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Triangles : Exercise - 6.3 (Mathematics NCERT Class 10th)


               CLICK HERE to watch the second part

Q.1    State which pairs of triangle 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 : 121314151617Sol.

(i) In \Delta s ABC and PQR, we observe that
\angle A = \angle P = 60^\circ , \angle B = \angle Q = 80^\circ and \angle C = \angle R = 40^\circ
Therefore by AAA criterion of similarity,
\Delta ABC \sim \Delta PQR
(ii) In \Delta s ABC and PQR, we observe that
{{AB} \over {QR}} = {{BC} \over {RP}} = {{CA} \over {PQ}} = {1 \over 2}
Therefore by SSS criterion of similarity, \Delta ABC \sim \Delta QRP
(iii) In \Delta s LMP and DEF, we observe that the ratio of the sides of these triangles are not equal.
So, the two triangles are not similar.
(iv) In \Delta s MNL and QPR, we observe that
\angle M = \angle Q = 70^\circ
But, {{MN} \over {PQ}} \ne {{ML} \over {QR}}
Therefore These two triangles are not similar as they do not satisfy SAS criterion of similarity.
(v) In \Delta s ABC and FDE, we have
\angle A = \angle F = 80^\circ
But, {{AB} \over {DF}} \ne {{AC} \over {EF}} [Because AC is not given]
So, by SAS criterion of similarity, these two triangless are not similar.
(vi) In \Delta s DEF and PQR
\angle D = \angle P = 70^\circ [Because \angle P = 180^\circ - 80^\circ - 30^\circ = 70^\circ ]
\angle E = \angle Q = 80^\circ
So, by AAA criterion of similarity,
\Delta DEF \sim \Delta PQR


Q.2      In figures, \Delta ODC \sim \Delta OBA, \angle BOC = 125^\circ and \angle CDO = 70^\circ . Find \angle DOC, \angle DCO and \angle OAB.
18Sol.

Since BD is a line and OC is a ray on it,
Therefore \angle DOC + \angle BOC = 180^\circ
 \Rightarrow \angle DOC + 125^\circ = 180^\circ
 \Rightarrow \angle DOC = 180^\circ - 125^\circ = 55^\circ
In \Delta CDO, we have
\angle CDO + \angle DOC + \angle DCO = 180^\circ
 \Rightarrow 70^\circ + 55^\circ + \angle DCO = 180^\circ
 \Rightarrow \angle DCO = 180^\circ - 125^\circ = 55^\circ
It is given that \Delta ODC \sim \Delta OBA
\angle OBA = \angle ODC,
\angle OAB = \angle OCD
 \Rightarrow \angle OBA = 70^\circ and \angle OAB = 55^\circ
Hence, \angle DOC = 55^\circ , \angle DCO = 55^\circ
and \angle OAB = 55^\circ


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 {{OA} \over {OC}} = {{OB} \over {OD}}.
Sol.

Given : ABCD is a trapezium in which AB || DC.
To prove : {{OA} \over {OC}} = {{OB} \over {OD}}
Proof : In \Delta s OAB and OCD, we have
\angle AOB = \angle COD [Vert. opp. \angle s]
\angle OAB = \angle OCD [Alternate \angle s]
and \angle OBA = \angle ODC [Alternate \angle s]
19Therefore by AAA criterion of similarity,
\Delta OAB \sim \Delta ODC
Hence, {{OA} \over {OC}} = {{OB} \over {OD}}
[Because in case of two similar triangles, the ratio of their corresponding sides are equal]


Q.4      In figure, {{QR} \over {QS}} = {{QT} \over {PR}} and \angle 1 = \angle 2. Show that \Delta PQS \sim \Delta TQR
20Sol.

We have, {{QR} \over {QS}} = {{QT} \over {PR}}
 \Rightarrow {{QT} \over {QR}} = {{PR} \over {QS}} .............. (1)
Also, \angle 1 = \angle 2 [Given]
 \Rightarrow PR = PQ .............. (2) [Because sides opp. to equal \angle s are equal]
From (1) and (2), we get
{{QT} \over {QR}} = {{PQ} \over {QS}} \Rightarrow {{PQ} \over {QT}} = {{QS} \over {QR}} ........... (3)
Therefore in \Delta s PQS and TQR, we have
{{PQ} \over {QT}} = {{QS} \over {QR}} and \angle PQS = \angle TQR = \angle Q
Therefore by SAS criterion of similarity,
\Delta PQS \sim \Delta TQR.


Q.5      S and T are points on sides PR and QR of \Delta PQR such that \angle P = \angle RTS. Show that \Delta RPQ \sim \Delta RTS.
Sol.
21

In \Delta s RPQ and RTS, we have
\angle RPQ = \angle RTS [Given]
\angle PRQ = \angle TRS [Common]
Therefore by AA criterion of similarity,
\Delta RPQ \sim \Delta RTS


Q.6      In figure, if \Delta ABE \cong \Delta ACD, show that \Delta ADE \sim \Delta ABC.
22Sol.

It is given that
\Delta ABE \cong \Delta ACD
Therefore AB = AC [Because corresponding parts of congruent triangles are equal]
and, AE = AD
 \Rightarrow {{AB} \over {AD}} = {{AC} \over {AE}}
 \Rightarrow {{AB} \over {AC}} = {{AD} \over {AE}} .......... (1)
Therefore In \Delta s ADE and ABC, we have
{{AB} \over {AC}} = {{AD} \over {AE}} [Because of (1)]
and, \angle BAC = \angle DAE [Common]
Thus, by SAS criterion of similarity,
\Delta ADE \sim \Delta ABC


Q.7       In figure, altitudes AD and CE of \Delta ABC intersect each other at the point P, Show that :
23            
(i) \Delta AEP \sim \Delta CDP
                (ii) \Delta ABD \sim \Delta CBE
                (iii) \Delta AEP \sim \Delta ADB
                (iv) \Delta PDC \sim \Delta BEC
Sol.

(i)   In \Delta s AEP and CDP, we have
       \angle AEP = \angle CDP = 90^\circ [Because CE \bot AB and AD \bot BC]
       \angle APE = \angle CPD [Vert. opp. \angle s]
       Therefore by AA - criterion of similarity, we have
       \Delta AEP \sim \Delta CDP
(ii)  In \Delta s ABD and CBE, we have
       \angle ABD = \angle CBE [Common angle]
       \angle ADB = \angle CEB = 90^\circ
       Therefore by AA-criterion of similarity, we have
       \Delta ABD \sim \Delta CBE
(iii)  In \Delta s AEP and ADB, we have
         \angle AEP = \angle ADB = 90^\circ [Becauese AD \bot BC and CE \bot AB]
         \angle PAE = \angle DAB [Common angle]
         Therefore by AA-criterion of similarity, we have
         \Delta AEP \sim \Delta ADB
(iv)   In \Delta s PDC and BEC, we have
         \angle PDC = \angle BEC = 90^\circ [Because AD \bot BC and CE \bot AB]
         \angle PCD = \angle ECB [Common angle]
         Therefore AA-criterion of similarity, we have
         \Delta PDC \sim \Delta BEC


Q.8      E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \Delta ABE \sim \Delta CFB.
Sol.
24

In \Delta s ABE and CFB, we have
\angle AEB = \angle CBF [Alt. \angle s]
\angle A = \angle C [Opp. \angle s of a ||gm]
Therefore by AA-criterion of similarity, we have
\Delta ADE \sim \Delta CFB


Q.9      In figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
               (i)  \Delta ABC \sim \Delta AMP
               (ii) {{CA} \over {PA}} = {{BC} \over {MP}}
25Sol.

(i)   In \Delta s ABC and AMP, we have
       \angle ABC = \angle AMP = 90^\circ [Given]
       and, \angle BAC = \angle MAP [Common \angle s]
       Therefore by AA, criterion of similarity, we have
       \Delta ABC \sim \Delta AMP
(ii)  We have, \Delta ABC \sim \Delta AMP [As proved above]
         \Rightarrow {{CA} \over {PA}} = {{BC} \over {MP}}


Q.10     CD and GH are respectively the bisectors of \angle ACB and \angle EGF such that D and H lie on sides AB and FE of \Delta ABC and \Delta EFG respectively. If \Delta ABC \sim \Delta FEG, show that :
                (i) {{CD} \over {GH}} = {{AC} \over {FG}}
                (ii) \Delta DCB \sim \Delta HGE
                (iii) \Delta DCA \sim \Delta HGF
Sol.
q10

We have, \Delta ABC \sim \Delta FEG
 \Rightarrow \angle A = \angle F .............. (1)
and, \angle C = \angle G
 \Rightarrow {1 \over 2}\angle C = {1 \over 2}\angle G
 \Rightarrow \angle 1 = \angle 3 and \angle 2 = \angle 4 .............. (2)
[Because CD and GH are bisectors of \angle C and \angle G respectively.
Therefore In \Delta s DCA and HGF, we have
\angle A = \angle F [From (1)]
\angle 2 = \angle 4 [From (2)]
Therefore by AA-criterion of similarity, we have
\Delta DCA \sim \Delta HGF
which proves the (iii) part.
We have, \Delta DCA \sim \Delta HGF [As proved above]d
 \Rightarrow {{AC} \over {FG}} = {{CD} \over {GH}}
 \Rightarrow {{CD} \over {GH}} = {{AC} \over {FG}}
which proves the (i) part.
In \Delta s DCB and HGE, we have
\angle 1 = \angle 3 [From (2)]
\angle B = \angle E [Because \Delta ABC \sim \Delta FEG]
Therefore by AA-criterion of similarity, we have
\Delta DCB \sim \Delta HGE
which proves, the (ii) part.


Q.11    In figure, E is a point on side CB produced of an isosceles triangle ABC with  AB = AC. If AD \bot BC and EF \bot AC, prove that \Delta ABD \sim \Delta ECF.
26Sol.

Here, \Delta ABC is isosceles with AB = AC
\angle B = \angle C
In \Delta s ABD and ECF, we have
\angle ABD = \angle ECF [Because \angle B = \angle C]
\angle ADB = \angle EFC = 90^\circ [Because AD \bot BC and EF \bot AC]
Therefore by AA-criterion of similarity,
\Delta ABD \sim \Delta 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 \Delta PQR (see figure). Show that \Delta ABC \sim \Delta PQR.
27Sol.

Given : AD is the median of \Delta ABC and PM is the median of \Delta PQR such that
{{AB} \over {PQ}} = {{BC} \over {QR}} = {{AD} \over {PM}}
To prove : \Delta ABC \sim \Delta PQR
Proof : BD = {1 \over 2}BC [Given]
and, QM = {1 \over 2}QR [Given]
Also, {{AB} \over {PQ}} = {{BC} \over {QR}} = {{AD} \over {PM}} [Given]
 \Rightarrow {{AB} \over {PQ}} = {{2BD} \over {2QM}} = {{AD} \over {PM}}
 \Rightarrow {{AB} \over {PQ}} = {{BC} \over {QR}} = {{AD} \over {PM}}
By SSS-criterion of similarity, we have
\Delta ABD \sim \Delta PQM
 \Rightarrow \angle B = \angle Q [Similar \Delta s have corresponding \angle s]
and, {{AB} \over {PQ}} = {{BC} \over {QR}} [Given]
Therefore by SAS-criterion of similarity, we have
\Delta ABC \sim \Delta PQR


Q.13    D is a point on the side BC of a triangle ABC such that \angle ADC = \angle BAC. Show that C{A^2} = CB.CD.
Sol.

In \Delta s ABC and DAC, we have
\angle ADC = \angle BAC and \angle C = \angle C
Therefore by AA-criterion of similarity, we have
\Delta ABC \sim \Delta DAC
28 \Rightarrow {{AB} \over {DA}} = {{BC} \over {AC}} = {{AC} \over {DC}}
 \Rightarrow {{CB} \over {CA}} = {{CA} \over {CD}}
 \Rightarrow C{A^2} = CB \times 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 \Delta ABC \sim \Delta PQR.
Sol.        Same as question 12.


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.

Let AB be the vertical pole and AC be the shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE = x metres.
We have, AB = 6 m, AC = 4 cm and DF = 28 m
In \Delta s ABC and DEF, we have
29\angle A = \angle D = 90^\circ ,
and \angle C = \angle F
[Because each is the angular elevation of the sun]
Therefore by AA-criterion of similarity,
\Delta ABC \sim \Delta DEF
30 \Rightarrow {{AB} \over {DE}} = {{AC} \over {DF}}
 \Rightarrow {6 \over x} = {4 \over {28}}
 \Rightarrow {6 \over x} = {1 \over 7}
 \Rightarrow x = 6 \times 7 = 42
Hence, the height of the tower is 42 metres.


Q.16    If AD and PM are medians of triangles ABC and PQR respectively, where \Delta ABC \sim \Delta PQR, prove that {{AB} \over {PQ}} = {{AD} \over {PM}}
Sol.

Given : AD and PM are the medians of \Delta s ABC and PQR respectively, where
\Delta ABC \sim \Delta PQR
To prove : {{AB} \over {PQ}} = {{AD} \over {PM}}
Proof : In \Delta s ABD and PQM, we have
\angle B = \angle Q [Because \Delta ABC \sim \Delta PQR]
{{AB} \over {PQ}} = {{{1 \over 2}BC} \over {{1 \over 2}QR}}  31[Since AD and PM are the median of BC and QR respectively]
 \Rightarrow {{AB} \over {PQ}} = {{BD} \over {QM}}
Therefore by SAS - criterion of similarity
32\Delta ABD \sim \Delta PQM
 \Rightarrow {{AB} \over {PQ}} = {{BD} \over {QM}} = {{AD} \over {PM}}
 \Rightarrow {{AB} \over {PQ}} = {{AD} \over {PM}}



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