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

NCERT Class 10th Maths Projects

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 :

Sol. :

`text{In }triangletext{ABC and }triangletext{PQR}`

`angletext{A}=angletext{P}`

`angletext{B}=angletext{Q}`

`angletext{C}=angletext{R}`

`triangletext{ABC}∼triangletext{PQR}``text{ [By AAA]}`

`text{Answer :}`

`text{Yes, AAA, }``triangletext{ABC}∼triangletext{PQR}`.

Sol. :

`text{In } triangletext{ABC and }triangletext{PQR}`

`(text{AB})/(text{QR})=2/4=1/2`

`(text{AC})/(text{PQ})=3/6=1/2`

`(text{BC})/(text{PR})=(2.5)/5=1/2`

`(text{AB})/(text{QR})=(text{AC})/(text{PQ})=(text{AB})/(text{QR})=1/2`

`triangletext{ABC}∼triangletext{PQR}``text{ [By SSS]}`

`text{Answer :}`

`text{Yes, SSS,}``triangletext{ABC}∼triangletext{PQR}.`

`text{Sol. :}`

`text{In }triangletext{LMP and }triangletext{DEF}`

`(text{MP})/(text{DE})=2/4=1/2`

`(text{PL})/(text{DF})=3/6=1/2`

`(text{ML})/(text{EF})=(2.7)/5`

`(text{MP})/(text{DE})=(text{PL})/(text{DE})ne(text{MP})/(text{EF})`

`triangletext{LMP}≁triangletext{DEF}`

`text{Answer :}`

`text{No, } triangletext{LMP}triangletext{DEF}`.

`text{Sol. :}`

`text{In } triangletext{LMN and }triangletext{PQR}`

`(text{MN})/(text{PQ})=(2.5)/5`

`angletext{M}=angletext{Q}=70^circ`

`(text{ML})/(text{EF})=5/10=1/2`

`(text{MN})/(text{PQ})=(text{ML})/(text{EF})=1/2 text{ and } angletext{M}=angletext{Q}=70^circ`

`triangletext{LMN}∼triangletext{PQR}``text{ [By SAS Similarity criterion]}`

`text{Answer :}`

`text{Yes, SAS, } triangletext{LMN}∼triangletext{PQR}`.

`text{Sol. :}`

`text{In } triangletext{ABC and }triangletext{DEF}`

`(text{AB})/(text{DF})=(2.5)/5=1/2`

`(text{BC})/(text{FE})=3/6=1/2`

`angletext{A}=angletext{F}=80^circ`

`text{where }angletext{F  is included but }angletext{A is not included.}`

`text{So that}`

`triangletext{ABC}≁triangletext{DEF}`

`text{Answer :}`

`text{No.}`

`text{Sol. :}`

`text{In } triangletext{DEF and }triangletext{PQR}`

`angletext{F}=180^circ-(angletext{F}+angletext{E})`

`angletext{F}=180^circ-(70^circ+80^circ)`

`angletext{F}=180^circ-150^circ`

`angletext{F}=30^circ`

`angletext{P}=180^circ-(angletext{Q}+angletext{R})`

`angletext{P}=180^circ-(80^circ+30^circ)`

`angletext{P}=180^circ-110^circ`

`angletext{P}=70^circ`

`text{then}`

`angletext{D}=angletext{P}=70^circ`

`angletext{E}=angletext{Q}=80^circ`

`angletext{F}=angletext{R}=30^circ`

`triangletext{DEF}∼triangletext{PQR}``text{  [By AAA Similarity criterion]}`

`text{Answer :}`

`text{Yes, AAA, } triangletext{DEF}∼triangletext{PQR}`.

Q2. In Figure, `triangleODC∼triangleOBA, angleBOC=125^circ` and `angle CDO=70^circ`. Find `angleDOC, angleDCO` and `angleOAB.`

`text{Sol.}`

`angle text{DOC}+125^circ = 180^circ``text{ [Linear pair]}`

`angle text{DOC} = 180^circ - 125^circ`

`angle text{DOC} = 55^circ`

`text{In} triangle text{DOC}`

`text{By angle sum property of }triangle`

`angle text{DCO} + 70^circ + 55^circ = 180^circ`

`angle text{DCO} = 180^circ - 125^circ`

`angle text{DCO} = 55^circ`

`text{And}`

`triangle text{ODC}∼triangletext{OBA},``text{  [Given]}`

`angle text{OAB} = angle text{DCO} = 55^circ`

`text{Answer:}`

`angle text{DOC}=55^circ,`

`angle text{DCO}=55^circ,`

`angle text{OAB}=55^circ`.

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 `(OA)/(OC)=(OB)/(OD)`.

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`.

