Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev

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Class 10 : Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev

The document Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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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: 
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
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

Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Hence, ΔABC ~ ΔQRP [SSS criterion]
(iii) In ΔLMP and ΔEFD
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴  ΔLMP is not similar to ΔEFD.
Since the three ratios are not same.
(iv) In ΔMNL and ΔQPR
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
(v) In ΔABC and ΔDEF
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
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.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev

Sol. 

Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒  Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
In ΔDOC,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒   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 Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. Given: Diagonals AC and BD intersect at O.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
AB || DC
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Proof: In ΔAOB and ΔCOD
∠1 = ∠2
∠3 = ∠4    [Alternate angles]
∴ ΔAOB ~ ΔCOD     [AA]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
[Corresponding sides of similar triangles]

Q.4. In the figure, Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol.  Given
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
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 Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
In ΔPQS and ΔTQR,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev    (Proved)
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∠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 ΔRPQ ~ ΔRTS.
Sol. In Δ RPQ ~ ΔRTS
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∠P = ∠RTS     [Given]
∠R = ∠R       [Common]
∴ ΔRPQ ~ ΔRTS    [AA]

Q.6. In the figure, if ΔABE ≌ ΔACD, show that ΔADE ~ ΔABC.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol.
Given: ΔABE ≌ ΔACD
To Prove: ΔADE ~ ΔABC
Proof: ΔABE ~ ΔACD
AB = AC and AE = AD
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
In ΔADE and ΔABC,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev   [Proved above]
∠A - ∠A   [Common]
∴ ΔADE ~ ΔABC     [SAS]

Q.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
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
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 ~ ΔCFB.
Sol.  In ΔABE and ΔCFB,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∠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) Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. (i) In ΔABC and ΔAMP,
∠B = ∠AMP [Each 90°]
∠A = ∠A [Common]
⇒ ΔABC ∼ ΔAMP [AA]
(ii) ΔABC ~ ΔAMP [proved above]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
[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 ΔABC ~ ΔFEG, show that:
 (i)
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev 
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF                       
Sol.  ΔABC ~ ΔFEG     [Given]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒ ∠A = ∠F
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
(i) In ΔACD and ΔFGH
∠A = ∠F    [Given]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴  ΔACD ~ ΔFGH
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
[Corresponding sides of similar triangles]
(ii) Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev    [Proved above]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev

In ΔDCB and ΔHGE,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴  ΔDCB ~ ΔHGE       [SAS]

(iii) In ΔDCA and ΔHGF,
∠1 = ∠2          [Bisectors]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒ 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 ΔABD ~ ΔECF.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
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 ~ ΔPQR.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. In ΔABC and ΔPQR
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
⇒ ΔABD ~ ΔPQM     [SAS]
∴ ∠B = ∠Q
[Corresponding angles of similar triangles]
In ΔABC and ΔPQR,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev      [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 CA2 = CB.CD.
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Sol. In ΔACB and ΔDCA,
∠BAC = ∠ADC     [Given]
∠C = ∠C       [Common]
ΔACB ~ ΔDCA
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
[Corresponding sides of similar triangles]
⇒ CA2 = CB x CD

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 ΔABC ~ ΔPQR.
Sol. Construction: Draw DE || AC and MS || PR
Proof: In ΔABC, D is mid-point of BC (given) and DE || AC (const.)
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴ E is mid-point of AB (by converse of mid-point theorem
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴ ΔADE ~ ΔPMS [SSS similarly]
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Now, in ΔABC and ΔPQR,
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev     [Given]
∠A = ∠P    [Proved above]
∴ ΔABC ~ ΔPQR   (SAS)

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.
Sol. DE is vertical stick of length = 6m
Length of shadow = 4 m
Let height of tower = h m
Length of shadow = 28 m
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev

Q.16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev.
Sol. When ΔABC ~ ΔPQR
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
In ΔABD and ΔPQM
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
∴  ΔABD ~ ΔPQM
Ex 6.3 NCERT Solutions- Triangles Class 10 Notes | EduRev
[Corresponding sides of similar triangles]

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