**NCERT Solution For Class 10, Maths, Chapter 6 Triangles, Exercise 6.3** solutions are given below. All the questions in exercise 6.3, chapter 6 are based on properties of triangles. NCERT class 10 maths chapter 6, ex. 6.3 clears all basic fundamentals related to congruence of triangles. This ex. 6.3 contains total fifteen questions to study.

Table of Contents

Toggle## Class 10, Maths, Chapter 6, Exercise 6.3 Solutions

**Q.1. State which pairs of triangles in Fig. 6.34 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:**

**Ans:**

**Q.2. In Fig. 6.35, Δ ODC ~ Δ OBA, ****∠**** BOC = 125° and ****∠**** CDO = 70°. Find ****∠**** DOC, ****∠**** DCO and ****∠**** OAB.**

**Ans:** Since BD is a line and OC is a ray on it,

∴ DOC + ∠BOC =180° [Linear pair]

⇒ DOC + 125° =180°

⇒ DOC =180° – 125° = 55°

In ∆ ODC, we have

∴ ∠CDO + ∠DOC + ∠DCO = 180° [Angle sum property]

⇒ 70° + 55° + ∠DCO = 180°

⇒ ∠DCO = 180° – 125° = 55°

It is given that ∆ ODC ~ ∆ OBA

∠DCO = ∠OBA,

∠OAB = 55^{0} (∵∠DCO = 55°)

Hence, ∠DOC = 55°, ∠DCO = 55° And ∠OAB = 55°

**Q.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 $\frac{OA}{OC}=\frac{OB}{OD}$**

**Ans: ****Given: **ABCD is a trapezium in which AB || DC

**To prove: $\frac{OA}{OC}=\frac{OB}{OD}$**

**Proof:** In ∆s AOB and COD, we have

∠AOB = ∠OCD [Alternate angles]

∠AOB = ∠COD [vertically opposite angles}

and ∠OBA = ∠ODC [Alternate angles]

∴ By AAA criterion of similarity,

∆ AOB ~ ∆ COD

Hence, **$\frac{OA}{OC}=\frac{OB}{OD}$**

[∵ In case of two similar triangles, the ration of their corresponding sides are equal]

**Q.4. In Figure, $\frac{QR}{QS}=\frac{QT}{PR}$ and ****∠**** 1 = ****∠**** 2. Show that ΔPQS ~ Δ TQR.**

** **

**Ans:** We have, in ΔPQR, we have

∠1 = ∠2 [Given]

⇒ PR = QP ………………. (i) [∵ Sides opp. to equal angles are equal]

In ΔPQS and ΔTQR, we have

⇒ ** $\frac{QR}{QS}=\frac{QT}{PR}$** [Given]

From (i), we get

⇒ ** $\frac{QR}{QS}=\frac{QT}{QP}$ (****∵ ****PR = QP)**

∠Q = ∠Q [common]

∴ ΔPQS ~ ΔTQR [SAS]

**Q.5. S and T are points on sides PR and QR of ΔPQR such that ****∠**** P = ****∠**** RTS. Show that Δ RPQ ~ Δ RTS.**

**Ans:**

In ∆s RPQ and RTS, we have

∠R = ∠R [common]

∠P = ∠RTS [Given]

∴ By AA-criterion of similarity

∆ RPQ ~ ∆ RTS [AA criteria]

**Q.6. In the given Figure, if Δ ABE ****≅**** Δ ACD, show that Δ ADE~ Δ ABC. **** **

** **

**Ans:** It is given that

∆ ABE ≅ ∆ ACD [Given]

∴ AB = AC [∵ CPCT]

AD = AE [∵ Corresponding parts of congruent triangles are equal]

In **Δ ADE and Δ ABC**

∠A = ∠A [Common]

⇒ $\frac{AD}{AB}=\frac{AE}{AC}$

⇒ $\frac{AB}{AC}=\frac{AD}{AE}$

Thus, by SAS criterion of similarity,

∆ADE ~ ∆ABC.

**Q.7. In Figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that**

**(i) Δ AEP ~ Δ CDP**

**(ii) Δ ABD ~ Δ CBE**

**(iii) Δ AEP ~ Δ ADB**

**(iv) Δ PDC ~ Δ BEC:**

** **

** **

** ****Ans: **** (i) In ****∆****s AEP and CDP, we have**

∠AEP = ∠CDP = 90° [∵ CE ⊥ AB and AD ⊥ BC]

∠APE = ∠CPD [Vertically opp. angles]

∴ By AA-criterion of similarity, we have

Δ AEP ~ Δ CDP

**(ii) In ****∆****s ABD and CBE, we have**

∠ABD = ∠CBE

∠B = ∠B [common angle]

∠ADB = ∠CEB = 90^{0}

∴ By AA-criterion of similarity, we have

Δ ABD ~ Δ CBE.

