDOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ∼ ΔOBA.

∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠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 AO/OC = OB/OD

Answer:

Given: ABCD is a trapezium in which AB DC.

In ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]

 ∴ .....  [Corresponding sides are proportional]

Hence,

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

Answer:

We have,

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR .....(i)

Given,

Using (i), we get,

In ΔPQS & In ΔTQR

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

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

Answer:

In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

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

Answer:

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] ...(i)
And, AD = AE [By cpct] ...(ii)
In ΔADE and ΔABC,
AD/AB = AE/AC [Dividing equation (ii) by (i)]
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]

7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Answer:

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

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

Answer:

In ΔABE and ΔCFB,we have,

∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴By AA-criterion of similarity, we have

∴ ΔABE ~ ΔCFB (By AA similarity criterion)

9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
 

Answer:

(i) In Δ ABC and AMP, we have,

∠ABC = ∠AMP =900 [Given]

∠BAC = ∠MAP [Common angles]

∴   Δ ABC ~ Δ AMP  [By AA-criterion of similarity, we have]

⇒ CA/PA = BC/MP ..... (Corresponding sides of similar trianlges are proportional)

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer:

We have, Δ ABC ~ Δ FEG

∴  ∠A = ∠F, ∠B = ∠E, & ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle Bisector)

And ∠DCB = ∠HGE (Angle Bisector)

In ΔACD & ΔFGH

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴  Δ ACD ~ Δ FGH [By AA similarity criterion]

⇒ [(CD)/ (GH)] = [(AC) / (FG)]

In ΔDCB & ΔHGE,

∠DCB= ∠HGE (Proved above)

∠B= ∠E (Proved above)

∴  Δ DCB~ Δ HGE [By AA similarity criterion]

In ΔDCA & ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴  Δ DCA~ Δ HGF [By AA similarity criterion]

11. In the following 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.

Answer:

Here Δ ABC is isosceles with AB = AC

∠B = ∠C

In Δ ABD and ECF, we have

∠ABD = ∠ECF[Each 90°]

∠ABD = ∠ECF = [Proved above] 

By AA-criterion of similarity, we have

Δ ABD ~ Δ ECF

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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 (see figure). Show that ΔABC ~ ΔPQR.

Answer:

Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

Median divides the opposite side.

∴   BD = BC / 2 [Given]

And QM = QR / 2 [Given]

Given that,

AB/PQ = BC/QR = AD/PM

⇒  AB/PQ =[( ½BC) / (½QR) ]= AD/PM

 ⇒   AB/PQ = BD/QM = AD/PM

In ΔABD and ΔPQM,

  AB/PQ = BD/QM = AD/PM [ Proved above]

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

AB / PQ = BC / QR

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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

Answer:

In triangles ABC and DAC,

∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB =CD/CA
⇒ CA2 = CB x 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.

Answer:

Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

⇒AB / PQ = AC / PR = AD / PM

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

⇒AB / PQ = AC / PR = AD / PM

⇒AB / PQ = BE / QL = [(2AD) / (2PM)]

⇒AB / PQ = BE / QL = AE / PL

∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC ∼ ΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3)

In ΔABC and ΔPQR,

AB / PQ = AC / PR

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By 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.

Answer:

Let AB the vertical pole and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Joined BC and EF.

Length of the vertical pole = 6m (Given)

Shadow of the pole = 4 m (Given)

Let Height of tower = h m

Length of shadow of the tower = 28 m (Given)

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (By AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒ h = 6 x 28/4

⇒ h = 6 x 7

⇒ h = 42 m

Hence, the height of the tower is 42 meters.

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Answer:

Given: AD and PM are the medians of triangles

ABC and PQR respectively, where

It is given that Δ ABC ~ Δ PQR

We know that the corresponding sides of similar triangles are in proportion.

