# NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Written by Team Trustudies
Updated at 2021-02-21

## NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.1

Q1 ) Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

(i) Similar
(ii) Similar
(iii)Equilateral
(iv)(a) Equal
(b) Proportional

Q2 ) Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

(i)Two Equilateral Triangle and Two Rectangle

(ii) 1)Triangle and rhombus

2)Rectangle and circle

Q3 ) State whether the following quadrilaterals are similar or not :

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

## NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.2

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

(i) Given, in Triangle ABC, DE II BC
$\frac{AD}{DB}=\frac{AE}{EC}$
[Using Basic proportionality theorem]
$?\frac{1.5}{3}=\frac{1}{EC}$
$?EC=\frac{3}{1.5}$
$?$EC = 3×$\frac{10}{15}$ = 2 cm
Hence, EC = 2 cm.

(ii) Given, in Triangle ABC, DE is parallel to BC
= $\frac{AD}{DB}=\frac{AE}{EC}$
[Using Basic proportionality theorem]
$?\frac{AD}{7.2}=\frac{1.8}{5.4}$
$?AD=1.8×\frac{7.2}{5.4}$
$=\left(\frac{18}{10}\right)×\left(\frac{72}{10}\right)×\left(\frac{10}{54}\right)$
$=\frac{24}{10}$

AD = 2.4 cm

Hence, AD = 2.4 cm.

Q2 ) 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.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, in $\mathrm{?}PQR$, E and F are two points on side PQ and PR respectively.

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,
$\frac{PE}{EQ}=\frac{3.9}{3}=\frac{39}{30}=\frac{13}{10}=1.3$
And
$\frac{PF}{FR}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}=1.5$
$?$ $\frac{PE}{EQ}?\frac{PF}{FR}$

Hence, EF is not parallel to QR.

(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm

Therefore, by using Basic proportionality theorem, we get,

$\frac{PE}{QE}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}$
And,
$\frac{PF}{RF}=\frac{8}{9}$

So, we get here,

$\frac{PE}{QE}=\frac{PF}{RF}$

Hence, EF is parallel to QR.

(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And,
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
$?$ $\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}$………….(i)
And,
$\frac{PE}{FR}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}$…………(ii)

So, we get here,
$\frac{PE}{EQ}=\frac{PF}{FR}$

Hence, EF is parallel to QR.

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

$\frac{AM}{AB}=\frac{AL}{AC}$…………………….. (i)

Similarly, given, LN || CD and using basic proportionality theorem,

$\frac{AN}{AD}=\frac{AL}{AC}$……………………………(ii)

From equation (i) and (ii), we get,
$\frac{AM}{AB}=\frac{AN}{AD}$

Hence, proved.

Q4 ) In the figure, DE||AC and DF || AE. Prove that $\frac{BF}{FE}=\frac{BE}{EC}$.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

In $\mathrm{?}ABC$, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

$\frac{BD}{DA}=\frac{BE}{EC}$ ………………………………………………(i)

In $\mathrm{?}ABC$, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

$\frac{BD}{DA}=\frac{BF}{FE}$ ………………………………………………(ii)

From equation (i) and (ii), we get
$\frac{BE}{EC}=\frac{BF}{FE}$

Hence, proved.

Q5 ) In the figure, DE||OQ and DF||OR, show that EF||QR.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given,
In $\mathrm{?}PQO$, DE || OQ

So by using Basic Proportionality Theorem,
$\frac{PD}{DO}=\frac{PE}{EQ}$ ………………. (i)

Again given, in $\mathrm{?}PQO$, DE || OQ ,

So by using Basic Proportionality Theorem,
$\frac{PD}{DO}=\frac{PF}{FR}$ ………………… (ii)

From equation (i) and (ii), we get,
$\frac{PE}{EQ}=\frac{PF}{FR}$

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in $\mathrm{?}PQR$.

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given here,
In $\mathrm{?}OPQ$, AB || PQ

By using Basic Proportionality Theorem,
$\frac{OA}{AP}=\frac{OB}{BQ}$…………….(i)

Also given,
In $\mathrm{?}OPR$, AC || PR

By using Basic Proportionality Theorem
$\frac{OA}{AP}=\frac{OC}{CR}$……………(ii)

From equation (i) and (ii), we get,
$\frac{OB}{BQ}=\frac{OC}{CR}$

Therefore, by converse of Basic Proportionality Theorem,

In $\mathrm{?}OQR$, BC || QR.

