NCERT Solutions for Class 10 Maths Chapter 6 Triangles

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# NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Edited By Ramraj Saini | Updated on Sep 08, 2023 07:12 PM IST | #CBSE Class 10th

## NCERT Solutions for Maths Chapter 6 Triangles Class 10

NCERT Solutions for Class 10 Maths Chapter 6, called Triangles are super important study materials for CBSE Class 10 students. This chapter aligns perfectly with the CBSE Syllabus for 2023-24, covering lots of rules and theorems. Sometimes, students get puzzled about which theorem to use. Here students will get NCERT solutions for Class 10 chapter wise. Practice these solutions for triangles class 10 to command the concepts.

NCERT solutions created at Careers360 are made to be crystal clear, explaining every step. Subject experts have created these solutions to help you prepare well for your board exams. They're not just for exams but also for tackling homework and assignments.

CBSE Class 10 exams often have questions from NCERT textbooks. So, Chapter 6's NCERT Solutions for Class 10 Maths are your best bet for getting ready and being able to tackle any kind of question from this chapter. We strongly recommend practicing these solutions regularly to ace your Class 10 board exams. Theorems of triangles class 10 pdf download are available freely, students can download using below link.

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## NCERT Solutions Class 10 Maths Chapter 6 - Important Points

Similar Triangles - A pair of triangles that have equal corresponding angles and proportional corresponding sides.

Equiangular Triangles:

• A pair of triangles that have corresponding angles equal.

• The ratio of any two corresponding sides in two equiangular triangles is always the same.

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Criteria for Triangle Similarity:

• Angle-Angle-Angle (AAA) Similarity

• Side-Angle-Side (SAS) Similarity

• Side-Side-Side (SSS) Similarity

Basic Proportionality Theorem:

• When a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.

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Converse of Basic Proportionality Theorem:

• In a pair of triangles when the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.

Free download NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF for CBSE Exam.

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Intext Questions and Exercise)

Maths Chapter 6 Triangles class 10 Excercise: 6.1

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.

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.

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.

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

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.

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

1.Rectangle and circle

2. A circle and a triangle.

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

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

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

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

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

$\therefore EC=2\, cm$

(ii)

Given: DE || BC

By using the proportionality theorem, we get

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

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

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

$\therefore AD=2.4\, cm$

(i)

Given :

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

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

We have

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

Hence, EF is not parallel to QR.

(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.

(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.

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.

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.

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.

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.

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.

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 $PQ||BC$

Draw a line EF passing through point O such that $EO||CD\, \, and\, \, FO||CD$

To prove :

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

In $\triangle ADC$ , we have $CD||EO$

So, by using basic proportionality theorem,

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

In $\triangle ABD$ , 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.

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

Given :

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

In $\triangle ABD$ , 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

(i) $\angle A=\angle P=60 \degree$

$\angle B=\angle Q=80 \degree$

$\angle C=\angle R=40 \degree$

$\therefore \triangle ABC \sim \triangle PQR$ (By AAA)

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

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

$\therefore \triangle ABC \sim \triangle PQR$ (By SSS)

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

(iv) $\triangle MNL \sim \triangle PQR$ by SAS similarity criteria.

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

(vi) In $\triangle DEF$ , we know that

$\angle D+\angle E+\angle F=180 \degree$

$\Rightarrow 70 \degree+80 \degree+\angle F=180 \degree$

$\Rightarrow 150 \degree+\angle F=180 \degree$

$\Rightarrow \angle F=180 \degree-150 \degree=30 \degree$

In $\triangle PQR$ , we know that

$\angle P+\angle Q+\angle R=180 \degree$

$\Rightarrow 110 \degree+\angle R=180 \degree$

$\Rightarrow \angle R=180 \degree-110 \degree=70 \degree$

$\angle Q=\angle P=70 \degree$

$\angle F=\angle R=30 \degree$

$\therefore \triangle DEF\sim \triangle PQR$ ( By AAA)

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

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

$\Rightarrow \angle DOC+125 \degree=180 \degree$

$\Rightarrow \angle DOC=180 \degree-125 \degree$

$\Rightarrow \angle DOC=55 \degree$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 \degree$

$\Rightarrow 55 \degree+ 70 \degree+\angle DCO=180 \degree$

$\Rightarrow \angle DCO+125 \degree=180 \degree$

$\Rightarrow \angle DCO=180 \degree-125 \degree$

$\Rightarrow \angle DCO=55 \degree$

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

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

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.

