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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 Feb 27, 2025 05:11 PM IST | #CBSE Class 10th
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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.

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

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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.

JEE Main Important Mathematics Formulas

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

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

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


Class 10 Maths Chapter 6 Triangles Excercise:6.4

Q1 Let ΔABCΔDEF and their areas be, respectively, 64 cm2 and 121 cm2 . If EF = 15.4 cm, find BC.

Answer:

ΔABCΔDEF ( Given )

ar(ABC) = 64 cm2 and ar(DEF)=121 cm2 .

EF = 15.4 cm (Given )

ar(ABC)ar(DEF)=AB2DE2=BC2EF2=AC2DF2

64121=BC2(15.4)2

811=BC15.4

8×15.411=BC

BC=11.2cm

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

Answer:

1635931770237

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

AB = 2 CD ( Given )

In AOBandCOD,

COD=AOB (vertically opposite angles )

OCD=OAB (Alternate angles)

ODC=OBA (Alternate angles)

AOBCOD (AAA similarity)

ar(AOB)ar(COD)=AB2CD2

ar(AOB)ar(COD)=(2CD)2CD2

ar(AOB)ar(COD)=4.CD2CD2

ar(AOB)ar(COD)=41

ar(AOB)=ar(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 ar(ABC)ar(DBC)=AODO

1635931784662

Answer:

1635931796737

Let DM and AP be perpendicular on BC.

areaoftriangle=12×base×perpendicular

ar(ABC)ar(BCD)=12×BC×AP12×BC×MD

In APOandDMO,

APO=DMO (Each 90 )

AOP=MOD (Vertically opposite angles)

APODMO, (AA similarity)

APDM=AODO

Since

ar(ABC)ar(BCD)=12×BC×AP12×BC×MD

ar(ABC)ar(BCD)=APMD=AODO

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

Answer:

Let ABCDEF, , therefore,

ar(ABC)=ar(DEF) (Given )

ar(ABC)ar(DEF)=AB2DE2=BC2EF2=AC2DF2................................1

ar(ABC)ar(DEF)=1

AB2DE2=BC2EF2=AC2DF2=1

AB=DE

BC=EF

AC=DF

ABCDEF (SSS )

Q5 D, E, and F are respectively the mid-points of sides AB, BC and CA of ΔABC . Find the ratio of the areas of ΔDEFandΔABC

Answer:

1635932236410

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

DE=12AC and DE||AC

In ΔBEDandΔABC ,

BED=BCA (corresponding angles )

BDE=BAC (corresponding angles )

ΔBEDΔABC (By AA)

ar(BED)ar(ABC)=DE2AC2

ar(BED)ar(ABC)=(12AC)2AC2

ar(BED)ar(ABC)=14

ar(BED)=14×ar(ABC)

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

ar(BED)=14×x

Similarly,

ar(CEF)=14×x and ar(ADF)=14×x

ar(ABC)=ar(ADF)+ar(BED)+ar(CEF)+ar(DEF)

x=x4+x4+x4+ar(DEF)

x=3x4+ar(DEF)

x3x4=ar(DEF)

4x3x4=ar(DEF)

x4=ar(DEF)


ar(DEF)ar(ABC)=x4x

ar(DEF)ar(ABC)=14

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.

ABCPQR

ABPQ=BCQR=ACPR............................1

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

BD=CD=12BCandQS=SR=12QR

Purring these value in 1,

ABPQ=BDQS=ACPR..........................3

In ABDandPQS,

B=Q (proved above)

ABPQ=BDQS (proved above)

ABDPQS (SAS )

Therefore,

ABPQ=BDQS=ADPS................4

ar(ABC)ar(PQR)=AB2PQ2=BC2QR2=AC2PR2

From 1 and 4, we get

ABPQ=BCQR=ACPR=ADPS

ar(ABC)ar(PQR)=AD2PS2

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 = 2a

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

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = 2a units

Both the triangles are equilateral triangles with each angle of 60 .

ABEDBF ( By AAA)

Using area theorem,

ar(ABC)ar(DBF)=(a2a)2=12

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 .

