NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 - Triangles

Q2 In Fig. 6.35, $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 ^\circ$ and $\angle CDO = 70 ^\circ$ . Find $\angle DOC , \angle DCO , \angle OAB$

Answer:

Given : $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 ^\circ$ and $\angle CDO = 70 ^\circ$

$\angle DOC+\angle BOC=180 ^\circ$ (DOB is a straight line)

$\Rightarrow \angle DOC+125 ^\circ=180 ^\circ$

$\Rightarrow \angle DOC=180 ^\circ-125 ^\circ$

$\Rightarrow \angle DOC=55 ^\circ$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 ^\circ$

$\Rightarrow 55 ^\circ+ 70 ^\circ+\angle DCO=180 ^\circ$

$\Rightarrow \angle DCO+125 ^\circ=180 ^\circ$

$\Rightarrow \angle DCO=180 ^\circ-125 ^\circ$

$\Rightarrow \angle DCO=55 ^\circ$

Since , $\Delta ODC \sim \Delta OBA$ , so

$\Rightarrow\angle OAB= \angle DCO=55 ^\circ$ ( Corresponding angles are equal in similar triangles).

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 $\frac{OA }{OC} = \frac{OB }{OD }$

Answer:

In $\triangle DOC\, and\, \triangle BOA$ , we have

$\angle CDO=\angle ABO$ ( Alternate interior angles as $AB||CD$ )

$\angle DCO=\angle BAO$ ( Alternate interior angles as $AB||CD$ )

$\angle DOC=\angle BOA$ ( Vertically opposite angles are equal)

$\therefore \triangle DOC\, \sim \, \triangle BOA$ ( By AAA)

$\therefore \frac{DO}{BO}=\frac{OC}{OA}$ ( corresponding sides are equal)

$\Rightarrow \frac{OA }{OC} = \frac{OB }{OD }$

Hence proved.

Q4 In Fig. 6.36, $\frac{QR }{QS } = \frac{QT}{PR}$ and $\angle 1 = \angle 2$ . Show that $\Delta PQS \sim \Delta TQR$

Answer:

Given : $\frac{QR }{QS } = \frac{QT}{PR}$ and $\angle 1 = \angle 2$

To prove : $\Delta PQS \sim \Delta TQR$

In $\triangle PQR$ , $\angle PQR=\angle PRQ$

$\therefore PQ=PR$

$\frac{QR }{QS } = \frac{QT}{PR}$ (Given)

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

In $\Delta PQS\, and\, \Delta TQR$ ,

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

$\angle Q=\angle Q$ (Common)

$\Delta PQS \sim \Delta TQR$ ( By SAS)

Q5 S and T are points on sides PR and QR of $\Delta$ PQR such that $\angle$ P = $\angle$ RTS. Show that $\Delta$ RPQ ~ $\Delta$ RTS.

Answer:

Given : $\angle$ P = $\angle$ RTS

To prove RPQ ~ $\Delta$ RTS.

In $\Delta$ RPQ and $\Delta$ RTS,

$\angle$ P = $\angle$ RTS (Given )

$\angle$ R = $\angle$ R (common)

$\Delta$ RPQ ~ $\Delta$ RTS. ( By AA)

Q6 In Fig. 6.37, if $\Delta$ ABE $\equiv$ $\Delta$ ACD, show that $\Delta$ ADE ~ $\Delta$ ABC.

Answer:

Given : $\triangle ABE \cong \triangle ACD$

To prove ADE ~ $\Delta$ ABC.

Since $\triangle ABE \cong \triangle ACD$

$AB=AC$ ( By CPCT)

$AD=AE$ (By CPCT)

In $\Delta$ ADE and $\Delta$ ABC,

$\angle A=\angle A$ ( Common)

and

$\frac{AD}{AB}=\frac{AE}{AC}$ ( $AB=AC$ and $AD=AE$ )

Therefore, $\Delta$ ADE ~ $\Delta$ ABC. ( By SAS criteria)

Q7 (1) In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta AEP \sim \Delta CDP$

Answer:

To prove : $\Delta AEP \sim \Delta CDP$

In $\Delta AEP \, \, and\, \, \Delta CDP$ ,

$\angle AEP=\angle CDP$ ( Both angles are right angle)

$\angle APE=\angle CPD$ (Vertically opposite angles )

$\Delta AEP \sim \Delta CDP$ ( By AA criterion)

Q7 (2) In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta ABD \sim \Delta CBE$

Answer:

To prove : $\Delta ABD \sim \Delta CBE$

In $\Delta ABD \, \, and\, \, \Delta CBE$ ,

$\angle ADB=\angle CEB$ ( Both angles are right angle)

$\angle ABD=\angle CBE$ (Common )

$\Delta ABD \sim \Delta CBE$ ( By AA criterion)

Q7 (3) In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta AEP \sim \Delta ADB$

Answer:

To prove : $\Delta AEP \sim \Delta ADB$

In $\Delta AEP \, \, \, and\, \, \Delta ADB$ ,

$\angle AEP=\angle ADB$ ( Both angles are right angle)

$\angle A=\angle A$ (Common )

$\Delta AEP \sim \Delta ADB$ ( By AA criterion)

Q7 (4) In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta PDC \sim \Delta BEC$

Answer:

To prove : $\Delta PDC \sim \Delta BEC$

In $\Delta PDC \, \, and\, \, \, \Delta BEC$ ,

$\angle CDP=\angle CEB$ ( Both angles are right angle)

$\angle C=\angle C$ (Common )

$\Delta PDC \sim \Delta BEC$ ( By AA criterion)

Q8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta ABE \sim \Delta CFB$

Answer:

To prove : $\Delta ABE \sim \Delta CFB$

In $\Delta ABE \, \, \, and\, \, \Delta CFB$ ,

$\angle A=\angle C$ ( Opposite angles of a parallelogram are equal)

$\angle AEB=\angle CBF$ ( Alternate angles of AE||BC)

$\Delta ABE \sim \Delta CFB$ ( By AA criterion )

Q9 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: $\Delta ABC \sim \Delta AMP$

Answer:

To prove : $\Delta ABC \sim \Delta AMP$

In $\Delta ABC \, \, and\, \, \Delta AMP$ ,

$\angle ABC=\angle AMP$ ( Each $90 ^\circ$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( By AA criterion )

Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that $\frac{CA }{PA } = \frac{BC }{MP}$

Answer:

To prove :

$\frac{CA }{PA } = \frac{BC }{MP}$

In $\Delta ABC \, \, and\, \, \Delta AMP$ ,

$\angle ABC=\angle AMP$ ( Each $90 ^\circ$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( By AA criterion )

$\frac{CA }{PA } = \frac{BC }{MP}$ ( corresponding parts of similar triangles )

Hence proved.

Q10 (1) 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 $\Delta ABC\: \: and\: \: \Delta EGF$ respectively. If $\Delta ABC \sim \Delta EGF$ , show that: $\frac{CD}{GH} = \frac{AC}{FG}$

Answer:

To prove :

$\frac{CD}{GH} = \frac{AC}{FG}$

Given : $\Delta ABC \sim \Delta EGF$

$\angle A=\angle F,\angle B=\angle E\, \, and \, \, \angle ACB=\angle FGE,\angle ACB=\angle FGE$

$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)

$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)

In $\Delta ACD \, \, and\, \, \Delta FGH$

$\therefore \angle ACD=\angle FGH$ ( proved above)

$\angle A=\angle F$ ( proved above)

$\Delta ACD \sim \Delta FGH$ ( By AA criterion)

$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$

Hence proved.

Q 10 (2) CD and GH are respectively the bisectors of $\angle ABC \: \:and \: \: \angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC \: \:and \: \: \Delta EGF$ respectively. If $\Delta ABC \sim \Delta EGF$ , show that: $\Delta DCB \sim \Delta HGE$

Answer:

To prove : $\Delta DCB \sim \Delta HGE$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCB \,\, \, and\, \, \Delta HGE$ ,

$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)

$\angle B=\angle E$ ( $\Delta ABC \sim \Delta EGF$ )

$\Delta DCB \sim \Delta HGE$ ( By AA criterion )

Q10 (3) CD and GH are respectively the bisectors of $\angle ABC \: \:and \: \: \angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC \: \:and \: \: \Delta EGF$ respectively. If $\Delta ABC\sim \Delta EGF$ , show that: $\Delta DCA \sim \Delta HGF$

Answer:

To prove : $\Delta DCA \sim \Delta HGF$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCA \, \, \, and\, \, \Delta HGF$ ,

$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)

$\angle A=\angle F$ ( $\Delta ABC \sim \Delta EGF$ )

$\Delta DCA \sim \Delta HGF$ ( By AA criterion )

Q11 In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD \perp BC$ and $EF \perp AC$ , prove that $\Delta ABD \sim \Delta ECF$

Answer:

To prove : $\Delta ABD \sim \Delta ECF$

Given: ABC is an isosceles triangle.