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, in $\mathrm{?}ABC$, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.
=>$\frac{AD}{DB}$ = 1 …………………………. (i)

In $\mathrm{?}ABC$, DE || BC,

By using Basic Proportionality Theorem,

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

From equation (i), we can write,
=> 1 = $\frac{AE}{EC}$
=> AE = EC

Hence, proved, E is the midpoint of AC.

Q8 ) Using Converse of basic proportionality theorem, 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).

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, in $\mathrm{?}ABC$, D and E are the mid points of AB and AC respectively, such that,

We have to prove that: DE || BC.

Since, D is the midpoint of AB
=>$\frac{AD}{BD}$ = 1……………………………….. (i)

Also given, E is the mid-point of AC.
AE=EC
=> $\frac{AE}{EC}$ = 1

From equation (i) and (ii), we get,
$\frac{AD}{BD}=\frac{AE}{EC}$

By converse of Basic Proportionality Theorem,
DE || BC

Hence, proved.

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, $\frac{AO}{BO}=\frac{CO}{DO}$

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In $\mathrm{?}ADC$, we have OE || DC

Therefore, By using Basic Proportionality Theorem
$\frac{AE}{ED}=\frac{AO}{CO}$ ……………..(i)

Now, In $\mathrm{?}ABD$, OE || AB

Therefore, By using Basic Proportionality Theorem
$\frac{DE}{EA}=\frac{DO}{BO}$…………….(ii)

From equation (i) and (ii), we get,
$\frac{AO}{CO}=\frac{BO}{DO}$
=>$\frac{AO}{BO}=\frac{CO}{DO}$

Hence, proved.

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
$\frac{AO}{BO}=\frac{CO}{DO}$.

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In $\mathrm{?}DAB$, EO || AB

Therefore, By using Basic Proportionality Theorem
$\frac{DE}{EA}=\frac{DO}{OB}$ ……………………(i)

Also, given,
$\frac{AO}{BO}=\frac{CO}{DO}$
=> $\frac{AO}{CO}=\frac{BO}{DO}$
=> $\frac{CO}{AO}=\frac{DO}{BO}$
=> $\frac{DO}{OB}=\frac{CO}{AO}$ …………………………..(ii)

From equation (i) and (ii), we get
$\frac{DE}{EA}=\frac{CO}{AO}$

Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
=> AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

## NCERT solutions for class 10 Maths Chapter 6 Triangles 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:

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

i) Given, in $\mathrm{?}$ ABC and $\mathrm{?}$ PQR,

$\mathrm{?}$A = $\mathrm{?}$P = 60°
$\mathrm{?}$B = $\mathrm{?}$Q = 80°
$\mathrm{?}$C = $\mathrm{?}$R = 40°

Therefore by AAA similarity criterion,

$\mathrm{?}$ ABC is similar to $\mathrm{?}$ PQR

(ii) Given, in $\mathrm{?}$ ABC and $\mathrm{?}$ PQR,
$\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$
By SSS similarity criterion,

$\mathrm{?}$ ABC is similar to $\mathrm{?}$ QRP

(iii) Given, in $\mathrm{?}$ LMP and $\mathrm{?}$ DEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
$\frac{MP}{DE}=\frac{2}{4}=\frac{1}{2}$
$\frac{PL}{DF}=\frac{3}{6}=\frac{1}{2}$
$\frac{LM}{EF}=\frac{2.7}{5}=\frac{27}{50}$
Here , $\frac{MP}{DE}=\frac{PL}{DF}$ $?$ $\frac{LM}{EF}$

Therefore, $\mathrm{?}$ LMP and $\mathrm{?}$ DEF are not similar.