ABC BDE (By AAA)

Let AB=BC=CA = x

then EB=BD=ED= x2

ar(ABC)ar(BDE)=(xx2)2=41

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

ar(ABC)ar(DEF)=AB2DE2=4292=1681

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,

h2=72+242

h2=49+576

h2=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,

h2=32+62

h2=9+36

h2=45

h=458

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,

h2=502+802

h2=2500+6400

h2=8900

h=8900100

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,

h2=52+122

h2=25+144

h2=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 PMQR . Show that PM2=QM.MR.

Answer:

1635933005578

Let MPR be x

In MPR ,

MRP=18090x

MRP=90x

Similarly,

In MPQ ,

MPQ=90MPR

MPQ=90x

MQP=18090(90x)=x

In QMPandPMR,

MPQ=MRP

PMQ=RMP

MQP=MPR

QMPPMR, (By AAA)

QMPM=MPMR

PM2=MQ×MR

Hence proved.

Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that AB2=BC.BD.

1635933025836

Answer:

In ADBandABC,

DAB=ACB(Each90)

ABD=CBA (common )

ADBABC (By AA)

ABBC=BDAB

AB2=BC.BD , hence prooved .

Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that AC2=BC.DC.

1635933077069

Answer:

Let CAB be x

In ABC ,

CBA=18090x

CBA=90x

Similarly,

In CAD ,

CAD=90CAB

CAD=90x

CDA=18090(90x)=x

In ABCandACD,

CBA=CAD

CAB=CDA

ACB=DCA ( Each right angle)

ABC,ACD (By AAA)

ACDC=BCAC

AC2=BC×DC

Hence proved

Q3 (3) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that AD2=BD.CD.

1635933091865

Answer:

In ACDandABD,

DCA=DAB(Each90)

CDA=ADB (common )

ACDABD (By AA)

CDAD=ADBD

AD2=BD×CD

Hence proved.

Q4 ABC is an isosceles triangle right angled at C. Prove that AB2=2AC2

Answer:

1635933142785

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

Let AC=BC

In ABC,

By Pythagoras theorem

AB2=AC2+BC2

AB2=AC2+AC2 (AC=BC)

AB2=2.AC2

Hence proved.

Q5 ABC is an isosceles triangle with AC = BC. If AB2=2AC2 , prove that ABC is a right triangle.

Answer:

1635933162430

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

In ABC,

AB2=2.AC2 (Given )

AB2=AC2+AC2 (AC=BC)

AB2=AC2+BC2

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 ADB,

By Pythagoras theorem,

AB2=AD2+BD2

(2a)2=AD2+a2

4a2=AD2+a2

4a2a2=AD2

3a2=AD2

AD=3a

The length of each altitude is 3a .

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 AOB, by Pythagoras theorem,

AB2=AO2+BO2..................1

In BOC, by Pythagoras theorem,

BC2=BO2+CO2..................2

In COD, by Pythagoras theorem,

CD2=CO2+DO2..................3

In AOD, by Pythagoras theorem,

AD2=AO2+DO2..................4

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

AB2+BC2+CD2+AD2=AO2+BO2+BO2+CO2+CO2+DO2+AO2+DO2

AB2+BC2+CD2+AD2=2(AO2+BO2+CO2+DO2)

AB2+BC2+CD2+AD2=2(2.AO2+2.BO2) (AO=CO and BO=DO)

AB2+BC2+CD2+AD2=4(AO2+BO2)

AB2+BC2+CD2+AD2=4((AC2)2+(BD2)2)

AB2+BC2+CD2+AD2=4((AC24)+(BD24))

AB2+BC2+CD2+AD2=AC2+BD2

Hence proved .

Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB. Show that OA2+OB2+OC2OD2OE2OF2=AF2+BD2+CE2,

1635933201520

Answer:


1635933215218

Join AO, BO, CO

In AOF, by Pythagoras theorem,

OA2=OF2+AF2..................1

In BOD, by Pythagoras theorem,

OB2=OD2+BD2..................2

In COE, by Pythagoras theorem,

OC2=OE2+EC2..................3

Adding equation 1,2,3,we get

OA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2 OA2+OB2+OC2OD2OE2OF2=AF2+BD2+EC2....................4

Hence proved

Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB. AF2+BD2+CE2=AE2+CD2+BF2.

1635933229014

Answer:

1635933270968

Join AO, BO, CO

In AOF, by Pythagoras theorem,

OA2=OF2+AF2..................1

In BOD, by Pythagoras theorem,

OB2=OD2+BD2..................2

In COE, by Pythagoras theorem,

OC2=OE2+EC2..................3

Adding equation 1,2,3,we get

OA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2 OA2+OB2+OC2OD2OE2OF2=AF2+BD2+EC2....................4

(OA2OE2)+(OC2OD2)+(OB2OF2)=AF2+BD2+EC2 AE2+CD2+BF2=AF2+BD2+EC2

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 AOB, by Pythagoras theorem

AB2=AO2+BO2

102=82+BO2

100=64+BO2

10064=BO2

36=BO2

BO=6m

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

Q10 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer:

1635933296399

OB is a pole.

In AOB, by Pythagoras theorem

AB2=AO2+BO2

242=182+AO2

576=324+AO2

576324=AO2

252=AO2

AO=67m

Hence, the distance of the stack from the base of the pole is 67 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 112 hours.

=1000×32=1500km

Distance travelled by second aeroplane due west in 112 hours.

=1200×32=1800km

OA and OB are the distance travelled.

By Pythagoras theorem,

AB2=OA2+OB2

AB2=15002+18002

AB2=2250000+3240000

AB2=5490000

AB2=30061km

Thus, the distance between the two planes is 30061km .

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 APC,

By Pythagoras theorem,

AP2+PC2=AC2

122+52=AC2

144+25=AC2

169=AC2

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 AE2+BD2=AB2+DE2.

Answer:

1635933368584

In ACE, by Pythagoras theorem,

AE2=AC2+CE2..................1

In BCD, by Pythagoras theorem,

DB2=BC2+CD2..................2

From 1 and 2, we get

AC2+CE2+BC2+CD2=AE2+DB2..................3

In CDE, by Pythagoras theorem,

DE2=CD2+CE2..................4

In ABC, by Pythagoras theorem,

AB2=AC2+CB2..................5

From 3,4,5 we get

DE2+AB2=AE2+DB2

Q14 The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2AB2=2AC2+BC2.

1635933387695

Answer:

In ACD, by Pythagoras theorem,

AC2=AD2+DC2

AC2DC2=AD2..................1

In ABD, by Pythagoras theorem,

AB2=AD2+BD2

AB2BD2=AD2.................2

From 1 and 2, we get

AC2CD2=AB2DB2..................3

Given : 3DC=DB, so

CD=BC4andBD=3BC4........................4

From 3 and 4, we get

AC2(BC4)2=AB2(3BC4)2

AC2(BC216)=AB2(9BC216)

16AC2BC2=16AB29BC2

16AC2=16AB28BC2

2AC2=2AB2BC2

2AB2=2AC2+BC2.

Hence proved.

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2=7AB2

Answer:

1635933424958

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

To prove : 9AD2=7AB2

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, BE=CE=a2

In AEB, by Pythagoras theorem

AB2=AE2+BE2

a2=AE2+(a2)2

a2(a24)=AE2

(3a24)=AE2

AE=(3a2)

Given : BD = 1/3 BC.

BD=a3

DE=BE=BD=a2a3=a6

In ADE, by Pythagoras theorem,

AD2=AE2+DE2

AD2=(3a2)2+(a6)2

AD2=(3a24)+(a236)

AD2=(7a29)

AD2=(7AB29)

9AD2=7AB2

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=a2

In AEB, by Pythagoras theorem

AB2=AE2+BE2

a2=AE2+(a2)2

a2(a24)=AE2

(3a24)=AE2

3a2=4AE2

4.(altitude)2=3.(side)2

Q17 Tick the correct answer and justify : In ΔABC AB = 63 cm, AC = 12 cm and BC = 6 cm.
The angle B is :

(A) 120°

(B) 60°

(C) 90°

(D) 45°

Answer:

In ΔABC AB = 63 cm, AC = 12 cm and BC = 6 cm.