$AB=AC \, \, and\, \, \angle B=\angle C$

In $\Delta ABD \, \, and\, \, \Delta ECF$ ,

$\angle ABD=\angle ECF$ ( $\angle ABD=\angle B=\angle C=\angle ECF$ )

$\angle ADB=\angle EFC$ ( Each $90 ^\circ$ )

$\Delta ABD \sim \Delta ECF$ ( By AA criterion)

Q12 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 Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$

Answer:

AD and PM are medians of triangles. So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$

Given :

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

In $\triangle ABD\, and\, \triangle PQM,$

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

$\therefore \triangle ABD\sim \triangle PQM,$ (SSS similarity)

$\Rightarrow \angle ABD=\angle PQM$ ( Corresponding angles of similar triangles )

In $\triangle ABC\, and\, \triangle PQR,$

$\Rightarrow \angle ABD=\angle PQM$ (proved above)

$\frac{AB}{PQ}=\frac{BC}{QR}$

Therefore, $\Delta ABC \sim \Delta PQR$ . ( SAS similarity)

Q13 D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$ . Show that $CA^2 = CB.CD.$

Answer:

In, $\triangle ADC \, \, and\, \, \triangle BAC,$

$\angle ADC = \angle BAC$ ( given )

$\angle ACD = \angle BCA$ (common )

$\triangle ADC \, \, \sim \, \, \triangle BAC,$ ( By AA rule)

$\frac{CA}{CB}=\frac{CD}{CA}$ ( corresponding sides of similar triangles )

$\Rightarrow CA^2=CB\times CD$

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 $\Delta ABC \sim \Delta PQR$

Answer:

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (given)

Produce AD and PM to E and L such that AD=DE and PM=DE. Now,

join B to E,C to E,Q to L and R to L.

AD and PM are medians of a triangle, therefore

QM=MR and BD=DC

AD = DE (By construction)

PM=ML (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (Given )

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$

$\Delta ABE \sim \Delta PQL$ (SSS similarity)

$\angle BAE=\angle QPL$ ...................1 (Corresponding angles of similar triangles)

Similarity, $\triangle AEC=\triangle PLR$

$\angle CAE=\angle RPL$ ........................2

Adding equation 1 and 2,

$\angle BAE+\angle CAE=\angle QPL+\angle RPL$

$\angle CAB=\angle RPQ$ ............................3

In $\triangle ABC\, and\, \, \triangle PQR,$

$\frac{AB}{PQ}=\frac{AC}{PR}$ ( Given )

$\angle CAB=\angle RPQ$ ( From above equation 3)

$\triangle ABC\sim \triangle PQR$ ( SAS similarity)

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.

Answer:

CD = pole

AB = tower

Shadow of pole = DF

Shadow of tower = BE

In $\triangle ABE\, \, and\, \triangle CDF,$

$\angle CDF=\angle ABE$ ( Each $90 ^\circ$ )

$\angle DCF=\angle BAE$ (Angle of sun at same place )

$\triangle ABE\, \, \sim \, \triangle CDF,$ (AA similarity)

$\frac{AB}{CD}=\frac{BE}{QL}$

$\Rightarrow \frac{AB}{6}=\frac{28}{4}$

$\Rightarrow AB=42$ cm

Hence, the height of the tower is 42 cm.

Q16 If AD and PM are medians of triangles ABC and PQR, respectively where $\Delta AB C \sim \Delta PQR$ , prove that $\frac{AB}{PQ} = \frac{AD }{PM}$

Answer:

$\Delta AB C \sim \Delta PQR$ ( Given )

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$ ............... ....1( corresponding sides of similar triangles )

$\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R$ ....................................2

AD and PM are medians of triangle.So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$ ..........................................3

From equation 1 and 3, we have

$\frac{AB}{PQ}=\frac{BD}{QM}$ ...................................................................4

In $\triangle ABD\, and\, \triangle PQM,$

$\angle B=\angle Q$ (From equation 2)

$\frac{AB}{PQ}=\frac{BD}{QM}$ (From equation 4)

$\triangle ABD\, \sim \, \triangle PQM,$ (SAS similarity)

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$