Ncert Solutions Class 10 Maths Triangles Exercise 6.3

The CBSE Class 10 Chapter 6 Triangles has a total of 6 exercises. The pdf containing the NCERT Solutions of the third exercise, Exercise 6.3, is available here. The solution, apart from the pdf format, is also available below. At BYJU'S, the subject experts solve the questions with utmost care, giving explanation for the steps that are difficult to understand.

There are plenty of questions in the NCERT textbook intended for the students to solve and practice. To score high marks in the Class 10 first and second term examination, solving and understanding all the NCERT solutions for class 10 Maths is a must.

Download PDF of NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.3

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Access other exercise solutions of class 10 Maths Chapter 6- Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Access Answers of Maths NCERT class 10 Chapter 6 – Triangles Exercise 6.3

1. State which pairs of triangles in 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:

Solution:

(i) Given, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore by AAA similarity criterion,

∴ ΔABC ~ ΔPQR

(ii) Given, in  ΔABC and ΔPQR,

AB/QR = 2/4 = 1/2,

BC/RP = 2.5/5 = 1/2,

CA/PA = 3/6 = 1/2

By SSS similarity criterion,

ΔABC ~ ΔQRP

(iii) Given, in ΔLMP and ΔDEF,

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

Therefore, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, it is given,

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

Therefore, by SAS similarity criterion

∴ ΔMNL ~ ΔQPR

(v) In ΔABC and ΔDEF, given that,

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, by sum of angles of triangles, we know that,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

Similarly, In ΔPQR,

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

Now, comparing both the triangles, ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Therefore, by AAA similarity criterion,

Hence, ΔDEF ~ ΔPQR

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

Solution:

As we can see from the figure, DOB is a straight line.

Therefore, ∠DOC + ∠ COB = 180°

⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)

= 55°

In ΔDOC, sum of the measures of the angles of a triangle is 180º

Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°

⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)

⇒ ∠DCO = 55°

It is given that, ΔODC ∝ ¼ ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Hence, Corresponding angles are equal in similar triangles

∠OAB = ∠OCD

⇒ ∠ OAB = 55°

∠OAB = ∠OCD

⇒ ∠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

Solution:

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles will be equal,

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.

4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Solution:

In ΔPQR,

∠PQR = ∠PRQ

∴ PQ = PR ………………………(i)

Given,

QR/QS = QT/PRUsing equation (i), we get

QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

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

Solution:

Given, S and T are point on sides PR and QR of ΔPQR

And ∠P = ∠RTS.

In ΔRPQ and ΔRTS,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (AA similarity criterion)

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

Solution:

Given, ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] ……………………………….(i)

And, AD = AE [By CPCT] ……………………………(ii)

In ΔADE and ΔABC, dividing eq.(ii) by eq(i),

AD/AB = AE/AC

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [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

Solution:

Given, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by 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.

Solution:

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

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

∴ ΔABE ~ ΔCFB (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

Solution:

Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP

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

Solution:

Given, 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.

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (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.

Solution:

Given, ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (using AA similarity criterion)

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 Fig 6.41). Show that ΔABC ~ ΔPQR.

Solution:

Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ΔABC ~ ΔPQR

As we know here,

AB/PQ = BC/QR = AD/PM

⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR ………………………….(i)

∠ABC = ∠PQR ……………………………(ii)

From equation(i) and(ii), we get,

ΔABC ~ ΔPQR [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

Solution:

Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Already given)

∠ACD = ∠BCA (Common angles)

∴ ΔADC ~ ΔBAC (AA similarity criterion)

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

∴ CA/CB = CD/CA

⇒ CA² = CB.CD.

Hence, proved.

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.

Solution:

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, ΔABC ~ ΔPQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In ΔABD and ΔCDE, we have

AD = DE  [By Construction.]

BD = DC [Since, AP is the median]

and, ∠ADB = ∠CDE [Vertically opposite angles]

∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]

⇒ AB = CE [By CPCT] …………………………..(i)

Also, in ΔPQM and ΔMNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [SAS criterion of congruence]

⇒ PQ = RN [CPCT] ………………………………(ii)

Now, AB/PQ = AC/PR = AD/PM

From equation (i) and(ii),

⇒CE/RN = AC/PR = AD/PM

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P …………………………………………….(iii)

Now, in ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

∠A = ∠P

∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

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

Solution:

Given, Length of the vertical pole = 6m

Shadow of the pole = 4 m

Let Height of tower =h m

Length of shadow of the tower = 28 m

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (AA similarity criterion)

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

∴ 6/h = 4/28

⇒h = (6×28)/4

⇒h = 6 × 7

⇒h= 42 m

Hence, the height of the tower is 42 m.

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

Solution:

Given, ΔABC ~ ΔPQR

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

∴AB/PQ = AC/PR = BC/QR……………………………(i)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)

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

∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)

From equations(i) and(iii), we get

AB/PQ = BD/QM ……………………….(iv)

In ΔABD and ΔPQM,

From equation (ii), we have

∠B = ∠Q

From equation(iv), we have,

AB/PQ = BD/QM

∴ ΔABD ~ ΔPQM (SAS similarity criterion)

⇒AB/PQ = BD/QM = AD/PM

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Topic similarity of Triangles is the base of the Exercise 6.3, Class 10 NCERT Maths Chapter 6, Triangles. There are 16 Questions in this exercise, that includes, 1 main question with 6 sub-questions, 12 Short Answer Questions and 3 Long Answer Questions. The Exercise 6.3 deals with 3 theorems, that contains AAA, SSS, SAS criterion. The three theorems that forms the base of the third exercise of Chapter 6 are:

• Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
• If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles
• Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
• Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

NCERT Solutions are one of the best study materials a student can make use to score great marks in the Class 10 term 1 and term 2 examination. Students who are thorough with the concepts and are well versed in solving the problems present in the NCERT textbook can easily score great marks in the class 10 first and second term exam as CBSE often asks questions, directly or indirectly, from the NCERT book.

Ncert Solutions Class 10 Maths Triangles Exercise 6.3

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