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

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

Answer:

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

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

$\Rightarrow \mathrm{\angle }DOC+{125}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DOC={180}^{\circ }-{125}^{\circ }$

$\Rightarrow \mathrm{\angle }DOC={55}^{\circ }$

In $\mathrm{\Delta }ODC,$

$\mathrm{\angle }DOC+\mathrm{\angle }ODC+\mathrm{\angle }DCO={180}^{\circ }$

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

$\Rightarrow \mathrm{\angle }DCO+{125}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DCO={180}^{\circ }-{125}^{\circ }$

$\Rightarrow \mathrm{\angle }DCO={55}^{\circ }$

Since , $\mathrm{\Delta }ODC\sim \mathrm{\Delta }OBA$ , so

$\Rightarrow \mathrm{\angle }OAB=\mathrm{\angle }DCO={55}^{\circ }$ ( 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 $\mathrm{△}DOCand\mathrm{△}BOA$ , we have

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

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

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

$\therefore \mathrm{△}DOC\sim \mathrm{△}BOA$ ( 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 $\mathrm{\angle }1=\mathrm{\angle }2$ . Show that $\mathrm{\Delta }PQS\sim \mathrm{\Delta }TQR$

Answer:

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

To prove : $\mathrm{\Delta }PQS\sim \mathrm{\Delta }TQR$

In $\mathrm{△}PQR$ , $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$

$\therefore PQ=PR$

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

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

In $\mathrm{\Delta }PQSand\mathrm{\Delta }TQR$ ,

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

$\mathrm{\angle }Q=\mathrm{\angle }Q$ (Common)

$\mathrm{\Delta }PQS\sim \mathrm{\Delta }TQR$ ( By SAS)

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

Answer:

Given : $\mathrm{\angle }$ P = $\mathrm{\angle }$ RTS

To prove RPQ ~ $\mathrm{\Delta }$ RTS.

In $\mathrm{\Delta }$ RPQ and $\mathrm{\Delta }$ RTS,

$\mathrm{\angle }$ P = $\mathrm{\angle }$ RTS (Given )

$\mathrm{\angle }$ R = $\mathrm{\angle }$ R (common)

$\mathrm{\Delta }$ RPQ ~ $\mathrm{\Delta }$ RTS. ( By AA)

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

Answer:

Given : $\mathrm{△}ABE\cong \mathrm{△}ACD$

To prove ADE ~ $\mathrm{\Delta }$ ABC.

Since $\mathrm{△}ABE\cong \mathrm{△}ACD$

$AB=AC$ ( By CPCT)

$AD=AE$ (By CPCT)

In $\mathrm{\Delta }$ ADE and $\mathrm{\Delta }$ ABC,

$\mathrm{\angle }A=\mathrm{\angle }A$ ( Common)

and

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

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

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

Answer:

To prove : $\mathrm{\Delta }AEP\sim \mathrm{\Delta }CDP$

In $\mathrm{\Delta }AEPand\mathrm{\Delta }CDP$ ,

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

$\mathrm{\angle }APE=\mathrm{\angle }CPD$ (Vertically opposite angles )

$\mathrm{\Delta }AEP\sim \mathrm{\Delta }CDP$ ( By AA criterion)

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

Answer:

To prove : $\mathrm{\Delta }ABD\sim \mathrm{\Delta }CBE$

In $\mathrm{\Delta }ABDand\mathrm{\Delta }CBE$ ,

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

$\mathrm{\angle }ABD=\mathrm{\angle }CBE$ (Common )

$\mathrm{\Delta }ABD\sim \mathrm{\Delta }CBE$ ( By AA criterion)

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

Answer:

To prove : $\mathrm{\Delta }AEP\sim \mathrm{\Delta }ADB$

In $\mathrm{\Delta }AEPand\mathrm{\Delta }ADB$ ,

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

$\mathrm{\angle }A=\mathrm{\angle }A$ (Common )

$\mathrm{\Delta }AEP\sim \mathrm{\Delta }ADB$ ( By AA criterion)

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

Answer:

To prove : $\mathrm{\Delta }PDC\sim \mathrm{\Delta }BEC$

In $\mathrm{\Delta }PDCand\mathrm{\Delta }BEC$ ,

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

$\mathrm{\angle }C=\mathrm{\angle }C$ (Common )

$\mathrm{\Delta }PDC\sim \mathrm{\Delta }BEC$ ( 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 $\mathrm{\Delta }ABE\sim \mathrm{\Delta }CFB$

Answer:

To prove : $\mathrm{\Delta }ABE\sim \mathrm{\Delta }CFB$

In $\mathrm{\Delta }ABEand\mathrm{\Delta }CFB$ ,

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

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

$\mathrm{\Delta }ABE\sim \mathrm{\Delta }CFB$ ( 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: $\mathrm{\Delta }ABC\sim \mathrm{\Delta }AMP$

Answer:

To prove : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }AMP$

In $\mathrm{\Delta }ABCand\mathrm{\Delta }AMP$ ,

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

$\mathrm{\angle }A=\mathrm{\angle }A$ ( common)

$\mathrm{\Delta }ABC\sim \mathrm{\Delta }AMP$ ( 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 $\mathrm{\Delta }ABCand\mathrm{\Delta }AMP$ ,

