Solutions:
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.
2. In Fig., ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
In Figure, ∆COD is similiar to ∆ AOB.
DOB is a straight line.
Therefore ∠DOC +∠COB = $180^0$
=> ∠DOC = $180^0$ - $125^0$ = $55^0$
In ∆ DOC = ∠DCO + ∠CDO +∠DOC = $180^0$
=> ∠DCO +$55^0$ + $70^0$ = $180^0$
=> ∠DCO = $180^0$ - $55^0$ - $70^0$
=> ∠DCO = $55^0$
Given that ∆ODC is similiar to ∆OBA .
Therefore ∠OAB = ∠OCD
∠OAB = $55^0$
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
${OA}/{OC}={OB}/{OD}$
In ∆DOC and ∆BOA,
∠DOC = ∠BOA (Vertically oposite angle)
∠CDO = ∠ABO (Alternate angles)
∠DCO = ∠BAO (Alternate angles)
Therefore by AAA similarity, ∆DOC and ∆BOA are similar.
Therefore ${DO}/{BO}={CO}/{AO}$
i.e ${OA}/{OC}={OB}/{OD}$ . Hence Proved
4. In Fig., ${QR}/{QS}={QT}/{PR}$ and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
In ∆PQR, ∠PQR = ∠PRQ
Therefore PQ=PR
Given, ${QR}/{QS}={QT}/{PR}$
therefore ${QR}/{QS}={QT}/{PQ}$ ..........(1)
In ∆PQS and ∆TQR ,
${QR}/{QS}={QT}/{PQ}$ .........from eqn (1)
and
∠PQS = ∠TQR (∠Q is common)
Therefore by SAS similarity,∆PQS and ∆TQR are similar.
5. S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆RPQ ~ ∆RTS.
In ∆RPQ and ∆RST,
∠RTS = ∠QPS (given)
∠R = ∠R (common)
Therefore by AA similarity , ∆RPQ ~ ∆RTS
6. In Fig., if ∆ ABE ≅ ∆ ACD, show that ∆ADE ~ ∆ABC.
Given ∆ ABE ≅ ∆ ACD,
therefore AB = AC
and AD = AE
In ∆ADE and ∆ABC,
${AD}/{AB}$ = ${AE}/{AC}$
and ∠A = ∠A (common angle)
Therefore by SAS similarity,∆ADE and ∆ABC are similar.
7. In Fig., 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
(i) In ∆AEP and ∆CDP,
∠APE = ∠CPD (vertically opposite angle)
∠AEP = ∠CDP (each 90 degree)
Therefore by AA similarity, ∆AEP ~ ∆CDP
(ii) In ∆ABD and ∆CBE
∠ADB = ∠CEB (each 90 degree)
∠ABD = ∠CBE (common)
Therefore by AA similarity, ∆ABD ~ ∆CBE
(iii) In ∆AEP and ∆ADB
∠AEP = ∠ADB (each 90 degree)
∠PAE = ∠DAB (Common)
Therefore by AA similarity, ∆AEP ~ ∆ADB
(iv) In ∆PDC and ∆BEC
∠PDC = ∠BEC (each 90 degree)
∠PCD = ∠BCE (Common)
Therefore by AA similarity, ∆PDC ~ ∆BEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
Show that ∆ABE ~ ∆CFB.
In ∆ABE and ∆CFB,
∠A = ∠C (opposite angle of parallelogram)
∠AEB = ∠CBF (Altenate angle as AE || BC)
Therefore by AA similarity,∆ABE ~ ∆CFB
9. In Fig., ABC and AMP are two right triangles, right angled at B and M respectively.
Prove that:
(i) ∆ABC ~ ∆AMP
(ii) ${CA}/{PA}={BC}/{MP}$
(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP (each 90 degree)
∠A =∠A (Common)
Therefore by AA similarity, ∆ABC ~ ∆AMP
(ii) Since ∆ABC ~ ∆AMP ,
=> ${AB}/{AM}$ = ${BC}/{MP}$= ${CA}/{PA}$
Hence Proved.
10. 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) ${CD}/{GH}={AC}/{FG}$
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
(i) Given,
∆ABC ~ ∆FEG,
$.^{.}.$ ∠A =∠F , ∠B =∠E
and ∠ACB = ∠FGE
$.^{.}.$ ∠ACD = ∠FGH (angle bisector)
and ∠DCB = ∠HGE (angle bisector)
In ∆ACD and ∆FGH,
∠A=∠F and ∠ACD = ∠FGH,
$.^{.}.$ by AA similarity, ∆ACD ~ ∆FGH ,
=> ${CD}/{GH} = {AC}/{FG}$
(ii) In ∆DCB and ∆HGE,
∠DCB = ∠HGE
∠B = ∠E
$.^{.}.$ by AA similarity,∆DCB ~ ∆HGE
(iii) In ∆DCA and ∆HGF,
∠ACD = ∠FGH,
∠A = ∠F
$.^{.}.$ by AA similarity,∆DCA ~ ∆HGF
11. In Fig. , 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.
Given that , ABC ia an isosceles triangle,
$.^{.}.$ AB=AC, ∠ABD = ∠ECF.
