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NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Edited By Ramraj Saini | Updated on Mar 16, 2025 06:06 PM IST | #CBSE Class 10th
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CBSE Class 10th  Exam Date : 18 Mar' 2025 - 18 Mar' 2025

NCERT Solutions for Class 10 Maths Chapter 6 is Triangle, an important study material for CBSE Class 10 students. This chapter aligns perfectly with the CBSE Syllabus for 2025-26, covering many rules and theorems. Sometimes, students get perplexed about which theorem to use. Here, students will get the NCERT solutions for Class 10 in chapter-wise. Practice these solutions for the triangles class 10, and sail the boat of the triangle.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF Free Download
  2. NCERT Solutions Class 10 Maths Chapter 6 - Important Points
  3. NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Intext Questions and Exercise)
  4. NCERT Class 10 Maths solutions chapter 6 - Topics
  5. Key Features Of NCERT Solutions for Class 10 Maths Chapter 6
  6. NCERT Solutions Of Class 10 - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. NCERT solutions for class 10 maths - chapter wise
  9. NCERT Exemplar solutions - Subject Wise
NCERT Solutions for Class 10 Maths Chapter 6 Triangles
NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT solutions created at Careers360 are made using the concept in the easiest way that a student can grasp. Subject experts created these solutions, keeping the ease and the board exam. NCERT class 10 triangle solution not just for exams but also for tackling homework, assignments, and to understand the relation of triangles with other concepts too.

Background wave

CBSE Class 10 exams usually have questions from NCERT textbooks. So, Chapter 6's NCERT Solutions for Class 10 Maths are the best bet for getting ready and tackling any kind of question from this chapter. We strongly recommend practicing these solutions regularly to ace the Class 10 board exams. Math triangle class 10 Theorems of triangles are available free for the students and can be downloaded using the below link.

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF Free Download

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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

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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.

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.

These concepts are used to solve Important questions of triangles class 10 with solutions.

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

ADDB=AEEC

1.53=1x

x=31.5=2cm

EC=2cm

(ii)

Let AD be x

Given: DE || BC

By using the proportionality theorem, we get

ADDB=AEEC

x7.2=1.85.4

x=7.23=2.4cm

AD=2.4cm

Q2 (1) E and F are points on the sides PQ and PR respectively of a 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

PEEQ=3.93=1.3cm and PFFR=3.62.4=1.5cm

We have

PEEQPFFR

Hence, EF is not parallel to QR.

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

Answer:

(ii)

1635923463637

Given :

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

PEEQ=44.5=89cm and PFFR=89cm


We have

PEEQ=PFFR

Hence, EF is parallel to QR.

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

Answer:

(iii)

1635923501505

Given :

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

PEPQ=0.181.28=964cm and PFPR=0.362.56=964cm


We have

PEEQ=PFFR

Hence, EF is parallel to QR.

Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that AMAB=ANAD

1635923547862

Answer:

Given : LM || CB and LN || CD

To prove :

AMAB=ANAD

Since , LM || CB so we have

AMAB=ALAC.............................................1

Also, LN || CD

ALAC=ANAD.............................................2


From equation 1 and 2, we have

AMAB=ANAD

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 :

BFFE=BEEC

Since , DE || AC so we have

BDDA=BEEC.............................................1

Also,DF || AE

BDDA=BFFE.............................................2

From equation 1 and 2, we have

BFFE=BEEC

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

PEEQ=PDDO.............................................1

Also, DF || OR

PFFR=PDDO.............................................2

From equation 1 and 2, we have

PEEQ=PFFR

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

OAAP=OBBQ.............................................1

Also, AC || PR

OAAP=OCCR.............................................2

From equation 1 and 2, we have

OBBQ=OCCR

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

APPB=AQQC..........................1

Since AP=PB

AQQC=11

AQ=QC

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,

APPB=11..........................1

AQQC=11...................................2

From equation 1 and 2, we get

AQQC=APPB

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 AOBO=CODO

Answer:

1635923757337

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

To prove :

AOBO=CODO

In ADC , we have CD||EO

So, by using basic proportionality theorem,

AEED=AOOC........................................1

In ABD , we have AB||EO

So, by using basic proportionality theorem,

DEEA=ODBO........................................2

Using equation 1 and 2, we get

AOOC=BOOD

AOBO=CODO

Hence proved.

Q10 The diagonals of a quadrilateral ABCD intersect each other at point O such that AOBO=CODO Show that ABCD is a trapezium.

Answer:

1635923771348

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

Given :

AOBO=CODO

In ABD , we have AB||EO

So, by using basic proportionality theorem,

AEED=BODO........................................1

However, its is given that

AOCO=BODO..............................2

Using equation 1 and 2 , we get

AEED=AOCO

EO||CD (By basic proportionality theorem)

AB||EO||CD

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) A=P=60

B=Q=80

C=R=40

ABCPQR (By AAA)

So , ABQR=BCRP=CAPQ


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

ABCPQR (By SSS)


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


(iv) MNLPQR by SAS similarity criteria.


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


(vi) In DEF , we know that

D+E+F=180

70+80+F=180

150+F=180

F=180150=30


In PQR , we know that

P+Q+R=180

30+80+R=180

110+R=180

R=180110=70

Q=P=70

E=Q=80

F=R=30

DEFPQR ( By AAA)

Q2 In Fig. 6.35, ΔODCΔOBA , BOC=125 and CDO=70 . Find DOC,DCO,OAB

1635924052755

Answer:

Given : ΔODCΔOBA , BOC=125 and CDO=70

DOC+BOC=180 (DOB is a straight line)

DOC+125=180

DOC=180125

DOC=55

In ΔODC,

DOC+ODC+DCO=180

55+70+DCO=180

DCO+125=180

DCO=180125

DCO=55

Since , ΔODCΔOBA , so

OAB=DCO=55 ( 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 OAOC=OBOD

Answer:

1635924090157

In DOCandBOA , we have

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

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

DOC=BOA ( Vertically opposite angles are equal)

DOCBOA ( By AAA)

DOBO=OCOA ( corresponding sides are equal)

OAOC=OBOD

Hence proved.

Q4 In Fig. 6.36, QRQS=QTPR and 1=2 . Show that ΔPQSΔTQR

1635924132584


Answer:

Given : QRQS=QTPR and 1=2

To prove : ΔPQSΔTQR

In PQR , PQR=PRQ

PQ=PR

QRQS=QTPR (Given)

QRQS=QTPQ


In ΔPQSandΔTQR ,

QRQS=QTPQ

Q=Q (Common)

ΔPQSΔTQR ( By SAS)

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

Answer:

1635924147463

Given : P = RTS

To prove RPQ ~ Δ RTS.

In Δ RPQ and Δ RTS,

P = RTS (Given )

R = R (common)

Δ RPQ ~ Δ RTS. ( By AA)

Q6 In Fig. 6.37, if Δ ABE Δ ACD, show that Δ ADE ~ Δ ABC.

1635924156441

Answer:

Given : ABEACD

To prove ADE ~ Δ ABC.

Since ABEACD

AB=AC ( By CPCT)

AD=AE (By CPCT)

In Δ ADE and Δ ABC,

A=A ( Common)

and

ADAB=AEAC ( AB=AC and AD=AE )

Therefore, Δ ADE ~ Δ ABC. ( By SAS criteria)

Q7 (1) In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: ΔAEPΔCDP

1635924178546

Answer:

To prove : ΔAEPΔCDP

In ΔAEPandΔCDP ,

AEP=CDP ( Both angles are right angle)

APE=CPD (Vertically opposite angles )

ΔAEPΔCDP ( By AA criterion)

Q7 (2) In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: ΔABDΔCBE

Answer:

To prove : ΔABDΔCBE

In ΔABDandΔCBE ,

ADB=CEB ( Both angles are right angle)

ABD=CBE (Common )

ΔABDΔCBE ( By AA criterion)

Q7 (3) In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: ΔAEPΔADB

Answer:

To prove : ΔAEPΔADB

In ΔAEPandΔADB ,

AEP=ADB ( Both angles are right angle)

A=A (Common )

ΔAEPΔADB ( By AA criterion)

Q7 (4) In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: ΔPDCΔBEC

Answer:

To prove : ΔPDCΔBEC

In ΔPDCandΔBEC ,

CDP=CEB ( Both angles are right angle)

C=C (Common )

ΔPDCΔBEC ( 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 ΔABEΔCFB

Answer:

1635924231460

To prove : ΔABEΔCFB

In ΔABEandΔCFB ,

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

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

ΔABEΔCFB ( By AA criterion )

Q9 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: ΔABCΔAMP

1635924254422

Answer:

To prove : ΔABCΔAMP

In ΔABCandΔAMP ,

ABC=AMP ( Each 90 )

A=A ( common)

ΔABCΔAMP ( By AA criterion )

Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that CAPA=BCMP

Answer:

To prove :

CAPA=BCMP

In ΔABCandΔAMP ,

ABC=AMP ( Each 90 )

A=A ( common)

ΔABCΔAMP ( By AA criterion )

CAPA=BCMP ( corresponding parts of similar triangles )

Hence proved.

Q10 (1) CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ΔABCandΔEGF respectively. If ΔABCΔEGF , show that: CDGH=ACFG

Answer:

1635924284479

To prove :

CDGH=ACFG

Given : ΔABCΔEGF

A=F,B=EandACB=FGE,ACB=FGE

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

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

In ΔACDandΔFGH

ACD=FGH ( proved above)

A=F ( proved above)

ΔACDΔFGH ( By AA criterion)

CDGH=ACFG

Hence proved.