∴ AB / PQ = AC / AD and BC / QR ....(1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = BC/2 &  QM = QR/2 ...(3)

From equations (1) and (3), we obtain

AB/PQ = BD/QM ... (4)

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

AB/PQ = BD/QM

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

AB/PQ = BD/QM  = AD/PM

Check out: CBSE Class 10th Sample Papers

Class 10 Maths Chapter 6 Triangles Ex 6.3 Summary

Exercise 6.3 focuses on understanding and applying the criteria for the similarity of triangles. The key aspects covered in this exercise are:

1. Angle-Angle (AA) Criterion:

• Definition: Two triangles are similar if two of their corresponding angles are congruent (equal in measure).

• Application: This criterion is useful when the measures of two angles of one triangle are known to be equal to two angles of another triangle. It is often used in problems where angle measures are given, and the goal is to prove similarity.

2. Side-Side-Side (SSS) Criterion:

• Definition: Two triangles are similar if the lengths of their corresponding sides are proportional.

• Application: This criterion is applied when the ratios of the lengths of corresponding sides are given or can be calculated. The triangles are similar if the ratio of each pair of corresponding sides is the same.

3. Side-Angle-Side (SAS) Criterion:

• Definition: Two triangles are similar if one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are proportional.

• Application: This criterion is useful when two sides of one triangle and the included angle are known to be proportional to two sides of another triangle and the included angle. It helps to establish similarity based on partial information about the sides and angles.

4. Proving Triangle Similarity:

• Definition: This involves demonstrating that two triangles meet one of the similarity criteria to establish that they are similar.

• Application: Students need to show that triangles satisfy one of the criteria (AA, SSS, or SAS) to prove their similarity. This often involves setting up proportions and solving equations.

5. Application of Properties:

• Definition: Using the properties of similar triangles to solve problems.

• Application: This includes using the fact that corresponding angles are equal and the ratios of corresponding sides are proportional to find unknown lengths or angles in geometric problems.

6. Ratio of Areas:

• Definition: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

• Application: If the ratio of the sides of two similar triangles is kkk, then the ratio of their areas is k2k^2k2. This is useful for solving problems involving the areas of similar triangles.

7. Use of Proportionality Theorems:

• Definition: Theorems related to proportional segments within similar triangles.

• Application: Students use proportionality theorems to determine lengths and solve problems involving the segments of similar triangles. For example, the Basic Proportionality Theorem (also known as Thales' Theorem) is often applied.

8. Geometric Constructions:

• Definition: Using geometric tools and methods to construct similar triangles and solve problems.

• Application: This involves drawing triangles that meet the similarity criteria and using geometric properties to find missing parts or verify similarity. Geometric constructions help visualize and solve problems.

9. Real-World Applications:

• Definition: Applying the concept of triangle similarity to practical and real-world situations.

• Application: This includes using triangle similarity in various contexts such as scaling models, navigation, and architectural design. Understanding how triangle similarity applies to real-life scenarios reinforces its importance and utility.

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Class 10 Maths Chapter 6 Triangles Ex 6.3 FAQs

Q1. How can I determine if two triangles are similar using the AA Criterion?

Ans. Two triangles are similar using the AA Criterion if two angles of one triangle are equal to two angles of another triangle. Since the third angle in each triangle will also be equal, this criterion is sufficient for proving similarity.

Q2. What does the Side-Side-Side (SSS) Criterion entail?

Ans. The SSS Criterion states that two triangles are similar if the lengths of their corresponding sides are in proportion. If the ratios of all three pairs of corresponding sides are equal, then the triangles are similar.

Q3. How do I use the Side-Angle-Side (SAS) Criterion to prove similarity?

Ans. The SAS Criterion requires that one angle of a triangle be equal to one angle of another triangle, and the sides that include these angles must be proportional. If these conditions are met, the triangles are similar.

Q4. What is the significance of the ratio of the areas of similar triangles?

Ans. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. For instance, if the ratio of the sides is k, then the ratio of the areas is

Q5.  How can proportionality theorems help in solving problems involving similar triangles?

Ans. Proportionality theorems, such as the Basic Proportionality Theorem (Thales' Theorem), are used to find unknown lengths in similar triangles. These theorems state that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.