In ∆ABD and ∆ECF,
∠ADB = ∠EFC (each 90 degree)
∠ABD = ∠ ECF
$.^{.}.$ by AA similarity,∆ABD ~ ∆ECF Hence proved
12. 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 Fig. 6.41). Show that ∆ABC ~ ∆PQR.
AD and PM are the median of triangle.Therefore,
$BD=1/2BC$ and $QM=1/2QR$
Given ,${AB}/{PQ}={BC}/{QR}={AD}/{PM}$
=> ,${AB}/{PQ}={2BD}/{2QM}={AD}/{PM}$
=> ,${AB}/{PQ}={BD}/{QM}={AD}/{PM}$
In ∆ABD and ∆PQM,
${AB}/{PQ}={BD}/{QM}={AD}/{PM}$
$.^{.}.$ ∆ABD ~ ∆PQM
=> ∠ABD = ∠PQM
In ∆ABC and ∆PQR,
∠ABD = ∠PQM
${AB}/{PQ}={BC}/{QR}$
$.^{.}.$ ∆ABC ~ ∆PQR
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
In ∆ADC = ∆BAC
∠ADC = ∠BAC,
∠ACD = ∠BCA,
$.^{.}.$ ∆ADC ~ ∆BAC
Hence, ${CA}/{CB}={CD}/{CA}$
$CA^2 = CB × CD$
Hence Proved
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 ∆ABC ~ ∆PQR.
Given,
${AB}/{PQ} = {AC}/{PR}={AD}/{PM}$
Extend AD to E and PM to L such that AD=DE and PM=ML .
Now join B to E and C to E and Q to L and R to L.
AD and PM are medians of triangle, therefore
BD=CD and QM =MR
So, the diagonals of AE and BC are bisecting at D , Therefore ABEC is a parallelogram.
Therefore AC=BE ,AB=CE ,
Similarily PQLR is a parallelogram.
Therefore PQ =RL and PR =QL
Given that ${AB}/{PQ} = {AC}/{PR}={AD}/{PM}$
=> ${AB}/{PQ} = {BE}/{QL} = {(AE)/2}/{PL}/2$
=> ${AB}/{PQ} = {BE}/{QL}={AE}/{PL}$
Therefore ∆ABE ~ ∆PQL
Similarily , ∆AEC ~ ∆PLR
Since the triangles are similar,
∠BAE = ∠QPL ...........(1)
and ∠CAE = ∠RPL ..........(2)
Adding eqn (1) and (2), we get
∠BAE + ∠CAE= ∠QPL + ∠RPL
=> ∠CAB = ∠RPQ
In ∆ABC and ∆PQR
${AB}/{PQ} = {AC}/{PR}$ .......(given)
and ∠CAB = ∠RPQ
Therefore by SAS similarity , ∆ABC ~ ∆PQR Hence Proved
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.
Let CD be the vertical pole and its shadow be DF
Let AB be the tower and its shadow be BE.
So, in ∆ABE and ∆CDF,
∠DCF = ∠BAE,
∠CDF = ∠ABE (each $90^0$)
By AA similarity,
$.^{.}.$ ∆ABE ~ ∆CDF
${AB}/{CD}$ = ${BE}/{DF}$
${AB}/{6}$ = ${28}/{4}$
=> $AB$ = 42 $m$
Ans: Height of tower is 42 $m$
16. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR,
prove that ${AB}/{PQ}={AD}/{PM}$.
Given that:∆ABC ~ ∆PQR,
therefore , ${AB}/{PQ}={BC}/{QR}={AC}/{PR}$ .......(1)
and ∠A=∠P , ∠B=∠Q , ∠C=∠R ........(2)
AD and PM are the median of triangle, therefore,
BD =${BC}/2$ and $QM = {QR}/2$ ..........(3)
From the eqn (1) and (3)
${AB}/{PQ}={BD}/{QM}$
In ∆ABD and ∆PQM
∠B=∠Q
${AB}/{PQ}={BD}/{QM}$
Therefore , by SAS similarity , ∆ABD ~ ∆PQM
${AB}/{PQ}={BD}/{QM}={AD}/{PM}$ Hence proved
