# Class 10 NCERT Solutions Maths Chapter 6 - Triangles

NCERT Solutions for CBSE Class 10 Maths make learning mathematics simple and scoring. Students can access NCERT Solutions for CBSE Class 10 Maths at any point in time. The following list of contents make a complete package of Topper Learning’s study material assuring perfect preparation:

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- CBSE Class 10 Maths Most Important Questions
- CBSE Class 10 Maths Past Year Papers
- CBSE Class 10 Maths Textbook Solutions
- CBSE Class 10 Maths Revision Notes

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NCERT Solutions for Class 10 Maths Chapter 6 - Triangles

In your earlier classes, you have learnt about Triangle and its properties. In this chapter, we shall be studying about two figures that may have the same shape but not the same size. This chapter will introduce you to the concept of the similarity of triangles. Here are the following key points discussed in this chapter:

- What are similar figures?
- Similarity of Triangles
- Criteria for Similarity of Triangles
- Areas of Similar Triangles
- Pythagoras Theorem

Students are often puzzled by complex questions from this chapter. Reduce all these complexities with in-depth and elaborate NCERT solutions for CBSE Class 10 Maths. TopperLearning's conceptual insight and stepwise solutions make Mathematics Chapter Triangles

very easy and scoring for students. Given below are solutions for each question from all the exercises in Class 10 Maths Chapter Triangles.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.1**

Understand what are similar figures with Class 10 Maths Chapter Triangles. This exercise from Maths Chapter Triangles tells us that all congruent figures are similar but the similar figures need not be congruent. There are some concept-based questions asked in this exercise answers to which are available on Triangles NCERT solutions. Such questions are popular in the category of multiple-choice questions. Explore and practise more questions with CBSE Class 10 Maths study material. TopperLearning’s CBSE Class 10 Maths Textbook solutions help you in accessing your solutions whenever you need them.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.2**

What can you say about the similarity of two triangles? What are the conditions for the similarity of two triangles? The answers to these questions are provided with apt methodology as accepted by the CBSE board by our subject experts in Triangles NCERT solutions. Refer to NCERT solutions for CBSE Class 10 Maths for answers to every question from the textbook. Students are advised to go through CBSE Class 10 Maths video lessons

CBSE Class 10 Maths sample papers, CBSE Class 10 Maths most important questions, CBSE Class 10 Maths past year papers and CBSE Class 10 Maths revision note for all-around preparation. Students can tally their solutions and use our doubt solver “Undoubt” feature in case of doubts.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.3**

Increase your knowledge in Maths Chapter Triangles by understanding the criteria for the similarity of triangles. Solve all the questions from Class 10 Maths Chapter Triangles Exercise 6.3 and refer to Triangles NCERT solutions to check your solutions. Learn all the basic theorems for similarity with TopperLearning’s CBSE Class 10 Maths video lessons. This exercise hand some important questions which can be checked in CBSE Class 10 Maths most important questions. Boost your confidence by taking tests and gain full clarity.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.4**

The concept of similarity is useful in calculating the area of triangles. Understand how extensively explained video solutions available at CBSE Class 10 Maths video lessons. Apply this concept in the questions from Mathematics Chapter Triangles Exercise Ex. 6.4 and check your solutions with CBSE Class 10 Maths Textbook solutions. Topperlearning’s study material explains the method of writing answers with proper conceptual insight. NCERT Solutions for CBSE Class 10 Maths helps not only in gaining clarity but also increases your scores, Assess your knowledge with CBSE Class 10 Maths past year papers and CBSE Class 10 Maths sample papers at your convenience.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.5**

You are already familiar with the Pythagoras Theorem from your earlier classes. You

had verified this theorem through some activities and made use of it in solving certain

problems. You have also seen proof of this theorem in Class IX. Now, we shall prove

this theorem using the concept of similarity of triangles. In proving this, we shall make use of a result related to the similarity of two triangles formed by the perpendicular to the hypotenuse from the opposite vertex of the right triangle.

TopperLearning’s Triangles NCERT solutions explain the theorem as well the examples associated with it in the simplest manner possible. Refer to CBSE Class 10 Maths study material anytime and get everything that you need. The questions from this exercise are often asked in the 3-4 mark category. Explore CBSE Class 10 Maths sample papers, CBSE Class 10 Maths most important questions, CBSE Class 10 Maths past year papers and attempt all the most important questions.