(iv) In $\mathrm{?}$ MNL and $\mathrm{?}$ QPR, it is given,
$\frac{MN}{QP}=\frac{ML}{QR}=\frac{1}{2}$
$\mathrm{?}$M = $\mathrm{?}$Q = 70°
Therefore, by SAS similarity criterion

$\mathrm{?}$ MNL is similar to $\mathrm{?}$ QPR

(v) In $\mathrm{?}$ ABC and $\mathrm{?}$ DEF, given that,
AB = 2.5, BC = 3, $\mathrm{?}$A = 80°, EF = 6, DF = 5, $\mathrm{?}$F = 80°
Here , $\frac{AB}{DF}=\frac{2.5}{5}=\frac{1}{2}$
And, $\frac{BC}{EF}=\frac{3}{6}=\frac{1}{2}$

$\mathrm{?}$B $?$ $\mathrm{?}$F

Hence, $\mathrm{?}$ ABC and $\mathrm{?}$ DEF are not similar.

(vi) In $\mathrm{?}$ DEF, by sum of angles of triangles, we know that,
$\mathrm{?}$D + $\mathrm{?}$E + $\mathrm{?}$F = 180°
70° + 80° + $\mathrm{?}$F = 180°
$\mathrm{?}$F = 180° – 70° – 80°
$\mathrm{?}$F = 30°

Similarly, In $\mathrm{?}$ PQR,
$\mathrm{?}$P + $\mathrm{?}$Q + $\mathrm{?}$R = 180 (Sum of angles of $\mathrm{?}$ )
$\mathrm{?}$P + 80° + 30° = 180°
$\mathrm{?}$P = 180° – 80° -30°
$\mathrm{?}$P = 70°

Now, comparing both the triangles, $\mathrm{?}$ DEF and $\mathrm{?}$ PQR, we have
$\mathrm{?}$D = $\mathrm{?}$P = 70°
$\mathrm{?}$E = $\mathrm{?}$Q = 80°
$\mathrm{?}$F = $\mathrm{?}$R = 30°

Therefore, by AAA similarity criterion,

Hence, $\mathrm{?}$ DEF is similar to $\mathrm{?}$ PQR

Q2 ) In the figure, $\mathrm{?}$ ODC $?$ $\mathrm{?}$ OBA, $\mathrm{?}$ BOC = 125° and $\mathrm{?}$ CDO = 70°. Find $\mathrm{?}$ DOC, $\mathrm{?}$ DCO and $\mathrm{?}$ OAB.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

As we can see from the figure, DOB is a straight line.

Therefore, $\mathrm{?}$ DOC + $\mathrm{?}$ COB = 180°
$\mathrm{?}$ DOC = 180° – 125° (Given, $\mathrm{?}$ BOC = 125°)
= 55°

In $\mathrm{?}$ DOC, Sum of the measures of the angles of a triangle is 180º

Therefore, $\mathrm{?}$ DCO + $\mathrm{?}$ CDO + $\mathrm{?}$ DOC = 180°
$\mathrm{?}$ DCO + 70º + 55º = 180°(Given, $\mathrm{?}$ CDO = 70°)
$\mathrm{?}$ DCO = 55°

It is given that, $\mathrm{?}$ ODC $?$ OBA,
Therefore, $\mathrm{?}$ ODC ~ OBA.

Hence, Corresponding angles are equal in similar triangles

$\mathrm{?}$ OAB = $\mathrm{?}$ OCD
$\mathrm{?}$ OAB = 55°

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 AO/OC = OB/OD

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

In $\mathrm{?}$DOC and $\mathrm{?}$BOA,

AB || CD, thus alternate interior angles will be equal,
$\mathrm{?}$CDO = $\mathrm{?}$ABO

Similarly,

$\mathrm{?}$DCO = $\mathrm{?}$BAO

Also, for the two triangles $\mathrm{?}$DOC and $\mathrm{?}$BOA, vertically opposite angles will be equal;
$\mathrm{?}$DOC = $\mathrm{?}$BOA
Hence, by AAA similarity criterion,
$\mathrm{?}$DOC ~ $\mathrm{?}$BOA

Thus, the corresponding sides are proportional.
DO/BO = OC/OA
OA/OC = OB/OD

Hence, proved.