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)

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)

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)

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)

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)

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)

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)

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 )

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

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

$\angle ABC=\angle AMP$ ( Each $90 \degree$ )

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

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

To prove :

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

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

$\angle ABC=\angle AMP$ ( Each $90 \degree$ )

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

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

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

Hence proved.

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.

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 )

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 )

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 $90 \degree$ )

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

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)

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 )

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

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

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

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

$\angle CAB=\angle RPQ$ ............................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)

CD = pole

AB = tower

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

$\angle CDF=\angle ABE$ ( Each $90 \degree$ )

$\angle DCF=\angle BAE$ (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.

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

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

Class 10 Maths Chapter 6 Triangles Excercise:6.4

$\Delta ABC \sim \Delta DEF$ ( Given )

ar(ABC) = 64 $cm^2$ and ar(DEF)=121 $cm^2$ .

EF = 15.4 cm (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}$

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

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

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

$\Rightarrow BC=11.2 cm$

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

AB = 2 CD ( Given )

In $\triangle AOB\, and\, \triangle COD,$

$\angle COD=\angle AOB$ (vertically opposite angles )

$\angle OCD=\angle OAB$ (Alternate angles)

$\angle ODC=\angle OBA$ (Alternate angles)

$\therefore \triangle AOB\, \sim \, \triangle COD$ (AAA similarity)

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{AB^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{(2CD)^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4.CD^2}{CD^2}$

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

$\Rightarrow ar(\triangle AOB)=ar(\triangle COD)=4:1$

Let DM and AP be perpendicular on BC.

$area\,\,of\,\,triangle=\frac{1}{2}\times base\times perpendicular$

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

In $\triangle APO\, and\, \triangle DMO,$

$\angle APO=\angle DMO$ (Each $90 \degree$ )

$\angle AOP=\angle MOD$ (Vertically opposite angles)

$\triangle APO\, \sim \, \triangle DMO,$ (AA similarity)

$\frac{AP}{DM}=\frac{AO}{DO}$

Since

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

$\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{AP}{ MD}=\frac{AO}{DO}$

Let $\triangle ABC\, \sim \, \triangle DEF,$ , therefore,

$ar(\triangle ABC\,) = \,ar( \triangle DEF)$ (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}................................1$

$\therefore \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=1$

$\Rightarrow \frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}=1$

$AB=DE$

$BC=EF$

$AC=DF$

$\triangle ABC\, \cong \, \triangle DEF$ (SSS )

D, E, and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$ . ( Given )

$DE=\frac{1}{2}AC$ and DE||AC

In $\Delta BED \: \:and \: \: \Delta ABC$ ,

$\angle BED=\angle BCA$ (corresponding angles )

$\angle BDE=\angle BAC$ (corresponding angles )

$\Delta BED \: \:\sim \: \: \Delta ABC$ (By AA)

$\frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{DE^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{(\frac{1}{2}AC)^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{1}{4}$

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times ar(\triangle ABC)$

Let ${ar(\triangle ABC)$ be x.

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times x$

Similarly,

$\Rightarrow ar(\triangle CEF)=\frac{1}{4}\times x$ and $\Rightarrow ar(\triangle ADF)=\frac{1}{4}\times x$

$ar(\triangle ABC)=ar(\triangle ADF)+ar(\triangle BED)+ar(\triangle CEF)+ar(\triangle DEF)$

$\Rightarrow x=\frac{x}{4}+\frac{x}{4}+\frac{x}{4}+ar(\triangle DEF)$

$\Rightarrow x=\frac{3x}{4}+ar(\triangle DEF)$

$\Rightarrow x-\frac{3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{4x-3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{x}{4}=ar(\triangle DEF)$

$\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{\frac{x}{4}}{x}$

$\Rightarrow \frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{1}{4}$

Let AD and PS be medians of both similar triangles.

