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**Q.1. State which pairs of triangles in Fig. 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: ****Sol.** (i) In Δ ABC and Δ PQR

We have:

∠A = ∠P [Each 60°]

∠B = ∠Q [Each 80°]

∠C = ∠R [Each 40°]

ΔABC ~ ΔPQR [AAA criterion]

(ii) In ΔABC and ΔQRP

Hence, ΔABC ~ ΔQRP [SSS criterion]

(iii) In ΔLMP and ΔEFD

∴ ΔLMP is not similar to ΔEFD.

Since the three ratios are not same.

(iv) In ΔMNL and ΔQPR

(v) In ΔABC and ΔDEF

ΔABC is not similar to ΔDEF

∵ Angles between two sides are not same

(vi) In ΔDEF and ΔPQR

∠E = ∠Q = 80°

∠F = ∠R = 30°

[∴ ∠F = 180° - (80° + 70°) = 30°]

∴ ΔDEF ~ ΔPQR [AA]**Q.2. In the figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.**

**Sol.**

⇒

In ΔDOC,

⇒ 125° + ∠DCO = 180°

⇒ ∠DCO = 180° - 125° = 55°

ΔODC ~ ΔOBA

∠OAB = ∠DCO = 55°

⇒ ∠DOC = ∠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 ****Sol.** Given: Diagonals AC and BD intersect at O.

AB **||** DC

Proof: In ΔAOB and ΔCOD

∠1 = ∠2

∠3 = ∠4 [Alternate angles]

∴ ΔAOB ~ ΔCOD [AA]

⇒

[Corresponding sides of similar triangles]**Q.4. In the figure, **** and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.****So**l. Given

To Prove: ΔPOS ∼ ΔTOR

To Prove: ΔPQS

Proof: In ΔPQR,

∠1 = ∠2 [Given]

PQ = PR [Sides opposite to equal angles]

OR/QS = QT/PR [Given]

Or

In ΔPQS and ΔTQR,

(Proved)

⇒

∠1 = ∠1 [Common]

∴ ∠PQS ~ ∠TQR [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.****Sol. **In Δ RPQ ~ ΔRTS

∠P = ∠RTS [Given]

∠R = ∠R [Common]

∴ ΔRPQ ~ ΔRTS [AA]**Q.6. In the figure, if **Δ**ABE ≌ ΔACD, show that **Δ**ADE ~ **Δ**ABC.****Sol.**

Given: ΔABE ≌ ΔACD

To Prove: ΔADE **~** ΔABC

Proof: ΔABE **~** ΔACD

AB = AC and AE = AD

⇒

In ΔADE and ΔABC,

[Proved above]

∠A - ∠A [Common]

∴ ΔADE ~ ΔABC [SAS]**Q.7. In the 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****Sol.** Given: AD and CE are altitudes of the ΔABC

(i) To Prove: ΔAEP ~ ΔCDP

Proof: In ΔAEP and ΔCDP,

∠AEP = ∠CDP [Each'90°]

∠APE = ∠CPD [Vertically opposite angles]

ΔAEP **~** ΔCDP [AA]

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB [Each 90°]

∠ABD = ∠CBE [Common]

ΔABD ~ ΔCBE [AA]

(iii) In Δ AEP and ΔADB,

∠AEP = ∠ADB [Each 90°]

∠A = ∠A [Common]

∴ ΔAEP ~ ΔADB [AA]

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC [Each 90°]

∠PCD = ∠BCE [Common]

ΔPDC ~ ΔBEC [AA]**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.****Sol.** In ΔABE and ΔCFB,

∠1 = ∠2

∠4 = ∠3 [Alternate angles]

⇒ ΔABE ∼ ΔCFB [AA]**Q.9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:****(i) **Δ** ABC ~ **Δ** AMP ****(ii) ****Sol.** (i) In ΔABC and ΔAMP,

∠B = ∠AMP [Each 90°]

∠A = ∠A [Common]

⇒ ΔABC ∼ ΔAMP [AA]

(ii) ΔABC ~ ΔAMP [proved above]

[Ratio of the Corresponding sides of similar Δs]**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 **Δ** ABC and **Δ** EFG respectively. If **Δ**ABC ~ **Δ**FEG, show that: (i)**

⇒ ∠A = ∠F

(i) In ΔACD and ΔFGH

∠A = ∠F [Given]

∴ ΔACD ~ ΔFGH

[Corresponding sides of similar triangles]

(ii) [Proved above]

In ΔDCB and ΔHGE,

∴ ΔDCB ~ ΔHGE [SAS]

(iii) In ΔDCA and ΔHGF,

∠1 = ∠2 [Bisectors]

⇒ ΔDCA ~ ΔHGF [SAS]**Q.11. In the 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.****Sol.** In ΔABD and ΔECF,

∠ADB = ∠EFC [Each 90°]

∠B = ∠C

[angles opposite to equal sides are equal]

⇒ ΔABD ~ ΔECF [AA]**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 figure). Show that **Δ** ABC ~ **Δ**PQR.****Sol. **In ΔABC and ΔPQR

⇒ ΔABD **~** ΔPQM [SAS]

∴ ∠B = ∠Q

[Corresponding angles of similar triangles]

In ΔABC and ΔPQR,

[Given]

∠B = ∠Q [As provecd]

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

∠BAC = ∠ADC [Given]

∠C = ∠C [Common]

ΔACB

⇒

[Corresponding sides of similar triangles]

⇒ CA

∴ E is mid-point of AB (by converse of mid-point theorem

∴ ΔADE ~ ΔPMS [SSS similarly]

Now, in ΔABC and ΔPQR,

[Given]

∠A = ∠P [Proved above]

∴ ΔABC ~ ΔPQR (SAS)

Length of shadow = 4 m

Let height of tower = h m

Length of shadow = 28 m

∴

In ΔABD and ΔPQM

∴ ΔABD

[Corresponding sides of similar triangles]

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