NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Maths Chapter 6 Triangles class 10 Excercise: 6.1

Q1 (1) Fill in the blanks using the correct word given in brackets: All circles are ______ . (congruent, similar)

Answer:

All circles are similar.

Since all the circles have a similar shape. They may have different radius but the shape of all circles is the same.

Therefore, all circles are similar.

Q1 (2) Fill in the blanks using the correct word given in brackets: All squares are ______. (similar, congruent)

Answer:

All squares are similar.

Since all the squares have a similar shape. They may have a different side but the shape of all square is the same.

Therefore, all squares are similar.

Q1 (3) Fill in the blanks using the correct word given in brackets: All ______triangles are similar. (isosceles, equilateral)

Answer:

All equilateral triangles are similar.

Since all the equilateral triangles have a similar shape. They may have different sides but the shape of all equilateral triangles is the same.

Therefore, all equilateral triangles are similar.

Q1 (4) Fill in the blanks using the correct word given in brackets : (iv) Two polygons of the same number of sides are similar if(a) there corresponding angles are _________and (b) their corresponding sides are________. (equal, proportional)

Answer:

Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.

Thus, (a) equal

(b) proportional

Q2 (1) Give two different examples of a pair of similar figures.

Answer:

The two different examples of a pair of similar figures are :

1. Two circles with different radii.

2. Two rectangles with different breadth and length.

Q2 (2) Give two different examples of a pair of non-similar figures.

Answer:

The two different examples of a pair of non-similar figures are :

1.Rectangle and circle

2. A circle and a triangle.

Q3 State whether the following quadrilaterals are similar or not:

Answer:

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. $1:2$ but their corresponding angles are not equal.

Class 10 Maths Chapter 6 Triangles Excercise: 6.2

Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer:

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{1.5}{3}=\frac{1}{x}$

$\Rightarrow x=\frac{3}{1.5}=2cm$

$\therefore EC=2cm$

(ii)

Let AD be x

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{x}{7.2}=\frac{1.8}{5.4}$

$\Rightarrow x=\frac{7.2}{3}=2.4cm$

$\therefore AD=2.4cm$

Q2 (1) E and F are points on the sides PQ and PR respectively of a $\mathrm{△}$ PQR. For each of the following cases, state whether EF || QR: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer:

(i)

Given :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3cm$ and $\frac{PF}{FR}=\frac{3.6}{2.4}=1.5cm$

We have

$\frac{PE}{EQ}\ne \frac{PF}{FR}$

Hence, EF is not parallel to QR.

Q2 (2) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Answer:

(ii)

Given :

PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}cm$ and $\frac{PF}{FR}=\frac{8}{9}cm$

We have

$\frac{PE}{EQ}=\frac{PF}{FR}$

Hence, EF is parallel to QR.

Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Answer:

(iii)

Given :

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

$\frac{PE}{PQ}=\frac{0.18}{1.28}=\frac{9}{64}cm$ and $\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}cm$

We have

$\frac{PE}{EQ}=\frac{PF}{FR}$

Hence, EF is parallel to QR.

Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that $\frac{AM}{AB}=\frac{AN}{AD}$

Answer:

Given : LM || CB and LN || CD

To prove :

$\frac{AM}{AB}=\frac{AN}{AD}$

Since , LM || CB so we have

$\frac{AM}{AB}=\frac{AL}{AC}.............................................1$

Also, LN || CD

$\frac{AL}{AC}=\frac{AN}{AD}.............................................2$

From equation 1 and 2, we have

$\frac{AM}{AB}=\frac{AN}{AD}$

Hence proved.

Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC

Answer:

Given : DE || AC and DF || AE.

To prove :

$\frac{BF}{FE}=\frac{BE}{EC}$

Since , DE || AC so we have

$\frac{BD}{DA}=\frac{BE}{EC}.............................................1$

Also,DF || AE

$\frac{BD}{DA}=\frac{BF}{FE}.............................................2$

From equation 1 and 2, we have

$\frac{BF}{FE}=\frac{BE}{EC}$

Hence proved.

Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Answer:

Given : DE || OQ and DF || OR.

To prove EF || QR.

Since DE || OQ so we have

$\frac{PE}{EQ}=\frac{PD}{DO}.............................................1$

Also, DF || OR

$\frac{PF}{FR}=\frac{PD}{DO}.............................................2$

From equation 1 and 2, we have

$\frac{PE}{EQ}=\frac{PF}{FR}$

Thus, EF || QR. (converse of basic proportionality theorem)

Hence proved.

Q6 In Fig. 6.21, A, B, and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:

Given : AB || PQ and AC || PR

To prove: BC || QR

Since, AB || PQ so we have

$\frac{OA}{AP}=\frac{OB}{BQ}.............................................1$

Also, AC || PR

$\frac{OA}{AP}=\frac{OC}{CR}.............................................2$

From equation 1 and 2, we have

$\frac{OB}{BQ}=\frac{OC}{CR}$

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Q7 Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.

i.e. $PQ||BC$ and $AP=PB$ .

Using basic proportionality theorem, we have

$\frac{AP}{PB}=\frac{AQ}{QC}..........................1$

Since $AP=PB$

$\frac{AQ}{QC}=\frac{1}{1}$

$\Rightarrow AQ=QC$

$\therefore$ Q is the midpoint of AC.

Q8 Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:

Let P is the midpoint of line AB and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of line AB and AC, respectively.

i.e. $AQ=QC$ and $AP=PB$ .

we have,

$\frac{AP}{PB}=\frac{1}{1}..........................1$

$\frac{AQ}{QC}=\frac{1}{1}...................................2$

From equation 1 and 2, we get

$\frac{AQ}{QC}=\frac{AP}{PB}$

$\therefore$ By basic proportionality theorem, we have $PQ||BC$

Q9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
that $\frac{AO}{BO}=\frac{CO}{DO}$

Answer:

Draw a line EF passing through point O such that $EO||CDandFO||CD$

To prove :

$\frac{AO}{BO}=\frac{CO}{DO}$

In $\mathrm{△}ADC$ , we have $CD||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{AO}{OC}........................................1$

In $\mathrm{△}ABD$ , we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{DE}{EA}=\frac{OD}{BO}........................................2$

Using equation 1 and 2, we get

$\frac{AO}{OC}=\frac{BO}{OD}$

$\Rightarrow \frac{AO}{BO}=\frac{CO}{DO}$

Hence proved.

Q10 The diagonals of a quadrilateral ABCD intersect each other at point O such that $\frac{AO}{BO}=\frac{CO}{DO}$ Show that ABCD is a trapezium.

Answer:

Draw a line EF passing through point O such that $EO||AB$

Given :

$\frac{AO}{BO}=\frac{CO}{DO}$

In $\mathrm{△}ABD$ , we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{BO}{DO}........................................1$

However, its is given that

$\frac{AO}{CO}=\frac{BO}{DO}..............................2$

Using equation 1 and 2 , we get

$\frac{AE}{ED}=\frac{AO}{CO}$

$\Rightarrow EO||CD$ (By basic proportionality theorem)

$\Rightarrow AB||EO||CD$

$\Rightarrow AB||CD$

Therefore, ABCD is a trapezium.

Class 10 Maths Chapter 6 Triangles Excercise: 6.3

Q1 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 :

Answer:

(i) $\mathrm{\angle }A=\mathrm{\angle }P={60}^{\circ }$

$\mathrm{\angle }B=\mathrm{\angle }Q={80}^{\circ }$

$\mathrm{\angle }C=\mathrm{\angle }R={40}^{\circ }$

$\therefore \mathrm{△}ABC\sim \mathrm{△}PQR$ (By AAA)

So , $\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$

(ii) As corresponding sides of both triangles are proportional.