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

$\mathrm{\angle }A=\mathrm{\angle }A$ ( common)

$\mathrm{\Delta }ABC\sim \mathrm{\Delta }AMP$ ( 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 $\mathrm{\angle }ACB$ and $\mathrm{\angle }EGF$ such that D and H lie on sides AB and FE of $\mathrm{\Delta }ABCand\mathrm{\Delta }EGF$ respectively. If $\mathrm{\Delta }ABC\sim \mathrm{\Delta }EGF$ , show that: $\frac{CD}{GH}=\frac{AC}{FG}$

Answer:

To prove :

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

Given : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }EGF$

$\mathrm{\angle }A=\mathrm{\angle }F,\mathrm{\angle }B=\mathrm{\angle }Eand\mathrm{\angle }ACB=\mathrm{\angle }FGE,\mathrm{\angle }ACB=\mathrm{\angle }FGE$

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

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

In $\mathrm{\Delta }ACDand\mathrm{\Delta }FGH$

$\therefore \mathrm{\angle }ACD=\mathrm{\angle }FGH$ ( proved above)

$\mathrm{\angle }A=\mathrm{\angle }F$ ( proved above)

$\mathrm{\Delta }ACD\sim \mathrm{\Delta }FGH$ ( By AA criterion)

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

Hence proved.

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

Answer:

To prove : $\mathrm{\Delta }DCB\sim \mathrm{\Delta }HGE$

Given : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }EGF$

In $\mathrm{\Delta }DCBand\mathrm{\Delta }HGE$ ,

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

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

$\mathrm{\Delta }DCB\sim \mathrm{\Delta }HGE$ ( By AA criterion )

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

Answer:

To prove : $\mathrm{\Delta }DCA\sim \mathrm{\Delta }HGF$

Given : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }EGF$

In $\mathrm{\Delta }DCAand\mathrm{\Delta }HGF$ ,

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

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

$\mathrm{\Delta }DCA\sim \mathrm{\Delta }HGF$ ( 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 $\mathrm{\Delta }ABD\sim \mathrm{\Delta }ECF$

Answer:

To prove : $\mathrm{\Delta }ABD\sim \mathrm{\Delta }ECF$

Given: ABC is an isosceles triangle.

$AB=ACand\mathrm{\angle }B=\mathrm{\angle }C$

In $\mathrm{\Delta }ABDand\mathrm{\Delta }ECF$ ,

$\mathrm{\angle }ABD=\mathrm{\angle }ECF$ ( $\mathrm{\angle }ABD=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }ECF$ )

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

$\mathrm{\Delta }ABD\sim \mathrm{\Delta }ECF$ ( 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 $\mathrm{\Delta }PQR$ (see Fig. 6.41). Show that $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$

Answer:

AD and PM are medians of triangles. So,

$BD=\frac{BC}{2}andQM=\frac{QR}{2}$

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 $\mathrm{△}ABDand\mathrm{△}PQM,$

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

$\therefore \mathrm{△}ABD\sim \mathrm{△}PQM,$ (SSS similarity)

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

In $\mathrm{△}ABCand\mathrm{△}PQR,$

$\Rightarrow \mathrm{\angle }ABD=\mathrm{\angle }PQM$ (proved above)

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

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

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

Answer:

In, $\mathrm{△}ADCand\mathrm{△}BAC,$

$\mathrm{\angle }ADC=\mathrm{\angle }BAC$ ( given )

$\mathrm{\angle }ACD=\mathrm{\angle }BCA$ (common )

$\mathrm{△}ADC\sim \mathrm{△}BAC,$ ( By AA rule)

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

$\Rightarrow C{A}^2=CB\times CD$

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 $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$

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 )

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

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

$\mathrm{\Delta }ABE\sim \mathrm{\Delta }PQL$ (SSS similarity)

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

Similarity, $\mathrm{△}AEC=\mathrm{△}PLR$

$\mathrm{\angle }CAE=\mathrm{\angle }RPL$ ........................2

Adding equation 1 and 2,

$\mathrm{\angle }BAE+\mathrm{\angle }CAE=\mathrm{\angle }QPL+\mathrm{\angle }RPL$

$\mathrm{\angle }CAB=\mathrm{\angle }RPQ$ ............................3

In $\mathrm{△}ABCand\mathrm{△}PQR,$

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

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

$\mathrm{△}ABC\sim \mathrm{△}PQR$ ( 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 $\mathrm{△}ABEand\mathrm{△}CDF,$

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

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

$\mathrm{△}ABE\sim \mathrm{△}CDF,$ (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 $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$ , prove that $\frac{AB}{PQ}=\frac{AD}{PM}$

Answer:

$\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$ ( Given )

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

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

AD and PM are medians of triangle.So,

$BD=\frac{BC}{2}andQM=\frac{QR}{2}$ ..........................................3

From equation 1 and 3, we have

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

In $\mathrm{△}ABDand\mathrm{△}PQM,$

$\mathrm{\angle }B=\mathrm{\angle }Q$ (From equation 2)

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

$\mathrm{△}ABD\sim \mathrm{△}PQM,$ (SAS similarity)

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