**CBSE Class 10 Mathematics Chapter 6 - Triangles Exercise Ex. 6.6**

TopperLearning’s study material is an asset for every 10th grader. We not only help you with the main exercise but option optional exercise like this one. All the solutions from this exercise are available at NCERT solutions for CBSE Class 10 Maths. Extend your limits and attempt questions from this exercise and check your solutions with Triangles NCERT solutions. Refer to video solutions and most important questions at any hour and let there be no hurdles in your preparation. Although this exercise is optional, it is always wise to have a look at the questions and tally your solutions. Clear all your doubts with our special feature “Undoubt” where all your doubts will be catered by our team of subject experts.

## Triangles Exercise Ex. 6.1

### Solution 1

(ii) All squares are SIMILAR.

(iii) All EQUILATERAL triangles are similar.

(iv) Two polygons of same number of sides are similar, if their corresponding angles are EQUAL and their corresponding sides are PROPORTIONAL.

### Solution 2

Two squares with sides 1 cm and 2 cm

(ii) Trapezium and Square

Triangle and Parallelogram

### Solution 3

## Triangles Exercise Ex. 6.2

### Solution 1

Let EC = x

Since DE || BC.

Therefore, by basic proportionality theorem,

(ii)

Let AD = x

Since DE || BC,

Therefore by basic proportionality theorem,

### Solution 2

Given that PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4

Now,

(ii)

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

(iii)

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

### Solution 3

In the given figure

Since LM || CB,

Therefore by basic proportionality theorem,

### Solution 4

In ABC,

Since DE || AC

### Solution 5

In POQ

Since DE || OQ

### Solution 6

### Solution 7

Consider the given figure

PQ is a line segment drawn through midpoint P of line AB such that PQ||BC

i.e. AP = PB

Now, by basic proportionality theorem

i.e. AQ = QC

Or, Q is midpoint of AC.

### Solution 8

Consider the given figure

PQ is a line segment joining midpoints P and Q of line AB and AC respectively.

i.e. AP = PB and AQ = QC

Now, we may observe that

And hence basic proportionality theorem is verified

So, PQ||BC

### Solution 9

### Solution 10

## Triangles Exercise Ex. 6.3

### Solution 1

(i) A = P = 60°

B = Q = 80°

C = R = 40°

Therefore ABC ~ PQR [by AAA rule]

(iii) Triangles are not similar as the corresponding sides are not proportional.

(iv) Triangles are not similar as the corresponding sides are not proportional.

(v) Triangles are not similar as the corresponding sides are not proportional.

(vi) In DEF

D + E + F = 180°

(Sum of measures of angles of a triangle is 180)

70° + 80° + F = 180°

F = 30°

Similarly in PQR

P + Q + R = 180°

(Sum of measures of angles of a triangle is 180)

P + 80° +30° = 180°

P = 70°

Now In DEF and PQR

D = P = 70°

E = Q = 80°

F = R = 30°

Therefore DEF ~ PQR [by AAA rule]

### Solution 2

Since DOB is a straight line

Therefore DOC + COB = 180°

Therefore DOC = 180° - 125°

= 55°

In DOC,

DCO + CDO + DOC = 180°

DCO + 70° + 55° = 180°

DCO = 55°

Since ODC ~ OBA,

Therefore OCD = OAB [corresponding angles equal in similar triangles]

Therefore OAB = 55°

### Solution 3

In DOC and BOA

AB || CD

Therefore CDO = ABO [Alternate interior angles]

DCO = BAO [Alternate interior angles]

DOC = BOA [Vertically opposite angles]

Therefore DOC ~ BOA [AAA rule)

### Solution 4

PQR = PRQ

Therefore PQ = PR (i)

Given,

### Solution 5

In RPQ and RST

RTS = QPS [given]

R = R [common angle]

RST = RQP [ Remaining angles]

Therefore RPQ ~ RTS [by AAA rule]

### Solution 6

Since ABE ACD

Therefore AB = AC (1)

AD = AE (2)

Now, in ADE and ABC,

Dividing equation (2) by (1)

### Solution 7

(i)

In AEP and CDP

Since CDP = AEP = 90°

CPD = APE (vertically opposite angles)

PCD = PAE (remaining angle)

Therefore by AAA rule,

AEP ~ CDP

(ii)

In ABD and CBE

ADB = CEB = 90°

ABD = CBE (common angle)

DAB = ECB (remaining angle)

Therefore by AAA rule

ABD ~ CBE

(iii)

In AEP and ADB

AEP = ADB = 90°

PAE = DAB (common angle)

APE = ABD (remaining angle)

Therefore by AAA rule

AEP ~ ADB

(iv)