Q 10 (2) CD and GH are respectively the bisectors of ABCandEGF such that D and H lie on sides AB and FE of ΔABCandΔEGF respectively. If ΔABCΔEGF , show that: ΔDCBΔHGE

Answer:

1635924301949

To prove : ΔDCBΔHGE

Given : ΔABCΔEGF

In ΔDCBandΔHGE ,

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

B=E ( ΔABCΔEGF )

ΔDCBΔHGE ( By AA criterion )

Q10 (3) CD and GH are respectively the bisectors of ABCandEGF such that D and H lie on sides AB and FE of ΔABCandΔEGF respectively. If ΔABCΔEGF , show that: ΔDCAΔHGF

Answer:

1635924370515

To prove : ΔDCAΔHGF

Given : ΔABCΔEGF

In ΔDCAandΔHGF ,

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

A=F ( ΔABCΔEGF )

ΔDCAΔHGF ( 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 ADBC and EFAC , prove that ΔABDΔECF

1635924369822

Answer:

To prove : ΔABDΔECF

Given: ABC is an isosceles triangle.

AB=ACandB=C

In ΔABDandΔECF ,

ABD=ECF ( ABD=B=C=ECF )

ADB=EFC ( Each 90 )

ΔABDΔECF ( 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 ΔPQR (see Fig. 6.41). Show that ΔABCΔPQR

1635924392444

Answer:

AD and PM are medians of triangles. So,

BD=BC2andQM=QR2

Given :

ABPQ=BCQR=ADPM

ABPQ=12BC12QR=ADPM

ABPQ=BDQM=ADPM

In ABDandPQM,

ABPQ=BDQM=ADPM

ABDPQM, (SSS similarity)

ABD=PQM ( Corresponding angles of similar triangles )

In ABCandPQR,

ABD=PQM (proved above)

ABPQ=BCQR

Therefore, ΔABCΔPQR . ( SAS similarity)

Q13 D is a point on the side BC of a triangle ABC such that ADC=BAC . Show that CA2=CB.CD.

Answer:

1635924451786

In, ADCandBAC,

ADC=BAC ( given )

ACD=BCA (common )

ADCBAC, ( By AA rule)

CACB=CDCA ( corresponding sides of similar triangles )

CA2=CB×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 ΔABCΔPQR

Answer:

1635924470215

ABPQ=ACPR=ADPM (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

ABPQ=ACPR=ADPM (Given )

ABPQ=BEQL=2.AD2.PM

ABPQ=BEQL=AEPL

ΔABEΔPQL (SSS similarity)

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

Similarity, AEC=PLR

CAE=RPL ........................2

Adding equation 1 and 2,

BAE+CAE=QPL+RPL

CAB=RPQ ............................3

In ABCandPQR,

ABPQ=ACPR ( Given )

CAB=RPQ ( From above equation 3)

ABCPQR ( 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 ABEandCDF,

CDF=ABE ( Each 90 )

DCF=BAE (Angle of sun at same place )

ABECDF, (AA similarity)

ABCD=BEQL

AB6=284

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 ΔABCΔPQR , prove that ABPQ=ADPM

Answer:

1635924504095

ΔABCΔPQR ( Given )

ABPQ=ACPR=BCQR ............... ....1( corresponding sides of similar triangles )

A=P,B=Q,C=R ....................................2

AD and PM are medians of triangle.So,

BD=BC2andQM=QR2 ..........................................3

From equation 1 and 3, we have

ABPQ=BDQM ...................................................................4

In ABDandPQM,

B=Q (From equation 2)

ABPQ=BDQM (From equation 4)

ABDPQM, (SAS similarity)

ABPQ=BDQM=ADPM


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: 10th class maths triangles solutions 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: Triangles questions 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 Questions (FAQs)

1. What are the important topics covered in NCERT Class 10 Maths Chapter 6?

NCERT Class 10 Maths Chapter 6, "Triangles," focuses on the similarity of triangles, including criteria for similarity (AA, SSS, SAS), and theorems like the Basic Proportionality Theorem and Pythagoras Theorem.

2. How to prove the Pythagoras theorem using similarity of triangles?

To prove the Pythagorean theorem (a2+b2=c2) using similar triangles, we can draw an altitude from the right angle to the hypotenuse, creating two smaller triangles that are similar to each other and the original triangle, then use the proportional sides of similar triangles to derive the theorem.

3. What is the Basic Proportionality Theorem (Thales Theorem) in Class 10?

Thales's Theorem or Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

4. What are the real-life applications of similarity of triangles?

Similar triangles have various real-life applications, including measuring inaccessible heights or distances, understanding how zoom works on cameras, and in fields like cartography, engineering, and construction.

5. What is the difference between congruence and similarity of triangles?

Congruent triangles are identical in both shape and size, while similar triangles have the same shape but can differ in size, with corresponding sides being proportional. These concepts are discussed in class 10 maths chapter 6 solutions, practice these to command the concepts.

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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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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