$\therefore \mathrm{△}ABC\sim \mathrm{△}PQR$ (By SSS)

(iii) Given triangles are not similar because corresponding sides are not proportional.

(iv) $\mathrm{△}MNL\sim \mathrm{△}PQR$ by SAS similarity criteria.

(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides

(vi) In $\mathrm{△}DEF$ , we know that

$\mathrm{\angle }D+\mathrm{\angle }E+\mathrm{\angle }F={180}^{\circ }$

$\Rightarrow {70}^{\circ }+{80}^{\circ }+\mathrm{\angle }F={180}^{\circ }$

$\Rightarrow {150}^{\circ }+\mathrm{\angle }F={180}^{\circ }$

$\Rightarrow \mathrm{\angle }F={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

In $\mathrm{△}PQR$ , we know that

$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }$

$\Rightarrow {30}^{\circ }+{80}^{\circ }+\mathrm{\angle }R={180}^{\circ }$

$\Rightarrow {110}^{\circ }+\mathrm{\angle }R={180}^{\circ }$

$\Rightarrow \mathrm{\angle }R={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

$\mathrm{\angle }Q=\mathrm{\angle }P={70}^{\circ }$

$\mathrm{\angle }E=\mathrm{\angle }Q={80}^{\circ }$

$\mathrm{\angle }F=\mathrm{\angle }R={30}^{\circ }$

$\therefore \mathrm{△}DEF\sim \mathrm{△}PQR$ ( By AAA)

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)

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

Class 10 Maths Chapter 6 Triangles Excercise:6.4

Q1 Let $\mathrm{\Delta }ABC\sim \mathrm{\Delta }DEF$ and their areas be, respectively, 64 $c{m}^2$ and 121 $c{m}^2$ . If EF = 15.4 cm, find BC.

Answer:

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

ar(ABC) = 64 $c{m}^2$ and ar(DEF)=121 $c{m}^2$ .

EF = 15.4 cm (Given )

$\frac{ar(\mathrm{△}ABC)}{ar(\mathrm{△}DEF)}=\frac{A{B}^2}{D{E}^2}=\frac{B{C}^2}{E{F}^2}=\frac{A{C}^2}{D{F}^2}$

$\frac{64}{121}=\frac{B{C}^2}{(15.4{)}^2}$

$\Rightarrow \frac{8}{11}=\frac{BC}{15.4}$

$\Rightarrow \frac{8\times 15.4}{11}=BC$

$\Rightarrow BC=11.2cm$

Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Answer:

Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

AB = 2 CD ( Given )

In $\mathrm{△}AOBand\mathrm{△}COD,$

$\mathrm{\angle }COD=\mathrm{\angle }AOB$ (vertically opposite angles )

$\mathrm{\angle }OCD=\mathrm{\angle }OAB$ (Alternate angles)

$\mathrm{\angle }ODC=\mathrm{\angle }OBA$ (Alternate angles)

$\therefore \mathrm{△}AOB\sim \mathrm{△}COD$ (AAA similarity)

$\frac{ar(\mathrm{△}AOB)}{ar(\mathrm{△}COD)}=\frac{A{B}^2}{C{D}^2}$

$\frac{ar(\mathrm{△}AOB)}{ar(\mathrm{△}COD)}=\frac{(2CD{)}^2}{C{D}^2}$

$\frac{ar(\mathrm{△}AOB)}{ar(\mathrm{△}COD)}=\frac{4.C{D}^2}{C{D}^2}$

$\Rightarrow \frac{ar(\mathrm{△}AOB)}{ar(\mathrm{△}COD)}=\frac{4}{1}$

$\Rightarrow ar(\mathrm{△}AOB)=ar(\mathrm{△}COD)=4:1$

Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $\frac{ar(ABC)}{ar(DBC)}=\frac{AO}{DO}$

Answer: