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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

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.

1635922757481

2. Two rectangles with different breadth and length.

1635922782211

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

1635922829572

2. A circle and a triangle.

1635922838551

Q3 State whether the following quadrilaterals are similar or not:

1635922858166


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).

1635923248436

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}=2\, cm

\therefore EC=2\, cm

(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}=2.4\, cm

\therefore AD=2.4\, cm

Q2 (1) 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 = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer:

(i)

1635923264991

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.

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

1635923547862

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

1635923629376

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.

1635923658934

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.

1635923722792

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:

1635923725135

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:

1635923738877

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:

1635923757337

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.

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:

1635923771348

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

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 :

1635924024150

Answer:

(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 30 \degree+80 \degree+\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 E=\angle Q=80 \degree

\angle F=\angle R=30 \degree

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

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

1635924052755

Answer:

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).

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:

1635924090157

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

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

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

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

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

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

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

Hence proved.

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

1635924132584


Answer:

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

To prove : \Delta PQS \sim \Delta TQR

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

\therefore PQ=PR

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

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


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

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

\angle Q=\angle Q (Common)

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

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

Answer:

1635924147463

Given : \angle P = \angle RTS

To prove RPQ ~ \Delta RTS.

In \Delta RPQ and \Delta RTS,

\angle P = \angle RTS (Given )

\angle R = \angle R (common)

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

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

1635924156441

Answer:

Given : \triangle ABE \cong \triangle ACD

To prove ADE ~ \Delta ABC.

Since \triangle ABE \cong \triangle ACD

AB=AC ( By CPCT)

AD=AE (By CPCT)

In \Delta ADE and \Delta ABC,

\angle A=\angle A ( Common)

and

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

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

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

1635924178546

Answer:

To prove : \Delta AEP \sim \Delta CDP

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

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

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

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

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

Answer:

To prove : \Delta ABD \sim \Delta CBE

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

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

\angle ABD=\angle CBE (Common )

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

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

Answer:

To prove : \Delta AEP \sim \Delta ADB

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

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

\angle A=\angle A (Common )

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

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

Answer:

To prove : \Delta PDC \sim \Delta BEC

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

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

\angle C=\angle C (Common )

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

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

Answer:

1635924231460

To prove : \Delta ABE \sim \Delta CFB

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

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

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

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

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

Answer:

To prove :

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

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

\angle ABC=\angle AMP ( Each 90 \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.

Q10 (1) CD and GH are respectively the bisectors of \angle ACB and \angle EGF such that D and H lie on sides AB and FE of \Delta ABC\: \: and\: \: \Delta EGF respectively. If \Delta ABC \sim \Delta EGF , show that: \frac{CD}{GH} = \frac{AC}{FG}

Answer:

1635924284479

To prove :

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

Given : \Delta ABC \sim \Delta EGF

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

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

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

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

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

\angle A=\angle F ( proved above)

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

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

Hence proved.

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

Answer:

1635924301949

To prove : \Delta DCB \sim \Delta HGE

Given : \Delta ABC \sim \Delta EGF

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

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

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

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

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

Answer:

1635924370515

To prove : \Delta DCA \sim \Delta HGF

Given : \Delta ABC \sim \Delta EGF

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

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

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

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

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

1635924369822

Answer:

To prove : \Delta ABD \sim \Delta ECF

Given: ABC is an isosceles triangle.

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

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

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

\angle ADB=\angle EFC ( Each 90 \degree )

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

Q12 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \Delta PQR (see Fig. 6.41). Show that \Delta ABC \sim \Delta PQR

1635924392444

Answer:

AD and PM are medians of triangles. So,

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

Given :

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

1635924451786

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

\angle ADC = \angle BAC ( given )

\angle ACD = \angle BCA (common )

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

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

\Rightarrow CA^2=CB\times CD

Q14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \Delta ABC \sim \Delta PQR

Answer:

1635924470215

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

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

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

AD and PM are medians of a triangle, therefore

QM=MR and BD=DC

AD = DE (By construction)

PM=ML (By construction)

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

Similarly, PQLR is also a parallelogram.

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

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

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

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

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

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

Similarity, \triangle AEC=\triangle PLR

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

Adding equation 1 and 2,

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

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

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

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

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

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

Q15 A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer:

1635924490568

CD = pole

AB = tower

Shadow of pole = DF

Shadow of tower = BE

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.

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

Answer:

1635924504095

\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

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

Answer:

\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

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:

1635931770237

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

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}

1635931784662

Answer:

1635931796737

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}

Q5 D, E, and F are respectively the mid-points of sides AB, BC and CA of \Delta ABC . Find the ratio of the areas of \Delta DEF \: \:and \: \: \Delta ABC

Answer:

1635932236410

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}

Q6 Proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:

1635932258592

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}

Q7 Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer:

1635932273637

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}

Q8 Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is

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

Answer:

1635932291018

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.

Q9 Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

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

Answer:

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

Q1 (1) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 7 cm, 24 cm, 25 cm

Answer:

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.

Q1 (2) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm

Answer:

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.

Q1 (3) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 50 cm, 80 cm, 100 cm

Answer:

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.

Q1 (4) Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm

Answer:

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.

Q2 PQR is a triangle right angled at P and M is a point on QR such that PM \perp QR . Show that PM ^2 = QM . MR .

Answer:

1635933005578

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.

Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC \perp BD. Show that AB^2 = BC . BD .

1635933025836

Answer:

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 .

Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC \perp BD. Show that AC^2 = BC . DC .

1635933077069

Answer:

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

Q4 ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC ^2

Answer:

1635933142785

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.

Q5 ABC is an isosceles triangle with AC = BC. If AB ^ 2 = 2 AC ^ 2 , prove that ABC is a right triangle.

Answer:

1635933162430

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.

Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:

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

1635933173577

AB=BC=AC=2a

AD is perpendicular to BC.

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 .

Q7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

1635933187293

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

Adding equation 1,2,3,4,we get

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 .

Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD \perp BC, OE \perp AC and OF \perp AB. Show that OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2,

1635933201520

Answer:


1635933215218

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

Adding equation 1,2,3,we get

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

Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD \perp BC, OE \perp AC and OF \perp AB. AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.

1635933229014

Answer:

1635933270968

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

Adding equation 1,2,3,we get

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

Q9 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Answer:

1635933282458

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.

Q11 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 1/2 hours?

Answer:

1635933311960

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 .

Q12 Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their top

Answer:

1635933349053

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.

Q13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2 + BD^2 = AB^2 + DE^2.

Answer:

1635933368584

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

Q14 The perpendicular from A on side BC of a \Delta ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB^2 = 2 AC^2 + BC^2.

1635933387695

Answer:

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.

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD^2 = 7 AB^2

Answer:

1635933424958

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

Q16 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:

1635933437532

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

Q17 Tick the correct answer and justify : In \Delta ABC AB = 6 \sqrt 3 cm, AC = 12 cm and BC = 6 cm.
The angle B is :

(A) 120°

(B) 60°

(C) 90°

(D) 45°

Answer:

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

Q1 In Fig. 6.56, PS is the bisector of \angle QPR \: \: of\: \: \Delta PQR . Prove that \frac{QS }{SR } = \frac{PQ }{PR }

1635933619787

Answer:

1635933634308

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.

Q2 In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD \perp AC, DM \perp BC and DN \perp AB. Prove that : DM^2 = DN . MC

1635933650488

Answer:

1635933658497

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

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD \perp AC, DM \perp BC and DN \perp AB. Prove that: DN^2 = DM . AN

1635933672815

Answer:

1635933684708

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.

Q3 In Fig. 6.58, ABC is a triangle in which \angle ABC > 90° and AD \perp CB produced. Prove that AC^2 = AB^2 + BC^2 + 2 BC . BD.

1635933712397

Answer:

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)

Q4 In Fig. 6.59, ABC is a triangle in which \angle ABC < 90° and AD \perp BC. Prove that AC^2 = AB^2 + BC^2 - 2 BC . BD.

1635933729274

Answer:

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.

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that : AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2

1635933758550

Answer:

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

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that : AB ^2 = AD ^2 - BC .DM + \left ( \frac{BC}{2} \right ) ^2

1635933791582

Answer:

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

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM \perp BC. Prove that: AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2

1635933807753

Answer:

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

Adding equation 1 and 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

Q6 Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:

1635933824435

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.

SO, AB=CD and BC=AD

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

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

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

AD=AD (common)

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

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

Adding 2 and, we get

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 \

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : \Delta APC \sim \Delta DPB

1635933839234

Answer:

1635933848323

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)

Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : AP . PB = CP . DP

1635933860992

Answer:

1635933929931

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

Q8 (1) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that \Delta PAC \sim \Delta PDB

1635933941642

Answer:

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)

Q8 (2) In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that PA. PB = PC. PD

1635933976947

Answer:

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

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that \frac{BD }{CD} = \frac{AB}{AC} Prove that AD is the bisector of \angle BAC.

1635933987426

Answer:

1635933998131

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

Thus, AD bisects angle BAC.

Q10 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

1635934015475

Answer:

1635934024120

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 Of Class 10 - Subject Wise

NCERT Books and NCERT Syllabus

NCERT solutions for class 10 maths - chapter wise

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.

  • After this, you can directly jump to practice exercises.

  • 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!

Frequently Asked Question (FAQs)

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 .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

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.

https://school.careers360.com/boards/cbse/cbse-date-sheet

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.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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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The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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