Well-designed AP 10th Class Maths Textbook Solutions Chapter 6 Triangles Exercise 6.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Triangles Class 10 Exercise 6.3 Solutions – 10th Class Maths 6.3 Exercise Solutions

Question 1.

State which pairs of triangles in given 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:

i)

Solution:

In ∆ABC and ∆PQR

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

By Angle-Angle-Angle criterion of similarity

∆ABC ~ ∆PQR.

ii)

Solution:

In ∆ABC and ∆QRP

By side-side-side criterion of similarity ∆ABC ~ ∆QRP.

iii)

Solution:

In ∆LMP and ∆DEF

\(\frac{\mathrm{LM}}{\mathrm{DE}}\) = \(\frac{2.7}{4}\)

\(\frac{\mathrm{MP}}{\mathrm{EF}}\) = \(\frac{2}{5}\)

\(\frac{\mathrm{PL}}{\mathrm{DF}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

\(\frac{2.7}{4}\) ≠ \(\frac{2}{5}\) ≠ \(\frac{1}{2}\)

So, \(\frac{\mathrm{LM}}{\mathrm{DE}}\) ≠ \(\frac{\mathrm{MP}}{\mathrm{EF}}\) ≠ \(\frac{\mathrm{PL}}{\mathrm{DF}}\)

Therefore, ∆LMP ~ ∆DEF.

iv)

Answer:

In ∆NML and ∆PQR

\(\frac{\mathrm{NM}}{\mathrm{PQ}}\) \(=\frac{2.5}{5}\) = \(\frac{1}{2}\)

∠M = ∠Q = 70°

\(\frac{\mathrm{NM}}{\mathrm{PQ}}\) = \(\frac{\mathrm{ML}}{\mathrm{QR}}\) = \(\frac{1}{2}\) and ∠M = ∠Q = 70°

Therefore, side-angle-side criterion of similarity, ∆NML ~ ∆PQR.

v)

Solution:

In ∆ABC and ∆DEF

∠A = ∠F = 80°

\(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{2.5}{7}\)

\(\frac{\mathrm{BC}}{\mathrm{EF}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

\(\frac{\mathrm{AC}}{\mathrm{DF}}\) = \(\frac{x}{5}\)

\(\frac{\mathrm{AB}}{\mathrm{DE}}\) ≠ \(\frac{\mathrm{BC}}{\mathrm{EF}}\) ≠ \(\frac{\mathrm{AC}}{\mathrm{DF}}\) and ∠A = ∠F = 80°

Therefore, ∆ABC ~ ∆DEF

vi)

Solution:

In ∆DEF,

∠D + ∠E + ∠F = 180°

70° + 80° + ∠F = 180°

∠F = 180° – 150° = 30°

In ∆PQR,

∠P + ∠Q + ∠R = 180°

∠P + 80° + 30° = 180°

∠P = 180° – 110° = 70°

∠D = ∠P = 70°

∠E = ∠Q = 80°

∠F = ∠R = 30°

Therefore, Angle-Angle-Angle criterion of similarity ∆DEF ~ ∆PQR.

Question 2.

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

Solution:

Given ∆ODC ~ ∆OBA

∠BOC = 125° and ∠CDO = 70°

∠BOC + ∠COD = 180°

∠COD = 180° – ∠BOC

∠COD = 180° – 125° = 55°

∠AOB = ∠COD = 55°

(Vertically opposite angles)

∠OAB + ∠OBA + ∠AOB = 180°

∠OAB + 70° + 55° = 180°

∠OAB= 180° – 125° = 55°

Therefore, ∠DOC = 55°,

∠DCO = ∠OAB = 55°

∠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 \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\).

Solution:

Given in ABCD trapezium, AB || DC and diagonals AC, BD intersect at ‘O’.

AB || DC.

So, ∠OAB = ∠OCD

(Alternate interior angles)

∠AOB = ∠COD (Vertically opposite angles)

∠OBA = ∠ODC (Alternate interior angles)

In ∆AOB and ∆COD,

∠OAB = ∠OCD

∠AOB = ∠COD

∠OBA = ∠ODC

By Angle-Angle-Angle criterion of similarity

∆MOB ~ ∆COD.

If two triangles are similar, then their corresponding sides are in proportion.

Therefore, \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\).

Question 4.

In given figure, \(\frac{\mathbf{Q R}}{\mathbf{Q S}}\) = \(\frac{\text { QT }}{\text { PR }}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Answer:

Given \(\frac{\mathrm{QR}}{\mathrm{QT}}\) = \(\frac{\mathrm{QS}}{\mathrm{PR}}\)

\(\frac{\mathrm{QR}}{\mathrm{QT}}\) = \(\frac{\mathrm{QS}}{\mathrm{PR}}\)

⇒ \(\frac{\mathrm{QT}}{\mathrm{QR}}\) = \(\frac{\mathrm{PR}}{\mathrm{QS}}\) → (1)

But we know, in ∆PQR, sides opposite to equal angles are equal.