In PDC and BEC

PDC = BEC = 90°

PCD = BCE (common angle)

CPD = CBE

Therefore by AAA rule

PDC ~ BEC

### Solution 8

ABE and CFB

A = C (opposite angles of a parallelogram)

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

ABE = CFB (remaining angle)

Therefore ABE ~ CFB (by AAA rule)

### Solution 9

In ABC and AMP

ABC = AMP = 90

A = A (common angle)

ACB = APM (remaining angle)

Therefore ABC ~ AMP (by AAA rule)

### Solution 10

Since ABC ~ FEG

Therefore A = F

B = E

As, ACB = FGE

Therefore ACD = FGH (angle bisector)

And DCB = HGE (angle bisector)

Therefore ACD ~ FGH (by AAA rule)

And DCB ~ HGE (by AAA rule)

and ACD = FGH

Therefore DCA ~ HGF (by SAS rule)

### Solution 11

In ABD and ECF,

Given that AB = AC (isosceles triangles)

So, ABD = ECF

ADB = EFC = 90

BAD = CEF

Therefore ABD ~ ECF (by AAA rule)

### Solution 12

Median divides opposite side.

Therefore ABD ~ PQM (by SSS rule)

Therefore ABD = PQM (corresponding angles of similar triangles)

Therefore ABC ~ PQR (by SAS rule)

### Solution 13

In ADC and BAC

Given that ADC = BAC

ACD = BCA (common angle)

CAD = CBA (remaining angle)

Hence, ADC ~ BAC [by AAA rule]

So, corresponding sides of similar triangles will be proportional to each other

### Solution 14

### Solution 15

Let AB be a tower

CD be a pole

Shadow of AB is BE

Shadow of CD is DF

The light rays from sun will fall on tower and pole at same angle and at the same time.

So, DCF = BAE

And DFC = BEA

CDF = ABE (tower and pole are vertical to ground)

Therefore ABE ~ CDF

So, height of tower will be 42 meters.

### Solution 16

Since ABC ~ PQR

So, their respective sides will be in proportion

Also, A = P, B = Q, C = R (2)

Since, AD and PM are medians so they will divide their opposite sides in equal halves.

From equation (1) and (3)

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal

Hence, ABD ~ PQM (by SAS rule)

## Triangles Exercise Ex. 6.4

### Solution 6

Let us assume two similar triangles as ABC ~ PQR. Let AD and PS be the medians of these triangles.

A = P, B = Q, C = R

Since, AD and PS are medians

### Solution 1

### Solution 2

Since AB || CD

OAB = OCD (Alternate interior angles)

OBA = ODC (Alternate interior angles)

AOB = COD (Vertically opposite angles)

Therefore AOB ~ COD (By AAA rule)

### Solution 3

Since ABC and DBC are one same base,

Therefore ratio between their areas will be as ratio of their heights.

Let us draw two perpendiculars AP and DM on line BC.

In APO and DMO,

APO = DMO = 90

AOP = DOM (vertically opposite angles)

OAP = ODM (remaining angle)

Therefore APO ~ DMO (By AAA rule)

### Solution 4

### Solution 5

Since D and E are mid points of ABC

### Solution 6

Let us assume two similar triangles as ABC ~ PQR. Let AD and PS be the medians of these triangles.

A = P, B = Q, C = R

Since, AD and PS are medians

### Solution 7

Let ABCD be a square of side a.

Therefore its diagonal

Two desired equilateral triangles are formed as ABE and DBF

Side of an equilateral triangle ABE described on one of its side = a

Side of an equilateral triangle DBF described on one of its diagonal

We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.

### Solution 8

We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.

Let side of ABC = x

Hence, (c)

### Solution 9

Given that sides are in the ratio 4:9.

Hence, (d).

## Triangles Exercise Ex. 6.5

### Solution 1

i.Given that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.

Clearly, 49 + 576 = 625 or 7^{2} + 24^{2} = 25^{2} .

Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 25 cm.

ii.Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle

iii.Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides we may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle.

iv.Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides we may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 12^{2} + 5^{2} = 13^{2}.

Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.

The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 13 cm.

### Solution 2

### Solution 3

iii. In DCA & DAB

DCA = DAB = 90Âº

CDA = ADB (common angle)

DAC = DBA (remaining angle)

### Solution 4

Given that ABC is an isosceles triangle.

Therefore AC = CB

Applying Pythagoras theorem in ABC (i.e. right angled at point C)

### Solution 5

### Solution 6

Let AD be the altitude in given equilateral triangle ABC.