**(iii) In ****∆****s AEP and ADB, we have**

∠AEP = ∠ADB = 90° [∵ AD ⊥ BC and CE ⊥ AB]

∠PAE = ∠DAB [Common angle]

∴ By AA-criterion of similarity, we have

Δ AEP ~ Δ ADB.

**(iv) In ****∆****s PDC and BEC, we have**

∠PDC = ∠BEC = 90° [∵ AD ⊥ BC and CE ⊥ AB]

∠PCD = ∠ECB [Common angle]

∴ By AA-criterion of similarity, we have

Δ PDC ~ Δ BEC.

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

** **

**Ans: ** In ∆s ABE and CFB, we have

∠A=∠C [Opp. Angles of parallelogram]

∠ABE = ∠CFB [Alternate interior angles]

∴ By AA-criterion of similarity, we have

Δ ABE ~ Δ CFB.

**Q.9. In Figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:**

**(i) Δ ABC ~ Δ AMP**

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

** **

**Ans:**

** (i) In ****∆****s ABC and AMP, we have**

∠ABC = ∠AMP = 90° [Given]

And, ∠A = ∠A [Common]

∴ By AA, criterion of similarity, we have

Δ ABC ~ Δ AMP.

**(ii) We have, Δ ABC ~ Δ AMP** [As proved above]

⇒ ** $\frac{CA}{PA}=\frac{BC}{MP}$** [corresponding parts of similar triangle]

**Q.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 Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:**

**(i) $\frac{CD}{GH}=\frac{AC}{FG}$**

**(ii) Δ DCB ~ Δ HGE**

**(iii) Δ DCA ~ Δ HGF**

**Ans: **Δ ABC ~ Δ FEG [Given]

⇒ ∠A = ∠F, ∠B = ∠E, ∠C = ∠G

**(i) In Δ ADC and FGH, we have**

∠A = ∠F

∠1 = ∠2 $\left[ \frac{1}{2}\angle C=\frac{1}{2}\angle G \right]$

∴ By AA-criterion of similarity,

⇒ ** $\frac{CD}{GH}=\frac{AC}{FG}$** [corresponding parts of similar triangle]

**(ii) in Δ DCB ~ Δ HGE **

∠B = ∠E [proved above]

∠3 = ∠4 $\left[ \frac{1}{2}\angle C=\frac{1}{2}\angle G \right]$

**Δ DCB ~ Δ HGE [AA]**

**(iii) In ****∆****s DCA and HGF, we have**

∠A = ∠F [Proved above]

∠1 = ∠2 [Proved above]

∴ By AA-criterion of similarity, we have

Δ DCA ~ Δ HGF

**Q.11. In 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 Δ ABD ~ Δ ECF.**

** **

**Ans:** Here, ∆ABC is isosceles with AB = AC

∴ ∠B = ∠C [Angle opp. to equal sides in triangle]

In ∆s ABD and ECF, we have

∠ABD = ∠ECF [∵ ∠B = ∠C]

∠ADB = ∠EFC = 90^{0} [∵ AD ⊥ BC and EF ⊥ AC]

∴ By AA-criterion of similarity.

∆ ABD ~ ∆ ECF.

**Q.12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see Fig. 6.41). Show that Δ ABC ~ Δ PQR.**

** **** **

**Ans:**

**Q.13. D is a point on the side BC of a triangle ABC such that ****∠**** ADC = ****∠**** BAC. Show that CA ^{2} = CB.CD.**

** **

**Ans:** In ΔADC and ΔBAC, we have

∠ADC = ∠BAC [Given]

∠C = ∠C [Common]

∴ By AA-criterion of similarity, we have

Δ ADC ~ Δ BAC

⇒ $\frac{CA}{CB}=\frac{CD}{CA}$

⇒ CA^{2} = CB × CD (proved)

**Q.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 Δ ABC ~ Δ PQR**

**Ans:**

**Q.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.**

**Ans:**

**Q.16. If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that $\frac{AB}{PQ}=\frac{AD}{PM}$**

**Ans:**