4. In the fig.6.36, QR/QS = QT/PR and $\mathrm{?}$ 1 = $\mathrm{?}$ 2. Show that $\mathrm{?}$ PQS ~ $\mathrm{?}$ TQR.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

In $\mathrm{?}$ PQR,
$\mathrm{?}$ PQR = $\mathrm{?}$ PRQ
PQ = PR ………………………(i)
Given,
QR/QS = QT/PR
Using equation (i), we get
QR/QS = QT/QP……………….(ii)
In $\mathrm{?}$ PQS and $\mathrm{?}$ TQR, by equation (ii),
QR/QS = QT/QP
$\mathrm{?}$ Q = $\mathrm{?}$ Q
$\mathrm{?}$ PQS ~ $\mathrm{?}$ TQR [By SAS similarity criterion]

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, S and T are point on sides PR and QR of $\mathrm{?}$ PQR

And $\mathrm{?}$ P = $\mathrm{?}$ RTS.

In $\mathrm{?}$ RPQ and $\mathrm{?}$ RTS,
$\mathrm{?}$ RTS = $\mathrm{?}$ QPS (Given)
$\mathrm{?}$ R = $\mathrm{?}$ R (Common angle)
$\mathrm{?}$ RPQ ~ $\mathrm{?}$ RTS (AA similarity criterion)

Q6 ) In the figure, if $\mathrm{?}$ ABE $?$ $\mathrm{?}$ ACD, show that $\mathrm{?}$ ADE ~ $\mathrm{?}$ ABC.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, $\mathrm{?}$ ABE $?$ $\mathrm{?}$ ACD.

AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)

In $\mathrm{?}$ ADE and $\mathrm{?}$ ABC, dividing eq.(ii) by eq(i),
$\mathrm{?}$ A = $\mathrm{?}$ A [Common angle]

$\mathrm{?}$ ADE ~ $\mathrm{?}$ ABC [SAS similarity criterion]

Q7 ) In the figure, altitudes AD and CE of $\mathrm{?}$ ABC intersect each other at the point P. Show that:
(i) $\mathrm{?}$ AEP ~ $\mathrm{?}$ CDP
(ii) $\mathrm{?}$ ABD ~ $\mathrm{?}$ CBE
(iii) $\mathrm{?}$ AEP ~ $\mathrm{?}$ ADB
(iv) $\mathrm{?}$ PDC ~ $\mathrm{?}$ BEC.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, altitudes AD and CE of $\mathrm{?}$ ABC intersect each other at the point P.
(i) In $\mathrm{?}$ AEP and $\mathrm{?}$ CDP,
$\mathrm{?}$ AEP = $\mathrm{?}$ CDP (90° each)
$\mathrm{?}$ APE = $\mathrm{?}$ CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
$\mathrm{?}$ AEP ~ $\mathrm{?}$ CDP

(ii) In $\mathrm{?}$ ABD and $\mathrm{?}$ CBE,
$\mathrm{?}$ ADB = $\mathrm{?}$ CEB ( 90° each)
$\mathrm{?}$ ABD = $\mathrm{?}$ CBE (Common Angles)
Hence, by AA similarity criterion,
$\mathrm{?}$ ABD ~ $\mathrm{?}$ CBE

(iii) In $\mathrm{?}$ AEP and $\mathrm{?}$ ADB,
$\mathrm{?}$ AEP = $\mathrm{?}$ ADB (90° each)
$\mathrm{?}$ PAE = $\mathrm{?}$ DAB (Common Angles)
Hence, by AA similarity criterion,
$\mathrm{?}$ AEP ~ $\mathrm{?}$ ADB

(iv) In $\mathrm{?}$ PDC and $\mathrm{?}$ BEC,
$\mathrm{?}$ PDC = $\mathrm{?}$ BEC (90° each)
$\mathrm{?}$ PCD = $\mathrm{?}$ BCE (Common angles)
Hence, by AA similarity criterion,
$\mathrm{?}$ PDC ~ $\mathrm{?}$ BEC
Hence, by AA similarity criterion,
$\mathrm{?}$ PDC ~ $\mathrm{?}$ BEC

Q8 ) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\mathrm{?}$ ABE ~ $\mathrm{?}$ CFB.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In $\mathrm{?}$ ABE and $\mathrm{?}$ CFB,

$\mathrm{?}$ A = $\mathrm{?}$ C
(Opposite angles of a parallelogram)
$\mathrm{?}$ AEB = $\mathrm{?}$ CBF
(Alternate interior angles as AE || BC)