`text{In }triangletext{ADC}`,

`text{EF∥DC}``text{  [By construction]}` 

`(text{AE})/(text{ED})=(text{AO})/(text{CO})`______(1)`text{[By Theroem 6.1]}`

`text{[Theorem 6.1: If a line is drawn parallel to one }``text{side of a triangle to intersect the other two sides }``text{in distinct points, the other two sides are }``text{divided in the same ratio.]}`

`text{In }triangletext{BAD}`

`text{EF∥AB}``text{      [By construction]}`

`(text{AE})/(text{ED})=(text{BO})/(text{DO})`______(2)`text{[By Theroem 6.1]}`

`text{from equation (1) and (2)}`

`(text{AO})/(text{CO})=(text{BO})/(text{DO})`

`(text{AO})/(text{BO})=(text{CO})/(text{DO})`

`text{Proved}`.

Q4. In Figure, `(QR)/(QS)=(QT)/(PR)` and `angle1=angle2`. Show that `trianglePQS∼triangleTQR.`

`text{Sol.}`

`text{In } triangletext{PQR,}`

`angle1=angle2``text{ [Given ]}`

`angletext{Q}=angletext{R}``text{ [According to Figure]}`

`text{PQ}=text{PR}`_____(1)`text{[Side opposite to equal angle is equal]}`

`(text{QR})/(text{QS})=(text{QT})/(text{PR})`_____(2)`text{ [Given ]}`

`text{from equation (1) and (2)}`

`(text{QR})/(text{QS})=(text{QT})/(text{PR})=(text{QT})/(text{QP})`

`(text{QS})/(text{QP})=(text{QR})/(text{QT})`______(3)

`text{In } triangletext{PQS and }triangle{TQR}`

`angletext{PQS}=angletext{TQR}`

`(text{QS})/(text{QP})=(text{QR})/(text{QT})``text{ [from equation (3)]}`

`triangletext{PQS}∼triangletext{TQR}``text{ [By SAS Similarity criterion]}` 

`text{Proved}`.

 Q5. `S` and `T` are points on sides `PR` and `QR` of `trianglePQR` such that `angleP=angleRTS.` Show that  `triangleRPQ∼triangleRTS.`

`text{Sol. :}`

`text{In }triangletext{RPQ and }triangletext{RTS}`

Point `S` and `T` are points on sides `PR` and `QR` and `angleP=angleRTS`.

`angletext{RPQ}=angletext{RTS}``text{  [ Given ]}`

`angletext{QRP}=angletext{SRT}``text{[ Common ]}`

`triangletext{RPQ}∼triangletext{RTS}``text{ [ By AA Similarity Criterion ]}`

`text{Proved}`.

Q6. In Figure, if `triangleABCcongtriangleACD`, show that `triangleADE∼triangleABC.`

`text{Sol. :}`

`text{In }triangletext{ABE and }triangletext{ACD}`

`triangletext{ABE}congtriangletext{ACD}``text{ [Given ]}`

`text{AB=AC}`_______(1) `text{ [CPCT]}`

`text{AE=AD}`_______(2) `text{  [CPCT]}`

`text{From equation (1) & (2)}`

`(text{AE})/(text{AB})=(text{AD})/(text{AC})`____(3)

`{In }triangletext{ADE and }triangletext{ABC}`

`angletext{EAD}=angletext{CAB}``text{ [Common]}`

`(text{AE})/(text{AB})=(text{AD})/(text{AC})``text{ [ from equation (3)]}`

`text{here corresponding sides are proportional }``text{and angle are equal.}`

`text{So that}`

`triangletext{ADE}∼triangletext{ABC}``text{ [By SAS Similarity Criterion]}`

`text{Proved}`.

Q7. In Figure, altitudes `AD` and `CE` of `triangleABC` intersect each other at the point `P`. Show that:

i) `triangleAEP∼triangleCDP`

ii) `triangleABD∼triangleCBE`

iii) `triangleAEP∼triangleADB`

iv) `trianglePDC∼triangleBEC`

`text{Sol. :}`

`angletext{D}=angletext{E}=90^circ``text{ [ Given]}`

i) `text{In }triangletext{AEP and }triangletext{CDP}`

`angletext{E}=angletext{D}``text{  [Given ]}`

`angletext{EPA}=angletext{DPC}``text{  [ Vertically opposite angles]}`

`triangletext{AEP}∼triangletext{CDP}``text{[By AA Similarity Criterion ]}`

`text{Proved}`.

ii) `text{In }triangletext{ABD and }triangletext{CBE}`

`angletext{DBA}=angletext{EBC}``text{ [Common]}`

`angletext{D}=angletext{E}``text{      [Given]}`

`triangletext{ABD}∼triangletext{CBE}``text{ [By AA Similarity Criterion]}`

`text{Proved}`.

