10th Maths 6.3
NCERT Class 10th solution of Exercise 6.1
NCERT Class 10th solution of Exercise 6.2
NCERT Class 10th solution of Exercise 6.4
NCERT Class 10th solution of Exercise 6.5
Exercise 6.3
Q1. 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 :
In △ABC and △PQR
∠A=∠P
∠B=∠Q
∠C=∠R
△ABC∼△PQR [By AAA]
Answer :
Yes, AAA, △ABC∼△PQR.
In △ABC and △PQR
ABQR=24=12
ACPQ=36=12
BCPR=2.55=12
ABQR=ACPQ=ABQR=12
△ABC∼△PQR [By SSS]
Sol. :
In △LMP and △DEF
In △RPQ and △RTS
Sol. :
Sol.
Sol.
Sol.
Sol.
Sol. 
Sol.
Sol.
Answer :
Yes, SSS,△ABC∼△PQR.
Sol. :
MPDE=24=12
PLDF=36=12
MLEF=2.75
MPDE=PLDE≠MPEF
△LMP≁△DEF
Answer :
No, △LMP△DEF.
Sol. :
In △LMN and △PQR
MNPQ=2.55
∠M=∠Q=70∘
MLEF=510=12
MNPQ=MLEF=12 and ∠M=∠Q=70∘
△LMN∼△PQR [By SAS Similarity criterion]
Answer :
Yes, SAS, △LMN∼△PQR.
Sol. :
In △ABC and △DEF
ABDF=2.55=12
BCFE=36=12
∠A=∠F=80∘
where ∠F is included but ∠A is not included.
So that
△ABC≁△DEF
Answer :
No.
Sol. :
In △DEF and △PQR
∠F=180∘-(∠F+∠E)
∠F=180∘-(70∘+80∘)
∠F=180∘-150∘
∠F=30∘
∠P=180∘-(∠Q+∠R)
∠P=180∘-(80∘+30∘)
∠P=180∘-110∘
∠P=70∘
then
∠D=∠P=70∘
∠E=∠Q=80∘
∠F=∠R=30∘
△DEF∼△PQR [By AAA Similarity criterion]
Answer :
Yes, AAA, △DEF∼△PQR.
Q2. In Figure, △ODC∼△OBA,∠BOC=125∘ and ∠CDO=70∘. Find ∠DOC,∠DCO and ∠OAB.
Sol.
∠DOC+125∘=180∘ [Linear pair]
∠DOC=180∘-125∘
∠DOC=55∘
In△DOC
By angle sum property of △
∠DCO+70∘+55∘=180∘
∠DCO=180∘-125∘
∠DCO=55∘
And
△ODC∼△OBA, [Given]
∠OAB=∠DCO=55∘
Answer:
∠DOC=55∘,
∠DCO=55∘,
∠OAB=55∘.
Q3. 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 OAOC=OBOD.
Sol. :
In trapezium ABCD which have AB∥DC and its diagonals AC and BD intersect each other at point O.
Construction: Draw a line EF∥AB∥DC passing through O.
In △ADC,
EF∥DC [By construction]
AEED=AOCO______(1)[By Theroem 6.1]
[Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.]
In △BAD
EF∥AB [By construction]
AEED=BODO______(2)[By Theroem 6.1]
from equation (1) and (2)
AOCO=BODO
AOBO=CODO
Proved.
Q4. In Figure, QRQS=QTPR and ∠1=∠2. Show that △PQS∼△TQR.
Sol.
In △PQR,
∠1=∠2 [Given ]
∠Q=∠R [According to Figure]
PQ=PR_____(1)[Side opposite to equal angle is equal]
QRQS=QTPR_____(2) [Given ]
from equation (1) and (2)
QRQS=QTPR=QTQP
QSQP=QRQT______(3)
In △PQS and △{TQR}
∠PQS=∠TQR
QSQP=QRQT [from equation (3)]
△PQS∼△TQR [By SAS Similarity criterion]
Proved.
Q5. S and T are points on sides PR and QR of △PQR such that ∠P=∠RTS. Show that △RPQ∼△RTS.
Sol. :
Point S and T are points on sides PR and QR and ∠P=∠RTS.
∠RPQ=∠RTS [ Given ]
∠QRP=∠SRT[ Common ]
△RPQ∼△RTS [ By AA Similarity Criterion ]
Proved.
Q6. In Figure, if △ABC≅△ACD, show that △ADE∼△ABC.
In △ABE and △ACD
△ABE≅△ACD [Given ]
AB=AC_______(1) [CPCT]
AE=AD_______(2) [CPCT]
From equation (1) & (2)
AEAB=ADAC____(3)
{In}△ADE and △ABC
∠EAD=∠CAB [Common]
AEAB=ADAC [ from equation (3)]
here corresponding sides are proportional and angle are equal.
So that
△ADE∼△ABC [By SAS Similarity Criterion]
Proved.
Q7. In 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
Sol. :
∠D=∠E=90∘ [ Given]
i) In △AEP and △CDP
∠E=∠D [Given ]
∠EPA=∠DPC [ Vertically opposite angles]
△AEP∼△CDP[By AA Similarity Criterion ]
Proved.
ii) In △ABD and △CBE
∠DBA=∠EBC [Common]
∠D=∠E [Given]
△ABD∼△CBE [By AA Similarity Criterion]
Proved.
iii) In △APE and △ADB
∠PAE=∠BAD [Common]
∠E=∠D [Given]
△APE∼△ADB [By AA Similarity Criterion]
Proved.
iv) In △PDC and △BEC
∠D=∠E [Given]
∠DCP=∠BCE [Common]
△PDC∼△BEC [By AA Similarity Criterion]
Proved.
Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ABE∼△CFB.
Sol.
In given figure, ABCD is parallelogram in which E is a point on side AD produced and BE intersect CD at F.
In △ABE and △ CFB
∠A=∠C [Opposite angle of parallelogram]
∠AEB=∠CBF [Alternat angle]
△ABE∼△CFB [By AA]
Proved.
Q9. In Figure, ABCandAMParetworight△s,right∠datBandM` respectively. Prove that:
i) △ABC∼△AMP
ii) CAPA=BCMP
i)
In △ABC and △{AMP}
∠ABC=∠AMP=90∘ [Given]
∠BAC=∠PAM [Common]
△ABC∼△AMP [By AA]
Proved.
ii)
△ABC∼△AMP [Proved]
CAPA=BCMP
[Ratio of the corresponding sides of similar triangles]
Proved.
Q10. CD and GH are respectively the bisectors 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) CDGH=ACFG
ii) △DCB∼△HGE
iii) △DCA∼△HGE
Sol.
i)
In △ACD and △FGH
∠A=∠F[△ABC∼△FEG]
∠1=∠2[12∠C=12∠G]
In △ACD∼△FGH[By AA]
CDGH=ACFG
[Ratio of the corresponding sides of similar triangles]
Proved.
ii)
CDGH=ACFG [Proved]
ACFG=BCEG
CDGH=BCEG
△DCB and △HGE
∠3=∠4[12∠C=12∠G]
CDGH=BCEG[Proved]
△DCB∼△HGE [By SAS]
Proved.
iii)
In △DCA and △HGE
∠1=∠2[12∠C=12∠G]
CDGH=ACFG [Proved]
△DCA∼△HGE [By SAS]
Proved.
Q11. In Figure, E is a point in side CB produced of an isosceles triangle ABC with AB=AC. If AD⊥BC and EF⊥AC, prove that △ABD∼△ECF.
In △ABC
AB=AC [Given]
∠ABC=∠ACB_____(1)
[Angles opposite to equal sides are equal.]
In △ABD and △ECF
∠ABD=∠ECF [From (1)]
∠ADB=∠EFC=90∘[AD⊥BC and EF⊥AC]
△ABD∼△ECF [By AA]
Proved.
Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △PQR ( See Figure ). Show that △ABC∼△PQR..
In △ABC and △{PQR}
ABPQ=BCQR=ADPM [Given]
ABPQ=12(BC)12(QR)=ADPM
ABPQ=BDQM=ADPM
△ABD∼△PQM [By SAS]
∠B=∠Q [CPCT]
In △ABC and △{PQR}
ABPQ=BCQR [Given]
∠B=∠Q [Proved]
△ABC∼△PQR [By SAS]
Proved.
Q13. D is a point on the side BC of a triangle ABC such that ∠ADC=∠BAC. Show that CA2=CB.CD.
Sol.
In △ABC and △DAC
∠BAC=∠ADC [Given]
∠ACB=∠ACD [Common]
△ABC∼△DAC [By AA]
CACD=CBCD[Corresponding sides are proportional]
CA2=CB.CD
Proved.
Q14. 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 △ABc∼△PQR..
Sol.
In △ABC and △PQR
ABPQ=ACQR=ADPM__(1)[Given]
[Here the medians are in the same ratio, so the side bisects the medians of the same ratio is also proportional.]
ADPM=BCQR____(2)
From (1) and (2)
ABPQ=ACRP=BCQR__(3)
ABPQ=BCQR=CARP[From (3)]
△ABc∼△PQR[By SSS]
Proved.
Q15. 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 28m long. Find the height of the tower.
Let PQ = h is tower and QR is the shadow of the tower
AB is pole and BC is the shadow of the pole
In △ABC and △PQR
∠C=∠R [Angle of elevation of the sun]
∠ABC=∠PQR [Given]
△ABC∼△PQR [By AA]
ABBC=PQQR [Sides of similar triangles]
64=h28
h=6×284
h=6×7
h=42 m
Answer
The height of the tower is 42 m.
Q16. If AD and PM are medians of triangles ABC and PQR, respectively where △ABC∼△PQR, prove that ABPQ=ADPM.
Sol.
△ABC∼△PQR [Given]
∠ABC=∠PQR__(1)[Angles of similar triangles]
ABPQ=BCQR [Sides of similar triangles]
ABPQ=12BC12QR
ABPQ=BDQM
In △ABD and △PQM
ABPQ=BDQM [Proved]
∠ABD=∠PQM[From (1)]
△ABD∼△PQM [By SAS]
ABPQ=ADPM [Sides of similar triangles]
Proved.
Comments