$\mathrm{?}$ ABE ~ $\mathrm{?}$ CFB (AA similarity criterion)

Q9 ) In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) $\mathrm{?}$ ABC ~ $\mathrm{?}$ AMP
(ii) CA/PA = BC/MP

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In $\mathrm{?}$ ABC and $\mathrm{?}$ AMP, we have,
$\mathrm{?}$ CAB = $\mathrm{?}$ MAP (common angles)
$\mathrm{?}$ ABC = $\mathrm{?}$ AMP = 90° (each 90°)
$\mathrm{?}$ ABC ~ $\mathrm{?}$ AMP (AA similarity criterion)

(ii) As, $\mathrm{?}$ ABC ~ $\mathrm{?}$ AMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP

Q10 ) CD and GH are respectively the bisectors of $\mathrm{?}$ ACB and $\mathrm{?}$ EGF such that D and H lie on sides AB and FE of $\mathrm{?}$ ABC and $\mathrm{?}$ EFG respectively. If $\mathrm{?}$ ABC ~ $\mathrm{?}$ FEG, Show that: (i) CD/GH = AC/FG (ii) $\mathrm{?}$ DCB ~ $\mathrm{?}$ HGE (iii) $\mathrm{?}$ DCA ~ $\mathrm{?}$ HGF

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, CD and GH are respectively the bisectors of $\mathrm{?}$ ACB and $\mathrm{?}$ EGF such that D and H lie on sides AB and FE of $\mathrm{?}$ ABC and $\mathrm{?}$ EFG respectively.

(i) From the given condition,

$\mathrm{?}$ ABC ~ $\mathrm{?}$ FEG.
$\mathrm{?}$ A = $\mathrm{?}$ F, $\mathrm{?}$ B = $\mathrm{?}$ E, and $\mathrm{?}$ ACB = $\mathrm{?}$ FGE

Since, $\mathrm{?}$ ACB = $\mathrm{?}$ FGE
$\mathrm{?}$ ACD = $\mathrm{?}$ FGH (Angle bisector)
And, $\mathrm{?}$ DCB = $\mathrm{?}$ HGE (Angle bisector)

In $\mathrm{?}$ ACD and $\mathrm{?}$ FGH,
$\mathrm{?}$ A = $\mathrm{?}$ F
$\mathrm{?}$ ACD = $\mathrm{?}$ FGH
$\mathrm{?}$ ACD ~ $\mathrm{?}$ FGH (AA similarity criterion)
CD/GH = AC/FG

(ii) In $\mathrm{?}$ DCB and $\mathrm{?}$ HGE,
$\mathrm{?}$ DCB = $\mathrm{?}$ HGE (Already proved)
$\mathrm{?}$ B = $\mathrm{?}$ E (Already proved)
$\mathrm{?}$ DCB ~ $\mathrm{?}$ HGE (AA similarity criterion)

(iii) In $\mathrm{?}$ DCA and $\mathrm{?}$ HGF,
$\mathrm{?}$ ACD = $\mathrm{?}$ FGH (Already proved)
$\mathrm{?}$ A = $\mathrm{?}$ F (Already proved)
$\mathrm{?}$ DCA ~ $\mathrm{?}$ HGF (AA similarity criterion)

Q11 ) In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $?$ BC and EF $?$ AC, prove that $\mathrm{?}$ ABD ~ $\mathrm{?}$ ECF.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, ABC is an isosceles triangle.

AB = AC
$\mathrm{?}$ ABD = $\mathrm{?}$ ECF

In $\mathrm{?}$ ABD and $\mathrm{?}$ ECF,
$\mathrm{?}$ ADB = $\mathrm{?}$ EFC (Each 90°)
$\mathrm{?}$ BAD = $\mathrm{?}$ CEF (Already proved)
$\mathrm{?}$ ABD ~ $\mathrm{?}$ ECF (using AA similarity 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 $\mathrm{?}$ PQR (see Fig 6.41). Show that $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, $\mathrm{?}$ ABC and $\mathrm{?}$ PQR, AB, BC and median AD of $\mathrm{?}$ ABC are proportional to sides PQ, QR and median PM of $\mathrm{?}$ PQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR

As we know here,
AB/PQ = BC/QR = AD/PM
$\frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$
AB/PQ = BC/QR = AD/PM
(D is the midpoint of BC. M is the midpoint of QR)
$\mathrm{?}$ ABD ~ $\mathrm{?}$ PQM [SSS similarity criterion]

$\mathrm{?}$ ABD = $\mathrm{?}$ PQM
[Corresponding angles of two similar triangles are equal]
$\mathrm{?}$ ABC = $\mathrm{?}$ PQR

In $\mathrm{?}$ ABC and $\mathrm{?}$ PQR
AB/PQ = BC/QR ………………………….(i)
$\mathrm{?}$ ABC = $\mathrm{?}$ PQR ……………………………(ii)

From equation (i) and (ii), we get,
$\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR [SAS similarity criterion]

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, D is a point on the side BC of a triangle ABC such that $\mathrm{?}$ ADC = $\mathrm{?}$ BAC.

In $\mathrm{?}$ ADC and $\mathrm{?}$ BAC,
$\mathrm{?}$ ADC = $\mathrm{?}$ BAC (Already given)
$\mathrm{?}$ ACD = $\mathrm{?}$ BCA (Common angles)
$\mathrm{?}$ ADC ~ $\mathrm{?}$ BAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
CA/CB = CD/CA
$C{A}^{2}=CB.CD$
Hence, 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 $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given: Two triangles $\mathrm{?}$ ABC and $\mathrm{?}$ PQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM
We have to prove, $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In $\mathrm{?}$ ABD and $\mathrm{?}$ CDE, we have
AD = DE [By Construction.]
BD = DC [Since, AP is the median]
and, $\mathrm{?}$ ADB = $\mathrm{?}$ CDE
[Vertically opposite angles]

$\mathrm{?}$ ABD $?$ $\mathrm{?}$ CDE [SAS criterion of congruence]
AB = CE [By CPCT] …………………………..(i)

Also, in $\mathrm{?}$ PQM and $\mathrm{?}$ MNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, $\mathrm{?}$ PMQ = $\mathrm{?}$ NMR
[Vertically opposite angles]
$\mathrm{?}$ PQM = $\mathrm{?}$ MNR [SAS criterion of congruence]
PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM

From equation (i) and (ii),
CE/RN = AC/PR = AD/PM
CE/RN = AC/PR = 2AD/2PM
CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
$\mathrm{?}$ ACE ~ $\mathrm{?}$ PRN [SSS similarity criterion]

Therefore, $\mathrm{?}$ 2 = $\mathrm{?}$ 4
Similarly, $\mathrm{?}$ 1 = $\mathrm{?}$ 3
$\mathrm{?}$ 1 + $\mathrm{?}$ 2 = $\mathrm{?}$ 3 + $\mathrm{?}$ 4
$\mathrm{?}$ A = $\mathrm{?}$ P …………………………………………….(iii)

Now, In $\mathrm{?}$ ABC and $\mathrm{?}$ PQR, we have
AB/PQ = AC/PR (Already given)

From equation (iii),
$\mathrm{?}$ A = $\mathrm{?}$ P
$\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR [ SAS similarity criterion]

Q15 ) A vertical pole of a 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.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m

In $\mathrm{?}$ ABC and $\mathrm{?}$ DEF,
$\mathrm{?}$ C = $\mathrm{?}$ E (angular elevation of sum)
$\mathrm{?}$ B = $\mathrm{?}$ F = 90°
$\mathrm{?}$ ABC ~ $\mathrm{?}$ DEF (AA similarity criterion)

AB/DF = BC/EF
(If two triangles are similar corresponding sides are proportional)
6/h = 4/28
h =( 6×28)/4
h = 6 × 7
h = 42 m

Hence, the height of the tower is 42 m.

If AD and PM are medians of triangles ABC and PQR, respectively where $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR prove that AB/PQ = AD/PM.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, $\mathrm{?}$ ABC ~ $\mathrm{?}$ PQR

We know that the corresponding sides of similar triangles are in proportion.