∴ PQ = QR → (2)

Put (2) in (1) ⇒ \(\frac{Q T}{Q R}\) = \(\frac{\mathrm{PQ}}{\mathrm{QS}}\)

⇒ \(\frac{\mathrm{PQ}}{\mathrm{QT}}\) = \(\frac{\mathrm{QS}}{\mathrm{QR}}\)

Thus in ∆PQR and ∆TQR,

\(\frac{\mathrm{PQ}}{\mathrm{QT}}\) = \(\frac{\mathrm{QS}}{\mathrm{QR}}\) and ∠PQS = ∠TQR

By Side-Angle-Side criterion similarity

∆PQS ~ ∆TQR.

Question 5.

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

Solution:

Given in ∆RPQ and ∆RTS

∠P = ∠RTS (Angle)

∠R = ∠PRQ (Angle)

By Angle-Angle-Angle similarity

∆RPQ ~ ∆RTS.

Question 6.

In given figure, if ∆ABE \(\cong\) ∆ACD, show that ∆ADE ~ ∆ABC.

Solution:

Given ∆ABE \(\cong\) ∆ACD

If two triangles are congruent, then their corresponding sides and angles are equal.

That is AB = AC (Side)

BE = CD (Side)

AE = AD (Side)

In ∆ADE and ∆ABC,

\(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{\mathrm{AD}}{\mathrm{DE}}\) and ∠BAC = ∠DAE.

By Side-Angle-Side similarity criterion

∆ADE ~ ∆ABC.

Question 7.

In the given figure, altItudes AD and CE of ∆ABC intersect each other at the point P. Show that:

i) ∆AEP ~ ∆CDP

ii) ∆ABD ~ ∆CBE

iii) ∆AEP ~ ∆ADB

iv) ∆PDC ~ ∆BEC

Solution:

Given AD and CE are altitudes of ∆ABC intersect at P.

i) In ∆AEP and ∆CDP,

∠AEP = ∠CDP = 90°

∠APE = ∠CPD (Vertically opposite angles)

By Angle-Angle criterion of similarity,

∆AEP ~ ∆CDP.

ii) In ∆ABD and ∆CBE

∠ADB = ∠CEB = 90°

∠ABD = ∠CBE (Common angle)

By Angle-Angle criterion of similarity,

∆ABD ~ ∆CBE.

iii) In ∆AEP ~ ∆ADB

∠AEP = ∠ADB = 90°

∠PAE = ∠DAB (Common angle)

By Angle-Angle criterion of similarity,

∆AEP ~ ∆ADB.

iv) In ∆PDC ~ ∆BEC

∠CDP = ∠BEC = 90°

∠PCD = ∠ECB (Common angle)

By Angle-Angle criterion of similarity,

∆APDC ~ ∆BEC.

Question 8.

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

Solution:

Given E is a point on the side AD produced of a parallelogram ABCD.

BE and CD intersect at F.

In ∆ABE and ∆CFB.

∠A = ∠C (Opposite angles are equal)

∠AEB = ∠CBF (Alternate interior angles)

By Angle-Angle criterion of similarity,

∆ABE ~ ∆CFB.

Question 9.

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

i) ∆ABC ~ ∆AMP

ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)

Solution:

i) Given ∆ABC and ∆AMP are two right triangles.

∠B = ∠M = 90°

In ∆ABC and ∆AMP,

∠ABC = ∠AMP = 90°

∠CAB = ∠MAP (Common angle)

By Angle-Angle criterion of similarity

∆ABC ~ ∆AMP

ii) If two triangles are similar, then their corresponding sides are in proportion.

Therefore, \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\).

Question 10.

CD and GH are respectively the bisec-tors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:

i) \(\frac{\mathbf{C D}}{\mathbf{G H}}\) = \(\frac{A C}{F G}\)

ii) ∆DCB ~ ∆HGE

iii) ∆DCA ~ ∆HGF

Solution:

Given CD and GH are bisectors of ∠ACB and ∠EGF and ∆ABC – ∆FEG.

If two triangles are similar, then their corresponding sides are in proportion and corresponding angles are equal.

i.e., ∠A = ∠F, ∠B = ∠E and ∠C = ∠G

i) If two triangles are similar, then their corresponding sides are in proportion.