$\triangle ABC\sim \triangle PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}............................1$

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

$BD=CD=\frac{1}{2}BC\, \, and\, QS=SR=\frac{1}{2}QR$

Purring these value in 1,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..........................3$

In $\triangle ABD\, and\, \triangle PQS,$

$\angle B=\angle Q$ (proved above)

$\frac{AB}{PQ}=\frac{BD}{QS}$ (proved above)

$\triangle ABD\, \sim \triangle PQS$ (SAS )

Therefore,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}................4$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}$

From 1 and 4, we get

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AD^2}{PS^2}$

Let ABCD be a square of side units.

Therefore, diagonal = $\sqrt{2}a$

Triangles form on the side and diagonal are $\triangle$ ABE and $\triangle$ DEF, respectively.

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = $\sqrt{2}a$ units

Both the triangles are equilateral triangles with each angle of $60 \degree$ .

$\triangle ABE\sim \triangle DBF$ ( By AAA)

Using area theorem,

$\frac{ar(\triangle ABC)}{ar(\triangle DBF)}=(\frac{a}{\sqrt{2}a})^2=\frac{1}{2}$

(A) 2: 1 (B) 1: 2 (C) 4 : 1 (D) 1: 4

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

All angles of the triangle are $60 \degree$ .

$\triangle$ ABC $\sim \triangle$ BDE (By AAA)

Let AB=BC=CA = x

then EB=BD=ED= $\frac{x}{2}$

$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}$

Option C is correct.

(A) 2 : 3 (B) 4: 9 (C) 81: 16 (D) 16: 81

Sides of two similar triangles are in the ratio 4: 9.

Let triangles be ABC and DEF.

We know that

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{4^2}{9^2}=\frac{16}{81}$

Option D is correct.

Class 10 Maths Chapter 6 Triangles Excercise: 6.5

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 7 cm, 24 cm

By Pythagoras theorem,

$h^2=7^2+24^2$

$h^2=49+576$

$h^2=625$

$h=25$ = given third side.

Hence, it is the right triangle with h=25 cm.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 3 cm, 6 cm

By Pythagoras theorem,

$h^2=3^2+6^2$

$h^2=9+36$

$h^2=45$

$h=\sqrt{45}\neq 8$

Hence, it is not the right triangle.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 50 cm, 80 cm

By Pythagoras theorem,

$h^2=50^2+80^2$

$h^2=2500+6400$

$h^2=8900$

$h=\sqrt{8900}\neq 100$

Hence, it is not a right triangle.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 5cm, 12 cm

By Pythagoras theorem,

$h^2=5^2+12^2$

$h^2=25+144$

$h^2=169$

$h=13$ = given third side.

Hence, it is a right triangle with h=13 cm.

Let $\angle MPR$ be x

In $\triangle MPR$ ,

$\angle MRP=180 \degree-90 \degree-x$

$\angle MRP=90 \degree-x$

Similarly,

In $\triangle MPQ$ ,

$\angle MPQ=90 \degree-\angle MPR$

$\angle MPQ=90 \degree-x$

$\angle MQP=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle QMP\, and\, \triangle PMR,$

$\angle MPQ\, =\angle MRP$

$\angle PMQ\, =\angle RMP$

$\angle MQP\, =\angle MPR$

$\triangle QMP\, \sim \triangle PMR,$ (By AAA)

$\frac{QM}{PM}=\frac{MP}{MR}$

$\Rightarrow PM^2=MQ\times MR$

Hence proved.

In $\triangle ADB\, and\, \triangle ABC,$

$\angle DAB\, =\angle ACB \, \, \, \, \, \, \, \, (Each 90 \degree)$

$\angle ABD\, =\angle CBA$ (common )

$\triangle ADB\, \sim \triangle ABC$ (By AA)

$\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}$

$\Rightarrow AB^2=BC.BD$ , hence prooved .