AB/PQ = AC/PR = BC/QR……………………………(i)
Also, $\mathrm{?}$ A = $\mathrm{?}$ P, $\mathrm{?}$ B = $\mathrm{?}$ Q, $\mathrm{?}$ C = $\mathrm{?}$ R ………….…..(ii)

Since AD and PM are medians, they will divide their opposite sides.
BD = $\frac{BC}{2}$ and QM = $\frac{QR}{2}$ ……………..………….(iii)

From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)

In $\mathrm{?}$ ABD and $\mathrm{?}$ PQM,

From equation (ii), we have
$\mathrm{?}$ B = $\mathrm{?}$ Q

From equation (iv), we have,
AB/PQ = BD/QM
$\mathrm{?}$ ABD ~ $\mathrm{?}$ PQM (SAS similarity criterion)
AB/PQ = BD/QM = AD/PM

## NCERT solutions for class 10 Maths Chapter 6 Triangles Exercise 6.4

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, $\mathrm{?}$ ABC ~ $\mathrm{?}$ DEF,
Area of $\mathrm{?}$ ABC = 64 $c{m}^{2}$
Area of $\mathrm{?}$ DEF = 121 $c{m}^{2}$
EF = 15.4 cm

$\frac{Areaof\mathrm{?}ABC}{Areaof\mathrm{?}DEF}=\frac{A{B}^{2}}{D{E}^{2}}$

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
$?$ $\frac{A{C}^{2}}{D{F}^{2}}=\frac{B{C}^{2}}{E{F}^{2}}$
$?$ $\frac{64}{121}=\frac{B{C}^{2}}{E{F}^{2}}$
$?$ $\left(\frac{8}{11}{\right)}^{2}=\left(\frac{BC}{15.4}{\right)}^{2}$
$?$ $\frac{8}{11}=\frac{BC}{15.4}$
$?$ BC = $8×\frac{15.4}{11}$
$?$ BC = 8 × 1.4
$?$ BC = 11.2 cm

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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In $\mathrm{?}$ AOB and $\mathrm{?}$ COD, we have
$\mathrm{?}$ 1 = $\mathrm{?}$ 2 (Alternate angles)
$\mathrm{?}$ 3 = $\mathrm{?}$ 4 (Alternate angles)
$\mathrm{?}$ 5 = $\mathrm{?}$ 6 (Vertically opposite angle)
$\mathrm{?}$ AOB ~ $\mathrm{?}$ COD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

$\frac{Areaof\left(\mathrm{?}AOB\right)}{Areaof\left(\mathrm{?}COD\right)}=\frac{A{B}^{2}}{C{D}^{2}}$
= $\frac{\left(2CD{\right)}^{2}}{C{D}^{2}}$ [ AB = CD]
$\frac{Areaof\left(\mathrm{?}AOB\right)}{Areaof\left(\mathrm{?}COD\right)}$
= $\frac{4C{D}^{2}}{CD}=\frac{4}{1}$

Hence, the required ratio of the area of $\mathrm{?}$ AOB and $\mathrm{?}$ COD = 4:1

Q3 ) In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area ($\mathrm{?}$ ABC)/area ($\mathrm{?}$ DBC) = AO/DO.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.

We have to prove: Area ($\mathrm{?}$ ABC)/Area ($\mathrm{?}$ DBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = $\frac{1}{2}$ × Base × Height
$\frac{ar\left(\mathrm{?}ABC\right)}{ar\left(\mathrm{?}DEF}=\frac{\frac{1}{2}BC\left(AP\right)}{\frac{1}{2}BC\left(DM\right)}=\frac{AP}{DM}$

In $\mathrm{?}$ APO and $\mathrm{?}$ DMO,
$\mathrm{?}$ APO = $\mathrm{?}$ DMO (Each 90°)
$\mathrm{?}$ AOP = $\mathrm{?}$ DOM
(Vertically opposite angles)
$\mathrm{?}$ APO ~ $\mathrm{?}$ DMO (AA similarity criterion)
$\frac{AP}{DM}=\frac{AO}{DO}$
$\frac{Area\left(\mathrm{?}ABC\right)}{Area\left(\mathrm{?}DBC\right)}=\frac{AO}{DO}$ .

Q4 ) If the areas of two similar triangles are equal, prove that they are congruent.

NCERT Soluti