\(\frac{\mathrm{AC}}{\mathrm{FG}}\) = \(\frac{\mathrm{CD}}{\mathrm{GH}}\)

ii) In ∆DCB and ∆HGE

\(\frac{\angle C}{2}\) = \(\frac{\angle G}{2}\)

∠BCD – ∠HGE (Angle)

∠B = ∠E

By Angle-Angle criterion of similarity

∆ADCB ~ ∆HGE.

iii) In ∆DCA and ∆HGF

∠A = ∠F (Angle)

∠ACD = ∠FGH (Angle)

By Angle-Angle criterion of similarity

∆DCA ~ ∆HGF.

Question 11.

In 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 ∆ABD ~ ∆ECF.

Solution:

Given in ∆ABC, AB = AC, AD ⊥ BC and EF ⊥ AC.

If AB = AC, then ∠B = ∠C.

Angles which are equal to the opposite sides are equal.

In ∆ABD, ∆ECF

∠ADB = ∠EFC = 90° (Angle)

∠ABD = ∠ECF (Angle)

By Angle-Angle criterion of similarity

∆ABD ~ ∆ECF.

Question 12.

Sides AB and BC and median AD of a triangle ABC are respectively propor-tional to sides PQ and QR and medi-an PM of ∆PQR (see figure). Show that ∆ABC ~ ∆PQR.

Solution:

Given \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

∆ABC and ∆PQR, where AD and PM are medians.

In ∆ABD and ∆PQM,

By side-side-side criterion of similarity

∆ABD ~ ∆PQM.

If two triangles are similar, then their corresponding sides are proportion and corresponding angles are equal.

∠ABD = ∠PQM

Now, in ∆ABC and ∆PQR

\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BD}}{\mathrm{QM}}\) and ∠B = ∠Q

By Side-Angle-Side criterion of similarity

∆ABC ~ ∆PQR.

Question 13.

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA^{2} = CB.CD.

Solution:

Given in ∆ABC, ∠ADC = ∠BAC.

In ∆ABC and ∆DAC

∠BAC = ∠CDA

∠ACB = ∠DCA (Common angle)

By Angle-Angle criterion of similarity

∆ABC ~ ∆DAC.

If two triangles are similar, then their corresponding sides are proportion.

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{\mathrm{AC}}{\mathrm{DC}}\)

AC∙AC = CB∙CD

Therefore, AC^{2} ≠ CB∙CD.

Question 14.

Sides AB and AC and median AD of a triangle ABC are respectively propor-tional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.

Solution:

In ∆ABC and ∆PQR,

\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

Extend AD to DE such that AD = DE and join BE and CE.

Similarly, extend PM to ML such that PM = ML and join QL and RL.

In quadrilateral ABEC diagonals bisect each other. So, ABEC, is a parallelogram. Similarly, quadrilateral PQLR also a parallelogram.

By side-side-side criterion of similarity

∆ABE ~ ∆PQL

So, ∠BAE = ∠QPL → (1)

Similarly, ∆CAE, ∆PLR

\(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\) = \(\frac{\mathrm{CE}}{\mathrm{LR}}\)

\(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\) = \(\frac{\mathrm{CE}}{\mathrm{LR}}\)

Therefore, \(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{A E}{P L}\) = \(\frac{\mathrm{CE}}{\mathrm{LR}}\)

By side-side-side criterion of similarity

∆CAE ~ ∆PLR

So, ∠CAE = ∠RPL → (2)

By adding (1) and (2)

∠BAE + ∠CAE = ∠QPL + ∠RPL

∠A = ∠P

In ∆ABC, ∆PQR,

\(\frac{A B}{P Q}\) = \(\frac{A C}{P R}\) and ∠A = ∠P

By Side-Angle-Side criterion of similarity

∆ABC ~ ∆PQR.

Question 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:

Let AB is the length of the pole is 6 m.

and length of its shadow BC = 4 m.

DE is the height of the tower = x m.

Length of its shadow EF = 24 m.

In ∆ABC, ∆DEF

∠B = ∠E = 90°

∠ACB = ∠DFE (sun rays can make equal angles)

By Angle-Angle criterion of similarity

∆ABC ~ ∆DEF

then, corresponding sides are in pro-portion.

⇒ x = 6 × 7 = 42 m.

Therefore, height of the tower is 42 m.

Question 16.

If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that \(\frac{A B}{P Q}\) = \(\frac{A D}{P M}\).

Solution:

Given ∆ABC ~ ∆PQR, AD and PM are medians of triangles ABC and PQR.

So, ∠B = ∠Q (Corresponding angles)

and \(\frac{A B}{P Q}\) = \(\frac{B C}{Q R}\) = \(\frac{A C}{P R}\)

By side-side-side criterion of similarity

∆ABD ~ ∆PQM

Corresponding sides are in proportion.

Therefore, \(\frac{A B}{P Q}\) = \(\frac{A D}{P M}\).