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Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.3

Last Updated : 03 Apr, 2024
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Question 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) We have, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

By AAA similarity criterion, we have

∴ ΔABC ~ ΔPQR

(ii) We have, in ΔABC and ΔPQR,

Computing the ratios of corresponding sides, 

AB/QR = 2.5/5 = 1/2

BC/RP = 2/4 = 1/2

 CA/PQ = 3/6 = 1/2

AB/QR = BC/RP = CA/PQ

Now,

By SSS similarity criterion, we have

ΔABC ~ ΔQRP

(iii) We have, in ΔLMP and ΔDEF,

Now,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

Computing the ratios of the corresponding sides, 

MP/DE = 2/4 = 1/2

PL/DF = 3/6 = 1/2

LM/EF = 2.7/5 = 27/50

Since, MP/DE = PL/DF ≠ LM/EF

Therefore, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, it is given,

Computing the ratios of the corresponding sides, 

MN/QP = ML/QR = 1/2

And,

∠M = ∠Q = 70°

By SAS similarity criterion

ΔMNL ~ ΔQPR

(v) In ΔABC and ΔDEF, we have,

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Computing the ratios of the corresponding sides, 

 AB/DF = 2.5/5 = 1/2

And, BC/EF = 3/6 = 1/2

⇒ ∠B ≠ ∠F

Therefore, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF, 

By angle sum property of a triangle,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

Similarly, In ΔPQR,

By angle sum property of a triangle,

∠P + ∠Q + ∠R = 180 

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

Now, in ΔDEF and ΔPQR, we have

∠D = ∠P = 70° [equal corresponding angles]

∠F = ∠Q = 80°

∠F = ∠R = 30°

Therefore,

By AAA similarity criterion,

ΔDEF ~ ΔPQR

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

Solution:

We know, DOB is a straight line, as given.

Therefore, ∠DOC + ∠COB = 180°

Also,

∠BOC = 125°

⇒ ∠DOC = 180° – 125°

= 55°

In ΔDOC

By angle sum property of a triangle, 

∠DCO + ∠CDO + ∠DOC = 180°

 ∠CDO = 70°

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

Also, given,

ΔODC ∝ ¼ ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Since, we know that the corresponding angles are equal in similar triangles,

Therefore, ∠OAB = ∠OCD

⇒ ∠ OAB = 55°

∠OAB = ∠OCD

∠OAB = 55°

Question 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,

Since, AB || CD, therefore the alternate interior angles are equal,

That is, ∠CDO = ∠ABO ….(i)

And,

∠DCO = ∠BAO …(ii)

Also, vertically opposite angles are equal;

That is,∠DOC = ∠BOA…(iii)

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA [From eq(i) , (ii) and (iii)]

Since the triangles are similar, therefore the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.

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

Solution:

We have,

In ΔPQR,

∠PQR = ∠PRQ

That is, PQ = PR .…(i)

Given,

QR/QS = QT/PR

By using eq(i), we obtain,

QR/QS = QT/QP …….(ii)

In ΔPQS and ΔTQR, 

QR/QS = QT/QP [By using eq(ii)]

∠Q = ∠Q

By SAS similarity criterion, therefore,

ΔPQS ~ ΔTQR 

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

Solution:

We know, 

S and T are the points on sides PR and QR of ΔPQR.

And, ∠P = ∠RTS.

In ΔRPQ and ΔRTS,

Given, 

∠RTS = ∠QPS 

∠R = ∠R (Common angle)

By AAA similarity criterion,

∴ ΔRPQ ~ ΔRTS 

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

Solution:

Given, ΔABE ≅ ΔACD.

By CPCT, we have

∴ AB = AC ……….(i)

And, AD = AE …………(ii)

In ΔADE and ΔABC, 

On dividing the eq.(ii) by eq(i), we obtain, 

AD/AB = AE/AC

Also, by common angle, 

∠A = ∠A 

By SAS similarity criterion, we have,

∴ ΔADE ~ ΔABC 

Question 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,

Since, both of the angles are 90° each.

∠AEP = ∠CDP 

Since,

∠APE = ∠CPD (Vertically opposite angles)

By AA similarity criterion, we have,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

Since, both of the angles are 90° each.

∠ADB = ∠CEB

∠ABD = ∠CBE (Common Angles)

By AA similarity criterion, we have,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

Since, both of the angles are 90° each.

∠AEP = ∠ADB 

∠PAE = ∠DAB (Common angles)

By AA similarity criterion, we have,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

Since, both of the angles are 90° each.

∠PDC = ∠BEC 

∠PCD = ∠BCE (Common angles)

By AA similarity criterion, we have,

ΔPDC ~ ΔBEC

Question 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. 

