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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 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.

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.

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

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.

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.

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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

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2. A circle and a triangle.

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Q3 State whether the following quadrilaterals are similar or not:

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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).

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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 }

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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

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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 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

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

  1. Starting Point:

    • Determine your starting point in Aligarh or the nearby area.

  2. Use Google Maps:

    • Open Google Maps on your phone or computer.
    • Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
    • Follow the navigation instructions provided by Google Maps.

  3. By Local Transport:

    • Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
    • Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.

  4. Landmarks to Look For:

    • As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.

  5. Ask for Directions:

    • If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.

  6. Final Destination:

    • Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

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

Read Complete Answer
SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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!

Read Complete Answer
Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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.

Read Complete Answer
sadafreen356 03 July,2024
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Popular CBSE Class 10th Questions

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

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