AB2+BC2=108+36

=144

=122

=AC2

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 QPRofΔPQR . Prove that QSSR=PQPR

1635933619787

Answer:

1635933634308

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

Given: PS is the bisector of QPRofΔPQR .

QPS=SPR.....................................1

By construction,

SPR=PRT.....................................2 (as PS||TR)

QPS=QTR.....................................3 (as PS||TR)

From the above equations, we get

PRT=QTR

PT=PR

By construction, PS||TR

In QTR, by Thales theorem,

QSSR=QPPT

QSSR=PQPR

Hence proved.

Q2 In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD AC, DM BC and DN AB. Prove that : DM2=DN.MC

1635933650488

Answer:

1635933658497

Join BD

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

CDB=90

2+3=90.............................1

In CDM, 1+2+DMC=180

1+2=90.......................2

In DMB, 3+4+DMB=180

3+4=90.......................3

From equation 1 and 2, we get 1=3

From equation 1 and 3, we get 2=4

In DCMandBDM,

1=3

2=4

DCMBDM, (By AA)

BMDM=DMMC

DNDM=DMMC (BM=DN)

DM2=DN.MC

Hence proved

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB. Prove that: DN2=DM.AN

1635933672815

Answer:

1635933684708

In DBN,

5+7=90.......................1

In DAN,

6+8=90.......................2

BD AC, ADB=90

5+6=90.......................3

From equation 1 and 3, we get 6=7

From equation 2 and 3, we get 5=8

In DNAandBND,

6=7

5=8

DNABND (By AA)

ANDN=DNNB

ANDN=DNDM (NB=DM)

DN2=AN.DM

Hence proved.

Q3 In Fig. 6.58, ABC is a triangle in which ABC > 90° and AD CB produced. Prove that AC2=AB2+BC2+2BC.BD.

1635933712397

Answer:

In ADB, by Pythagoras theorem

AB2=AD2+DB2.......................1

In ACD, by Pythagoras theorem

AC2=AD2+DC2.......................2

AC2=AD2+(BD+BC)2

AC2=AD2+(BD)2+(BC)2+2.BD.BC

AC2=AB2+BC2+2BC.BD. (From 1)

Q4 In Fig. 6.59, ABC is a triangle in which ABC < 90° and AD BC. Prove that AC2=AB2+BC22BC.BD.

1635933729274

Answer:

In ADB, by Pythagoras theorem

AB2=AD2+DB2

AD2=AB2DB2...........................1

In ACD, by Pythagoras theorem

AC2=AD2+DC2

AC2=AB2BD2+DC2 (From 1)

AC2=AB2BD2+(BCBD)2

AC2=AB2BD2+(BC)2+(BD)22.BD.BC

AC2=AB2+BC22BC.BD.

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that : AC2=AD2+BCDM+(BC2)2

1635933758550

Answer:

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

In AMD, by Pythagoras theorem

AD2=AM2+MD2.......................1

In AMC, by Pythagoras theorem

AC2=AM2+MC2

AC2=AM2+(MD+DC)2

AC2=AM2+(MD)2+(DC)2+2.MD.DC

AC2=AD2+DC2+2DC.MD. (From 1)

AC2=AD2+(BC2)2+2(BC2).MD. (BC=2 DC)

AC2=AD2+BCDM+(BC2)2

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that : AB2=AD2BC.DM+(BC2)2

1635933791582

Answer:

In ABM, by Pythagoras theorem

AB2=AM2+MB2

AB2=(AD2DM2)+MB2

AB2=(AD2DM2)+(BDMD)2

AB2=AD2DM2+(BD)2+(MD)22.BD.MD

AB2=AD2+(BD)22.BD.MD

AB2=AD2+(BC2)22(BC2).MD.=AC2 (BC=2 BD)