Now,

In ΔABE and ΔCFB,

Since, the opposite angles of a parallelogram are equal, we have,

∠A = ∠C 

Also, since AE || BC

∠AEB = ∠CBF (Alternate interior angles)

By AA similarity criterion,

∴ ΔABE ~ ΔCFB 

Question 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)

Since, both of the angles are equivalent to 90°

∠ABC = ∠AMP 

By AA similarity criterion, we have,

∴ ΔABC ~ ΔAMP

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion) [proved in part (i)]

Therefore, we know, 

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

Hence, CA/PA = BC/MP

Question 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) We have,

ΔABC ~ ΔFEG.

Therefore,

∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Now, since,

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE 

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

By AA similarity criterion, we have,

∴ ΔACD ~ ΔFGH 

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

Proved in part (i)

∠DCB = ∠HGE

∠B = ∠E

By AA similarity criterion,

∴ ΔDCB ~ ΔHGE 

(iii) In ΔDCA and ΔHGF,

Since, we have already proved, 

∠ACD = ∠FGH 

∠A = ∠F 

By AA similarity criterion,

∴ ΔDCA ~ ΔHGF 

Question 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.

Since, the sides of an isosceles triangle are equal, we have,

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

Since, each of the following angles are 90°.

∠ADB = ∠EFC 

Since, we have already proved, 

∠BAD = ∠CEF 

By AA similarity criterion, we have,

∴ ΔABD ~ ΔECF 

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 (see Fig 6.41). Show that ΔABC ~ ΔPQR.

Solution:

Given that in ΔABC and ΔPQR, 

AB is proportional to PQ

BC is proportional to QR

AD is proportional to PM

That is, AB/PQ = BC/QR = AD/PM

We know,

AB/PQ = BC/QR = AD/PM

Since, D is the midpoint of BC and M is the midpoint of QR

⇒AB/PQ = BC/QR = AD/PM 

By SSS similarity criterion, we have,

⇒ ΔABD ~ ΔPQM 

Now, since the corresponding angles of two similar triangles are equal, we obtain,

∴ ∠ABD = ∠PQM 

⇒ ∠ABC = ∠PQR

Now,

In ΔABC and ΔPQR

AB/PQ = BC/QR ………….(i)

∠ABC = ∠PQR ……………(ii)

From equation (i) and (ii), we get,

By SAS similarity criterion, we have,

ΔABC ~ ΔPQR 

Question 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

∠ACD = ∠BCA (Common angles)

By AA similarity criterion, we have,

∴ ΔADC ~ ΔBAC 

Since, the corresponding sides of similar triangles are in proportion, we obtain,

∴ CA/CB = CD/CA

That is, CA2 = CB.CD.

Hence, proved.

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 ΔABC ~ ΔPQR.

Solution:

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

Now,

AB/PQ = AC/PR = AD/PM

Construction, 

Produce AD further to E so that AD = DE. Now, join CE.

Similarly, produce PM further until N such that PM = MN. Join RN.

In ΔABD and ΔCDE, we have

From the construction done,

AD = DE 

Now, since AP is the median,

BD = DC

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

By SAS criterion,

∴ ΔABD ≅ ΔCDE 

By CPCT, we have,

⇒ AB = CE……………..(i)

In ΔPQM and ΔMNR,

From the construction done,

PM = MN

Now, since PM is the median,

QM = MR 

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

By SAS criterion,

∴ ΔPQM = ΔMNR 

By CPCT, 

⇒ PQ = RN …………………(ii)

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

From equation (i) and (ii), we conclude,

⇒CE/RN = AC/PR = AD/PM

⇒ CE/RN = AC/PR = 2AD/2PM

We know, 

2AD = AE and 2PM = PN.

⇒ CE/RN = AC/PR = AE/PN 

By SSS similarity criterion,

∴ ΔACE ~ ΔPRN 

Thus,

∠2 = ∠4

And, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ………………….(iii)

Now, in ΔABC and ΔPQR, we have

AB/PQ = AC/PR [Given]

From equation (iii), we have, 

∠A = ∠P

By SAS similarity criterion,

∴ ΔABC ~ ΔPQR 

Question 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

Length of shadow of the tower = 28 m

Shadow of the pole = 4 m

Let us assume the height of tower to be h m.

In ΔABC and ΔDEF,

∠C = ∠E (By angular elevation of sum)

Since, the following angles are equivalent to 90°

∠B = ∠F 

By AA similarity criterion, we have,

∴ ΔABC ~ ΔDEF 

Since, if two triangles are similar corresponding sides are proportional

∴ AB/DF = BC/EF 

Substituting values, 

∴ 6/h = 4/28

⇒h = (6×28)/4

⇒ h = 6 × 7

⇒ h = 42 m

Hence, the height of the tower specified is 42 m.

Question 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

Since, the corresponding sides of similar triangles are in proportion.

∴ AB/PQ = AC/PR = BC/QR……………(i)

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

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

∴ BD = BC/2 and QM = QR/2 ………….(iii)

From equations (i) and (iii), we obtain,

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

By SAS similarity criterion, we have,

∴ ΔABD ~ ΔPQM 

That is, AB/PQ = BD/QM = AD/PM



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