Let $\angle CAB$ be x

In $\triangle ABC$ ,

$\angle CBA=180 \degree-90 \degree-x$

$\angle CBA=90 \degree-x$

Similarly,

In $\triangle CAD$ ,

$\angle CAD=90 \degree-\angle CAB$

$\angle CAD=90 \degree-x$

$\angle CDA=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle ABC\, and\, \triangle ACD,$

$\angle CBA\, =\angle CAD$

$\angle CAB\, =\angle CDA$

$\angle ACB\, =\angle DCA$ ( Each right angle)

$\triangle ABC\, \sim \triangle ,ACD$ (By AAA)

$\frac{AC}{DC}=\frac{BC}{AC}$

$\Rightarrow AC^2=BC\times DC$

Hence proved

In $\triangle ACD\, and\, \triangle ABD,$

$\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 \degree)$

$\angle CDA\, =\angle ADB$ (common )

$\triangle ACD\, \sim \triangle ABD$ (By AA)

$\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}$

$\Rightarrow AD^2=BD\times CD$

Hence proved.

Given: ABC is an isosceles triangle right angled at C.

Let AC=BC

In $\triangle$ ABC,

By Pythagoras theorem

$AB^2=AC^2+BC^2$

$AB^2=AC^2+AC^2$ (AC=BC)

$AB^2=2.AC^2$

Hence proved.

Given: ABC is an isosceles triangle with AC=BC.

In $\triangle$ ABC,

$AB^2=2.AC^2$ (Given )

$AB^2=AC^2+AC^2$ (AC=BC)

$AB^2=AC^2+BC^2$

These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.

Hence proved.

Given: ABC is an equilateral triangle of side 2a.

AB=BC=AC=2a

We know that the altitude of an equilateral triangle bisects the opposite side.

So, BD=CD=a

In $\triangle$ ADB,

By Pythagoras theorem,

$AB^2=AD^2+BD^2$

$\Rightarrow (2a)^2=AD^2+a^2$

$\Rightarrow 4a^2=AD^2+a^2$

$\Rightarrow 4a^2-a^2=AD^2$

$\Rightarrow 3a^2=AD^2$

$\Rightarrow AD=\sqrt{3}a$

The length of each altitude is $\sqrt{3}a$ .

In $\triangle$ AOB, by Pythagoras theorem,

$AB^2=AO^2+BO^2..................1$

In $\triangle$ BOC, by Pythagoras theorem,

$BC^2=BO^2+CO^2..................2$

In $\triangle$ COD, by Pythagoras theorem,

$CD^2=CO^2+DO^2..................3$

In $\triangle$ AOD, by Pythagoras theorem,

$AD^2=AO^2+DO^2..................4$

$AB^2+BC^2+CD^2+AD^2=AO^2+BO^2+BO^2+CO^2+CO^2+DO^2+AO^2+DO^2$

$AB^2+BC^2+CD^2+AD^2=2(AO^2+BO^2+CO^2+DO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=2(2.AO^2+2.BO^2)$ (AO=CO and BO=DO)

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4(AO^2+BO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC}{2})^2+(\frac{BD}{2})^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC^2}{4})+(\frac{BD^2}{4}))$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=AC^2+BD^2$

Hence proved .

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$ $\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

Hence proved

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$ $\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

$\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2$ $\Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2$

OA is a wall and AB is a ladder.

In $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 10^2=8^2+BO^2$

$\Rightarrow 100=64+BO^2$

$\Rightarrow 100-64=BO^2$

$\Rightarrow 36=BO^2$

$\Rightarrow BO=6 m$

Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

OB is a pole.

In $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 24^2=18^2+AO^2$

$\Rightarrow 576=324+AO^2$

$\Rightarrow 576-324=AO^2$

$\Rightarrow 252=AO^2$

$\Rightarrow AO=6\sqrt{7} m$

Hence, the distance of the stack from the base of the pole is $6\sqrt{7}$ m.

Distance travelled by the first aeroplane due north in $1\frac{1}{2}$ hours.

$=1000\times \frac{3}{2}=1500 km$

Distance travelled by second aeroplane due west in $1\frac{1}{2}$ hours.

$=1200\times \frac{3}{2}=1800 km$

OA and OB are the distance travelled.

By Pythagoras theorem,

$AB^2=OA^2+OB^2$

$\Rightarrow AB^2=1500^2+1800^2$

$\Rightarrow AB^2=2250000+3240000$

$\Rightarrow AB^2=5490000$

$\Rightarrow AB^2=300\sqrt{61}km$

Thus, the distance between the two planes is $300\sqrt{61}km$ .

Let AB and CD be poles of heights 6 m and 11 m respectively.