iii) In `triangle text{APE and }triangletext{ADB}`

`angletext{PAE}=angletext{BAD}``text{ [Common]}`

`angletext{E}=angletext{D}``text{      [Given]}`

`triangletext{APE}∼triangletext{ADB}``text{ [By AA Similarity Criterion]}`

`text{Proved.}`

iv) `text{In }triangletext{PDC and }triangletext{BEC}`

`angletext{D}=angletext{E}``text{      [Given]}`

`angletext{DCP}=angletext{BCE}``text{ [Common]}`

`triangletext{PDC}∼triangletext{BEC}``text{ [By AA Similarity Criterion]}`

`text{Proved}`.

Q8. `E` is a point on the side `AD` produced of a parallelogram `ABCD` and `BE` intersects `CD` at `F`. Show that  `triangleABE∼triangleCFB`.

`text{Sol.}`

`text{In given figure, ABCD is parallelogram in }``text{which E is a point on side AD produced and BE }``text{intersect CD at F}.`

`text{In }triangle text{ABE and } triangle text{ CFB}`

`angle text{A} = angle text{C}``text{  [Opposite angle of parallelogram]}`

`angle text{AEB} = angle text{CBF}``text{   [Alternat angle]}`

`triangle text{ABE}∼triangle text{CFB}``text{ [By AA]}`

`text{Proved}`.

Q9. In Figure, `ABC and `AMP` are two right triangles, right angled at `B` and `M` respectively. Prove that:

i) `triangleABC∼triangleAMP`

ii) `(CA)/(PA)=(BC)/(MP)`

`text{Sol.}`

i)

`text{In }triangle text{ABC and }triangle{AMP}`

`angletext{ABC}=angletext{AMP}=90^circ``text{     [Given]}`

`angle text{BAC}=angle text{PAM}``text{       [Common]}`

`triangle text{ABC}∼triangle text{AMP}``text{   [By AA]}`

`text{Proved}`.

ii)

`triangle text{ABC}∼triangle text{AMP}``text{   [Proved]}`

`(text{CA})/(text{PA})=(text{BC})/(text{MP})`

`text{[Ratio of the corresponding sides of similar triangles]}`

`text{Proved}`.

Q10. `CD` and `GH` are respectively the bisectors of `angleACB` and `angleEGF` such that D and `H` lie on sides `AB` and `FE` of `triangleABC` and `triangleEFG` respectively. If  `triangleABC ∼triangleFEG`, show that

i) `(CD)/(GH)=(AC)/(FG)`

ii) `triangleDCB∼triangleHGE`

iii) `triangleDCA∼triangleHGE`

`text{Sol.}`

i)

`text{In }triangle text{ACD and }triangle text{FGH}`

`angle text{A}=angle text{F}``[triangletext{ABC}∼triangle text{FEG}]`

`angle 1 = angle 2``   [1/2angletext{C}=1/2angletext{G}]`

`text{In }triangle text{ACD}∼triangle text{FGH}``text{[By AA]}`

`(text{CD})/(text{GH})=(text{AC})/(text{FG})`

`text{[Ratio of the corresponding sides of similar }``text{triangles]}`

`text{Proved}`.

ii)

`(text{CD})/(text{GH})=(text{AC})/(text{FG})``text{   [Proved]}`

`(text{AC})/(text{FG})=(text{BC})/(text{EG})`

`(text{CD})/(text{GH})=(text{BC})/(text{EG})`

`triangle text{DCB and }triangle text{HGE}`

`angle 3 = angle 4``   [1/2angletext{C}=1/2angletext{G}]`

`(text{CD})/(text{GH})=(text{BC})/(text{EG})``text{[Proved]}`

`triangletext{DCB}∼triangletext{HGE}``text{ [By SAS]}`

`text{Proved}`.

iii) 

`text{In }triangletext{DCA and }triangletext{HGE}`

`angle 1 = angle 2``   [1/2angletext{C}=1/2angletext{G}]`

`(text{CD})/(text{GH})=(text{AC})/(text{FG})``text{   [Proved]}`

`triangletext{DCA}∼triangletext{HGE}``text{  [By SAS]}`

`text{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 `triangleABD∼triangleECF.`

`text{Sol.}`

`text{In }triangle text{ABC}`

`text{AB=AC}``text{ [Given]}`

`angle text{ABC}=angle text{ACB}`_____(1)

`text{[Angles opposite to equal sides are equal.]}`

`text{In }triangle text{ABD and }triangle text{ECF}`

`angle text{ABD}=angle text{ECF}``text{  [From (1)]}`

`angle text{ADB}=angle text{EFC}=90^circ``text{[AD⊥BC and EF⊥AC]}`

`triangle text{ABD}∼triangle text{ECF}``text{  [By AA]}`

`text{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 `trianglePQR` ( See Figure ). Show that `triangleABC∼trianglePQR.`.