We know that altitude bisects the opposite side.

So, BD = DC = a

Since in an equilateral triangle, all the altitudes are equal in length.

So, length of each altitude will be

### Solution 7

In AOB, BOC, COD, AOD

Applying Pythagoras theorem

### Solution 8

### Solution 9

### Solution 10

Let OB be the pole and AB be the wire.

Therefore by Pythagoras theorem,

### Solution 11

Let these distances are represented by OA and OB respectively.

Now applying Pythagoras theorem

### Solution 12

Let CD and AB be the poles of height 11 and 6 m.

Therefore CP = 11 - 6 = 5 m

From the figure we may observe that AP = 12m

In APC, by applying Pythagoras theorem

Therefore distance between their tops = 13 m.

### Solution 13

In ACE,

### Solution 14

### Solution 15

### Solution 16

Let side of equilateral triangle be a. And AE be the altitude of ABC

Now, in ABE by applying Pythagoras theorem

AB

^{2}= AE

^{2}+ BE

^{2}

### Solution 17

Given that AB = cm, AC = 12 cm and BC = 6 cm

We may observe that

AB

^{2}= 108

AC

^{2}= 144

And BC

^{2}= 36

AB

^{2}+BC

^{2}= AC

^{2}

Thus the given triangle ABC is satisfying Pythagoras theorem

Therefore triangle is a right angled triangle right angled at B

Therefore B = 90°.

Hence, (c).

## Triangles Exercise Ex. 6.6

### Solution 1

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that PS is angle bisector of QPR.

QPS = SPR (1)

SPR = PRT (As PS || TR) (2)

QPS = QTR (As PS || TR) (3)

Using these equations we may find

PRT = QTR from (2) and (3)

So, PT = PR (Since PTR is isosceles triangle)

Now in QPS and QTR

QSP = QRT (As PS || TR)

QPS = QTR (As PS || TR)

Q is common

QPS ~ QTR

### Solution 2

(i). Let us join DB.

DN || CB

DM || AB

So, DN = MB

DM = NB

The condition to be proved is the case when DNBM is a square or D is the midpoint of side AC.

Then CDB = ADB = 90°

2 + 3 = 90° (1)

In CDM

1 + 2 + DMC = 180°

1 + 2 = 90° (2)

In DMB

3 + DMB + 4 = 180°

3 + 4 = 90° (3)

From equation (1) and (2)

1 = 3

From equation (1) and (3)

2 = 4

BDM ~ DCM

(ii). Similarly in DBN

4 + 3 = 90° (4)

In DAN

5 + 6 = 90° (5)

In DAB

4 + 5 = 90° (6)

From equation (4) and (6)

3 = 5

From equation (5) and (6)

4 = 6

DNA ~ BND

### Solution 3

AB

^{2}= AD

^{2}+ DB

^{2}(1)

In ACD applying Pythagoras theorem

AC

^{2}= AD

^{2}+ DC

^{2}

AC

^{2}= AD

^{2}+ (DB + BC)

^{2}

AC

^{2}= AD

^{2}+ DB

^{2}+ BC

^{2}+ 2DB x BC

Now using equation (1)

AC

^{2}= AB

^{2}+ BC

^{2}+ 2BC . BD

### Solution 4

AD

^{2}+ DB

^{2}= AB

^{2}

AD

^{2}= AB

^{2}- DB

^{2}(1)

In ADC applying Pythagoras theorem

AD

^{2}+ DC

^{2}= AC

^{2}(2)

Now using equation (1)

AB

^{2}- BD

^{2}+ DC

^{2}= AC

^{2}

AB

^{2}- BD

^{2}+ (BC - BD)

^{2}= AC

^{2}

AC

^{2}= AB

^{2}- BD

^{2}+ BC

^{2}+ BD

^{2}- 2BC. BD

= AB

^{2}+ BC

^{2}- 2BC. BD

### Solution 5

AM

^{2}+ MD

^{2}= AD

^{2 }(1)

In AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

AM

^{2}+ (MD + DC)

^{2}= AC

^{2}

(AM

^{2}+ MD

^{2}) + DC

^{2}+ 2MD.DC = AC

^{2}

Using equation (1) we may get

AD

^{2}+ DC

^{2}+ 2MD.DC = AC

^{2}

(ii). In ABM applying Pythagoras theorem

AB

^{2}= AM

^{2}+ MB

^{2}

= (AD

^{2}- DM

^{2}) + MB

^{2}

= (AD

^{2}- DM

^{2}) + (BD - MD)