CP=11-6=5 m and AP= 12 m

In $\triangle$ APC,

By Pythagoras theorem,

$AP^2+PC^2=AC^2$

$\Rightarrow 12^2+5^2=AC^2$

$\Rightarrow 144+25=AC^2$

$\Rightarrow 169=AC^2$

$\Rightarrow AC=13m$

Hence, the distance between the tops of two poles is 13 m.

In $\triangle$ ACE, by Pythagoras theorem,

$AE^2=AC^2+CE^2..................1$

In $\triangle$ BCD, by Pythagoras theorem,

$DB^2=BC^2+CD^2..................2$

From 1 and 2, we get

$AC^2+CE^2+BC^2+CD^2=AE^2+DB^2..................3$

In $\triangle$ CDE, by Pythagoras theorem,

$DE^2=CD^2+CE^2..................4$

In $\triangle$ ABC, by Pythagoras theorem,

$AB^2=AC^2+CB^2..................5$

From 3,4,5 we get

$DE^2+AB^2=AE^2+DB^2$

In $\triangle$ ACD, by Pythagoras theorem,

$AC^2=AD^2+DC^2$

$AC^2-DC^2=AD^2..................1$

In $\triangle$ ABD, by Pythagoras theorem,

$AB^2=AD^2+BD^2$

$AB^2-BD^2=AD^2.................2$

From 1 and 2, we get

$AC^2-CD^2=AB^2-DB^2..................3$

Given : 3DC=DB, so

$CD=\frac{BC}{4}\, \, and\, \, BD=\frac{3BC}{4}........................4$

From 3 and 4, we get

$AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2$

$AC^2-(\frac{BC^2}{16})=AB^2-(\frac{9BC^2}{16})$

$16AC^2-BC^2=16AB^2- 9BC^2$

$16AC^2=16AB^2- 8BC^2$

$\Rightarrow 2AC^2=2AB^2- BC^2$

$2 AB^2 = 2 AC^2 + BC^2.$

Hence proved.

Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove : $9 AD^2 = 7 AB^2$

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

In $\triangle$ AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

$a^2=AE^2+(\frac{a}{2})^2$

$\Rightarrow a^2-(\frac{a^2}{4})=AE^2$

$\Rightarrow (\frac{3a^2}{4})=AE^2$

$\Rightarrow AE=(\frac{\sqrt{3}a}{2})$

Given : BD = 1/3 BC.

$BD=\frac{a}{3}$

$DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$

In $\triangle$ ADE, by Pythagoras theorem,

$AD^2=AE^2+DE^2$

$\Rightarrow AD^2=(\frac{\sqrt{3}a}{2})^2+(\frac{a}{6})^2$

$\Rightarrow AD^2=(\frac{3a^2}{4})+(\frac{a^2}{36})$

$\Rightarrow AD^2=(\frac{7a^2}{9})$

$\Rightarrow AD^2=(\frac{7AB^2}{9})$

$\Rightarrow 9AD^2=7AB^2$

Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

In $\triangle$ AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

$a^2=AE^2+(\frac{a}{2})^2$

$\Rightarrow a^2-(\frac{a^2}{4})=AE^2$

$\Rightarrow (\frac{3a^2}{4})=AE^2$

$\Rightarrow 3a^2=4AE^2$

$\Rightarrow 4.(altitude)^2=3.(side)^2$

(B) 60°

(C) 90°

(D) 45°

In $\Delta ABC$ AB = $6 \sqrt 3$ cm, AC = 12 cm and BC = 6 cm.

$AB^2+BC^2=108+36$

$=144$

$=12^2$

$=AC^2$

It satisfies the Pythagoras theorem.

Hence, ABC is a right-angled triangle and right-angled at B.

Option C is correct.

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.6

A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of $\angle QPR \: \: of\: \: \Delta PQR$ .

$\angle QPS=\angle SPR.....................................1$

By construction,

$\angle SPR=\angle PRT.....................................2$ (as PS||TR)

$\angle QPS=\angle QTR.....................................3$ (as PS||TR)

From the above equations, we get

$\angle PRT=\angle QTR$

$\therefore PT=PR$

By construction, PS||TR

In $\triangle$ QTR, by Thales theorem,

$\frac{QS}{SR}=\frac{QP}{PT}$

$\frac{QS }{SR } = \frac{PQ }{PR }$

Hence proved.