AD2+(BC2)2BC.MD.=AC2

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that: AC2+AB2=2AD2+12BC2

1635933807753

Answer:

In ABM, by Pythagoras theorem

AB2=AM2+MB2.......................1

In AMC, by Pythagoras theorem

AC2=AM2+MC2 ..................................2

Adding equation 1 and 2,

AB2+AC2=2AM2+MB2+MC2

AB2+AC2=2AM2+(BDDM)2+(MD+DC)2

AB2+AC2=2AM2+(BD)2+(DM)22.BD.DM+(MD)2+(DC)2+2.MD.DC

AB2+AC2=2AM2+2.(DM)2+BD2+(DC)2+2.MD.(DCBD) AB2+AC2=2(AM2+(DM)2)+(BC2)2+(BC2)2+2.MD.(BC2BC2) AB2+AC2=2(AM2+(DM)2)+(BC2)2+(BC2)2

AC2+AB2=2AD2+12BC2

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 DEA, by Pythagoras theorem

DA2=DE2+EA2.......................1

In DEB, by Pythagoras theorem

DB2=DE2+EB2

DB2=DE2+(EA+AB)2

DB2=DE2+(EA)2+(AB)2+2.EA.AB

DB2=DA2+(AB)2+2.EA.AB ....................................2

In ADF, by Pythagoras theorem

DA2=AF2+FD2

In AFC, by Pythagoras theorem

AC2=AF2+FC2=AF2+(DCFD)2

AC2=AF2+(DC)2+(FD)22.DC.FD

AC2=(AF2+FD2)+(DC)22.DC.FD

AC2=AD2+(DC)22.DC.FD.......................3

Since ABCD is a parallelogram.

SO, AB=CD and BC=AD

In DEAandADF,

DEA=AFD(each90)

DAE=ADF (AE||DF)

AD=AD (common)

DEAADF, (ASA rule)

EA=DF.......................6

Adding 2 and, we get

DA2+AB2+2.EA.AB+AD2+DC22.DC.FD=DB2+AC2 DA2+AB2+AD2+DC2+2.EA.AB2.DC.FD=DB2+AC2

BC2+AB2+AD2+2.EA.AB2.AB.EA=DB2+AC2 (From 4 and 6)

BC2+AB2+CD2=DB2+AC2 \

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : ΔAPCΔDPB

1635933839234

Answer:

1635933848323

Join BC

In APCandDPB,

APC=DPB ( vertically opposite angle)

CAP=BDP (Angles in the same segment)

APCDPB (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 APCandDPB,

APC=DPB ( vertically opposite angle)

CAP=BDP (Angles in the same segment)

APCDPB (By AA)

APDP=PCPB=CABD (Corresponding sides of similar triangles are proportional)

APDP=PCPB

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 ΔPACΔPDB

1635933941642

Answer:

In ΔPACandΔPDB,

P=P (Common)

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

So, ΔPACΔPDB ( 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 ΔPACandΔPDB,

P=P (Common)

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

So, ΔPACΔPDB ( By AA rule)

24440 APDP=PCPB=CABD (Corresponding sides of similar triangles are proportional)

APDP=PCPB

AP.PB=PC.DP

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that BDCD=ABAC Prove that AD is the bisector of BAC.

1635933987426

Answer:

1635933998131

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

BDCD=ABAC (Given )

BDCD=APAC

Using converse of Thales theorem,

AD||PC BAD=APC............1 (Corresponding angles)

DAC=ACP............2 (Alternate angles)

By construction,

AP=AC

APC=ACP............3

From equation 1,2,3, we get

BAD=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 ABC, using Pythagoras theorem

AC2=AB2+BC2

AC2=(1.8)2+(2.4)2

AC2=3.24+5.76

AC2=9.00

AC=3m

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.

= 12×5=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 ADB,

AB2+BD2=AD2

(1.8)2+BD2=(2.4)2

BD2=5.763.24=2.52m2

BD=1.587m

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 Questions (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 ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

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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