^{2}

= AD

^{2}- DM

^{2}+ BD

^{2}+ MD

^{2 - }2BD.MD

= AD

^{2}+ BD

^{2 - }2BD.MD

(iii). InAMB

AM

^{2}+ MB

^{2}= AB

^{2}(1)

In AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

Adding equation (1) and (2)

2AM

^{2}+ MB

^{2}+ MC

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+ (BD - DM)

^{2}+ (MD + DC)

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+BD

^{2}+ DM

^{2}- 2BD.DM + MD

^{2}+ DC

^{2}+ 2MD.DC = AB

^{2}+ AC

^{2}

2AM

^{2}+ 2MD

^{2}+ BD

^{2}+ DC

^{2}+ 2MD (-BD + DC) = AB

^{2}+ AC

^{2}

### Solution 6

Let ABCD be a parallelogram

Let us draw perpendicular DE on extended side AB and AF on side DC.

In DEA

DE

^{2}+ EA

^{2}= DA

^{2}(i)

In DEB

DE

^{2}+ EB

^{2}= DB

^{2}

DE

^{2}+ (EA + AB)

^{2}= DB

^{2}

(DE

^{2}+ EA

^{2}) + AB

^{2}+ 2EA. AB = DB

^{2}

DA

^{2}+ AB

^{2}+ 2EA.AB = DB

^{2}(ii)

In ADF

AD

^{2}= AF

^{2}+ FD

^{2}

In AFC

AC

^{2}= AF

^{2}+ FC

^{2}

= AF

^{2}+ (DC - FD)

^{2}

= AF

^{2}+ DC

^{2}+ FD

^{2}- 2DC - FD

= (AF

^{2}+ FD

^{2}) + DC

^{2}- 2DC . FD

AC

^{2}= AD

^{2}+ DC

^{2}- 2DC FD (iii)

Since ABCD is a parallelogram

AB = CD (iii)

And BC = AD (iv)

In DEA and ADF

DEA = AFD

EAD = FDA (EA || DF)

EDA = FAD (AF || ED)

AD is common in both triangles.

Since respective angles are same and respective sides are same

DEA AFD

So EA = DF (v)

Adding equation (ii) and (iii)

DA

^{2}+ AB

^{2}+ 2EA.AB + AD

^{2}+ DC

^{2}- 2DC.FD = DB

^{2}+ AC

^{2}

DA

^{2}+ AB

^{2}+ AD

^{2 }+ DC

^{2}+ 2EA.AB - 2DC.FD = DB

^{2}+ AC

^{2}

BC

^{2}+ AB

^{2}+ AD

^{2}+ DC

^{2}+ 2EA.AB-2AB.EA = DB

^{2}+ AC

^{2}

AB

^{2}+ BC

^{2}+ CD

^{2}+ DA

^{2}= AC

^{2}+ BD

^{2}

### Solution 7

Let us join CB

(i) In APC and DPB

APC = DPB {Vertically opposite angles}

CAP = BDP {Angles in same segment for chord CB}

APC ~ DPB {BY AA similarly criterion}

(ii) We know that corresponding sides of similar triangles are proportional

### Solution 8

(i) In PAC and PDB

P = P (common)

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

PCA = PBD

PAC ~ PDB

(ii) We know that corresponding sides of similar triangles are proportional.

### Solution 9

AD = AD (common)

So, DBA ~ DCA (By SSS)

Now, corresponding angles of similar triangles will be equal.

BAD = CAD

AD is angle bisector of BAC

### Solution 10

Let AB be the height of tip of fishing rod from water surface. Let BC be the horizontal distance of fly from the tip of fishing rod.

Then, AC is the length of string.

AC can be found by applying Pythagoras theorem in ABC

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (1.8)^{2} + (2.4)^{2}

AC^{2} = 3.24 + 5.76

AC^{2} = 9.00

AC = = 3

Now, she pulls string at rate of 5 cm per second.

So, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m

Let after 12 second Fly be at point D.

Length of string out after 12 second is AD

AD = AC - string pulled by Nazima in 12 seconds

= 3.00 - 0.6

= 2.4

In ADB

AB

^{2}+ BD

^{2}= AD

^{2}

(1.8)

^{2}+ BD

^{2}= (2.4)

^{2}

BD

^{2}= 5.76 - 3.24 = 2.52

BD = 1.587

Horizontal distance of fly = BD + 1.2

= 1.587 + 1.2

= 2.787

= 2.79 m

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