Join BD

Given : D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB.Also DN || BC, DM||NB

$\angle CDB=90 \degree$

$\Rightarrow \angle 2+\angle 3=90 \degree.............................1$

In $\triangle$ CDM, $\angle 1+\angle 2+\angle DMC=180 \degree$

$\angle 1+\angle 2=90 \degree.......................2$

In $\triangle$ DMB, $\angle 3+\angle 4+\angle DMB=180 \degree$

$\angle 3+\angle 4=90 \degree.......................3$

From equation 1 and 2, we get $\angle 1=\angle 3$

From equation 1 and 3, we get $\angle 2=\angle 4$

In $\triangle DCM\, \, and\, \, \triangle BDM,$

$\angle 1=\angle 3$

$\angle 2=\angle 4$

$\triangle DCM\, \, \sim \, \, \triangle BDM,$ (By AA)

$\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}$

$\Rightarrow \frac{DN}{DM}=\frac{DM}{MC}$ (BM=DN)

$\Rightarrow$ $DM^2 = DN . MC$

Hence proved

In $\triangle$ DBN,

$\angle 5+\angle 7=90 \degree.......................1$

In $\triangle$ DAN,

$\angle 6+\angle 8=90 \degree.......................2$

BD $\perp$ AC, $\therefore \angle ADB=90 \degree$

$\angle 5+\angle 6=90 \degree.......................3$

From equation 1 and 3, we get $\angle 6=\angle 7$

From equation 2 and 3, we get $\angle 5=\angle 8$

In $\triangle DNA\, \, and\, \, \triangle BND,$

$\angle 6=\angle 7$

$\angle 5=\angle 8$

$\triangle DNA\, \, \sim \, \, \triangle BND$ (By AA)

$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$

$\Rightarrow \frac{AN}{DN}=\frac{DN}{DM}$ (NB=DM)

$\Rightarrow$ $DN^2 = AN . DM$

Hence proved.

In $\triangle$ ADB, by Pythagoras theorem

$AB^2=AD^2+DB^2.......................1$

In $\triangle$ ACD, by Pythagoras theorem

$AC^2=AD^2+DC^2.......................2$

$AC^2=AD^2+(BD+BC)^2$

$\Rightarrow AC^2=AD^2+(BD)^2+(BC)^2+2.BD.BC$

$AC^2 = AB^2 + BC^2 + 2 BC . BD.$ (From 1)

In $\triangle$ ADB, by Pythagoras theorem

$AB^2=AD^2+DB^2$

$AD^2=AB^2-DB^2...........................1$

In $\triangle$ ACD, by Pythagoras theorem

$AC^2=AD^2+DC^2$

$AC^2=AB^2-BD^2+DC^2$ (From 1)

$\Rightarrow AC^2=AB^2-BD^2+(BC-BD)^2$

$\Rightarrow AC^2=AB^2-BD^2+(BC)^2+(BD)^2-2.BD.BC$

$AC^2 = AB^2 + BC^2 - 2 BC . BD.$

Given: AD is a median of a triangle ABC and AM $\perp$ BC.

In $\triangle$ AMD, by Pythagoras theorem

$AD^2=AM^2+MD^2.......................1$

In $\triangle$ AMC, by Pythagoras theorem

$AC^2=AM^2+MC^2$

$AC^2=AM^2+(MD+DC)^2$

$\Rightarrow AC^2=AM^2+(MD)^2+(DC)^2+2.MD.DC$

$AC^2 = AD^2 + DC^2 + 2 DC . MD.$ (From 1)

$AC^2 = AD^2 + (\frac{BC}{2})^2 + 2(\frac{BC}{2}). MD.$ (BC=2 DC)

$AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2$

In $\triangle$ ABM, by Pythagoras theorem

$AB^2=AM^2+MB^2$

$AB^2=(AD^2-DM^2)+MB^2$

$\Rightarrow AB^2=(AD^2-DM^2)+(BD-MD)^2$

$\Rightarrow AB^2=AD^2-DM^2+(BD)^2+(MD)^2-2.BD.MD$

$\Rightarrow AB^2=AD^2+(BD)^2-2.BD.MD$

$\Rightarrow AB^2 = AD^2 + (\frac{BC}{2})^2 -2(\frac{BC}{2}). MD.=AC^2$ (BC=2 BD)