`text{Sol.}`

`text{In }triangle text{ABC and }triangle{PQR}`

`(text{AB})/(text{PQ})=(text{BC})/(text{QR})=(text{AD})/(text{PM})``text{  [Given]}`

`(text{AB})/(text{PQ})=(1/2(text{BC}))/(1/2(text{QR}))=(text{AD})/(text{PM})`

`(text{AB})/(text{PQ})=(text{BD})/(text{QM})=(text{AD})/(text{PM})`

`triangle text{ABD}∼triangle text{PQM}``text{  [By SAS]}`

`angle text{B}=angle text{Q}``text{  [CPCT]}`

`text{In }triangle text{ABC and }triangle{PQR}`

`(text{AB})/(text{PQ})=(text{BC})/(text{QR})``text{  [Given]}`

`angle text{B}=angle text{Q}``text{  [Proved]}`

`triangle text{ABC}∼triangle text{PQR}``text{  [By SAS]}`

`text{Proved}`.

Q13. `D` is a point on the side `BC` of a triangle `ABC` such that `angleADC=angleBAC.` Show that `CA^2=CB.CD.`

`text{Sol.}`

`text{In }triangletext{ABC and }triangle text{DAC}`

`angletext{BAC}=angletext{ADC}``text{  [Given]}`

`angletext{ACB}=angletext{ACD}``text{  [Common]}`

`triangletext{ABC}∼triangle text{DAC}``text{ [By AA]}`

`(text{CA})/(text{CD})=(text{CB})/(text{CD})``text{[Corresponding sides are proportional]}`

`text{CA}^2=text{CB.CD}`

`text{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  `triangleABc∼trianglePQR.`.

`text{Sol.}`

`text{In }triangletext{ABC and }triangletext{PQR}`

`(text{AB})/(text{PQ})=(text{AC})/(text{QR})=(text{AD})/(text{PM})`__(1)`text{[Given]}`

`text{[Here the medians are in the same ratio, so }``text{the side bisects the medians of the same ratio}` `text{is also proportional.]}`

`(text{AD})/(text{PM})=(text{BC})/(text{QR})`____(2)

`text{From (1) and (2)}`

`(text{AB})/(text{PQ})=(text{AC})/(text{RP})=(text{BC})/(text{QR})`__(3)

`(text{AB})/(text{PQ})=(text{BC})/(text{QR})=(text{CA})/(text{RP})``text{[From (3)]}`

`triangleABc∼trianglePQR``text{[By SSS]}`

`text{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 `28`m long. Find the height of the tower.

`text{Sol.}`

`text{Let PQ = h is tower and QR is the shadow of the tower}`

`text{AB is pole and BC is the shadow of the pole}`

`text{In } triangle text{ABC and }triangle text{PQR}`

`angle text{C} = angle text{R}``text{ [Angle of elevation of the sun]}` 

`angle text{ABC}=angletext{PQR}``text{  [Given]}`

`triangle text{ABC}∼triangle text{PQR}``text{ [By AA]}`

`(text{AB})/(text{BC})=(text{PQ})/(text{QR})``text{ [Sides of similar triangles]}`

`6/4=(text{h})/28`

`text{h}=(6times28)/4`

`text{h}=6times7`

`text{h=42 m}`

`text{Answer}`

`text{The height of the tower is 42 m.}`

Q16. If `AD` and `PM` are medians of triangles `ABC` and `PQR`, respectively where `triangleABC∼trianglePQR`, prove that `(AB)/(PQ)=(AD)/(PM). ` 

`text{Sol.}`

`triangletext{ABC}∼triangletext{PQR}``text{ [Given]}`

`angle text{ABC}=angle text{PQR}`__(1)`text{[Angles of similar triangles]}`

`(text{AB})/(text{PQ})=(text{BC})/(text{QR})``text{ [Sides of similar triangles]}`

`(text{AB})/(text{PQ})=(1/2text{BC})/(1/2text{QR})`

`(text{AB})/(text{PQ})=(text{BD})/(text{QM})`

`text{In } triangle text{ABD and }triangle text{PQM}`

`(text{AB})/(text{PQ})=(text{BD})/(text{QM})``text{ [Proved]}`

`angletext{ABD}=angletext{PQM}``text{[From (1)]}`

`triangle text{ABD}∼triangle text{PQM}``text{ [By SAS]}`

`(text{AB})/(text{PQ})=(text{AD})/(text{PM})``text{ [Sides of similar triangles]}`

`text{Proved}`.