$\Rightarrow AD^2 + (\frac{BC}{2})^2 -BC. MD.=AC^2$

In $\triangle$ ABM, by Pythagoras theorem

$AB^2=AM^2+MB^2.......................1$

In $\triangle$ AMC, by Pythagoras theorem

$AC^2=AM^2+MC^2$ ..................................2

$AB^2+AC^2=2AM^2+MB^2+MC^2$

$\Rightarrow AB^2+AC^2=2AM^2+(BD-DM)^2+(MD+DC)^2$

$\Rightarrow AB^2+AC^2=2AM^2+(BD)^2+(DM)^2-2.BD.DM+(MD)^2+(DC)^2+2.MD.DC$

$\Rightarrow AB^2+AC^2=2AM^2+2.(DM)^2+BD^2+(DC)^2+2.MD.(DC-BD)$ $\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2.MD.(\frac{BC}{2}-\frac{BC}{2})$ $\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2$

$AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2$

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In $\triangle$ DEA, by Pythagoras theorem

$DA^2=DE^2+EA^2.......................1$

In $\triangle$ DEB, by Pythagoras theorem

$DB^2=DE^2+EB^2$

$DB^2=DE^2+(EA+AB)^2$

$DB^2=DE^2+(EA)^2+(AB)^2+2.EA.AB$

$DB^2=DA^2+(AB)^2+2.EA.AB$ ....................................2

In $\triangle$ ADF, by Pythagoras theorem

$DA^2=AF^2+FD^2$

In $\triangle$ AFC, by Pythagoras theorem

$AC^2=AF^2+FC^2=AF^2+(DC-FD)^2$

$\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD$

$\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD$

$\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3$

Since ABCD is a parallelogram.

In $\triangle DEA\, and\, \triangle ADF,$

$\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 \degree)$

$\angle DAE=\angle ADF$ (AE||DF)

$\triangle DEA\, \cong \, \triangle ADF,$ (ASA rule)

$\Rightarrow EA=DF.......................6$

$DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2$ $\Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2$

$\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2$ (From 4 and 6)

$\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2$ \

Join BC

In $\triangle APC\, \, and\, \triangle DPB,$

$\angle APC\, \, = \angle DPB$ ( vertically opposite angle)

$\angle CAP\, \, = \angle BDP$ (Angles in the same segment)

$\triangle APC\, \, \sim \triangle DPB$ (By AA)

Join BC

In $\triangle APC\, \, and\, \triangle DPB,$

$\angle APC\, \, = \angle DPB$ ( vertically opposite angle)

$\angle CAP\, \, = \angle BDP$ (Angles in the same segment)

$\triangle APC\, \, \sim \triangle DPB$ (By AA)

$\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

In $\Delta PAC \,and \,\Delta PDB,$

$\angle P=\angle P$ (Common)

$\angle PAC=\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, $\Delta PAC \sim \Delta PDB$ ( By AA rule)

In $\Delta PAC \,and \,\Delta PDB,$

$\angle P=\angle P$ (Common)

$\angle PAC=\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, $\Delta PAC \sim \Delta PDB$ ( By AA rule)

24440 $\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

Produce BA to P, such that AP=AC and join P to C.

$\frac{BD }{CD} = \frac{AB}{AC}$ (Given )

$\Rightarrow \frac{BD }{CD} = \frac{AP}{AC}$

Using converse of Thales theorem,

AD||PC $\Rightarrow \angle BAD=\angle APC............1$ (Corresponding angles)

$\Rightarrow \angle DAC=\angle ACP............2$ (Alternate angles)

By construction,

AP=AC

$\Rightarrow \angle APC=\angle ACP............3$

From equation 1,2,3, we get

$\Rightarrow \angle BAD=\angle APC$

Let AB = 1.8 m

BC is a horizontal distance between fly to the tip of the rod.

Then, the length of the string is AC.

In $\triangle$ ABC, using Pythagoras theorem

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(1.8)^2+(2.4)^2$

$\Rightarrow AC^2=3.24+5.76$

$\Rightarrow AC^2=9.00$

$\Rightarrow AC=3 m$

Hence, the length of the string which is out is 3m.

If she pulls in the string at the rate of 5cm/s, then the distance travelled by fly in 12 seconds.

= $12\times 5=60cm=0.6m$

Let D be the position of fly after 12 seconds.

Hence, AD is the length of the string that is out after 12 seconds.

Length of string pulled in by nazim=AD=AC-12

=3-0.6=2.4 m

In $\triangle$ ADB,

$AB^2+BD^2=AD^2$

$\Rightarrow (1.8)^2+BD^2=(2.4)^2$

$\Rightarrow BD^2=5.76-3.24=2.52 m^2$

$\Rightarrow BD=1.587 m$

Horizontal distance travelled by fly = BD+1.2 m

=1.587+1.2=2.787 m

= 2.79 m

## NCERT Class 10 Maths solutions chapter 6 - Topics

• Similarity of triangles

• Theorems based on similar triangles

• Areas of similar triangles

• Theorems related to Trapezium

• Pythagoras theorem

Also get the solutions of individual exercises-

## Key Features Of NCERT Solutions for Class 10 Maths Chapter 6

Comprehensive Coverage: NCERT Solutions for class 10 triangles cover all the topics and concepts included in the CBSE syllabus for this chapter.

Detailed Explanations: The class 10 triangles solutions provide step-by-step and clear explanations for each problem, making it easy for students to understand the concepts and solutions.

CBSE-Aligned: These triangle solutions class 10 are closely aligned with the CBSE curriculum for Class 10, ensuring that students are well-prepared for their board exams.

Illustrative Examples: The triangle solutions class 10 often include illustrative examples to help students grasp the application of mathematical concepts.

Clarity: Concepts are explained in a simple and straightforward manner, ensuring that students can follow along and build a strong foundation in mathematics.

## NCERT solutions for class 10 maths - chapter wise

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

## NCERT Exemplar solutions - Subject Wise

### How to use NCERT Solutions for Class 10 Maths Chapter 6 Triangles?

• First of all, go through all the concepts, theorems and examples given in the chapter.

• This chapter needs so much use of theorems. So you have to memorize these conditions and theorems to solve the problems.

• While solving the practising the exercises, if you face any problem in a question then you take the help of NCERT solutions for class 10 maths chapter 6.

• Once you have done the practise exercises you can move to previous year questions.

Keep working hard & happy learning!

1. What are the benefits of understanding all the concepts in chapter 6 maths class 10 NCERT solutions?

Understanding all the concepts of NCERT textbook class 10 chapter 6 maths can provide several benefits including Improved problem-solving skills, greater understanding of mathematical concepts, better performance on tests and exams, stronger foundation for further studies, and many. Hence students should practise NCERT solutions class 10 maths ch 6 concepts conprenhesively.

2. Can you list the key topics of maths chapter 6 class 10?

Class 10 chapter 6 maths contains multiple concepts including similar figures, similarity of triangles, criteria for similarity of triangles, areas of similar triangles, pythagoras theorem, and many more. Students can practise these concepts using problems provided in different exercises and they can also use class 10 triangles solutions.

3. What is theorem 6.1 in Class 10 maths triangles?

Triangle chapter class 10 theorem 6.1 states that: "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio." This theorem is often referred to as the "Alternate Interior Angles Theorem" or "AA Similarity Theorem". Students should practise triangles class 10 solutions  to get an in-depth understanding of the concepts.

4. How many exercise problems are included in NCERT Solutions for Class 10 Maths Chapter 6?

Class 10 chapter 6 maths triangle contains six exercises. These contain different typology of questions which are repeatedly asked in boards as well as competitive exams. Therefore students should  learn CBSE class 10 maths triangle  chapter in detail and practise lots of questions provided in different exercises.

5. What is theorem 6.2 in Class 10 maths triangle?

Theorem 6.2 in maths class 10, specifically in Triangles, states that:

"If a line is drawn parallel to one side of a triangle and it intersects the other two sides in distinct points, the corresponding angles formed by the line and the other two sides of the triangle will be equal." students can download these exercises in the form of  triangles class 10 pdf.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

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