NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles

NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles

Edited By Komal Miglani | Updated on Apr 29, 2025 05:42 PM IST | #CBSE Class 10th

Geometry consists of an important portion dedicated to triangles, which gains clarity through studying congruence rules for triangle properties. The concepts of triangular inequalities receive attention in this exercise. The lesson demonstrates the connections between sides and angles and shows how we can derive important findings through such comparisons. Basic triangle properties enable us to foretell various side-angle relationships without performing physical measurements.

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  1. NCERT Solutions Class 10 Maths Chapter 6: Exercise 6.3
  2. Download PDF
  3. Access Solution of Triangles Class 10 Chapter 6 Exercise: 6.3
  4. Topics covered in Chapter 10 Triangles: Exercise 6.3
  5. NCERT Solutions of Class 10 Subject Wise
  6. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles
NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles
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The content of Exercise 6.3 within the NCERT Solutions allows students to study questions concerning triangle inequalities and side–angle relationships. Such a study enables students to build their logical thinking combined with proof-writing competency by analysing relationships between "the side opposite to a larger angle is longer" and "the angle opposite to a longer side is greater." Through problem-solving activities in the NCERT Books, students gain confidence, which enables them to handle triangle inequality properties both in problems of daily life and competitive testing.

NCERT Solutions Class 10 Maths Chapter 6: Exercise 6.3


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Access Solution of Triangles Class 10 Chapter 6 Exercise: 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) $\angle A=\angle P=60 ^\circ$

$\angle B=\angle Q=80 ^\circ$

$\angle C=\angle R=40 ^\circ$

$\therefore \triangle ABC \sim \triangle PQR$ (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

$\angle D+\angle E+\angle F=180 ^\circ$

$\Rightarrow 70 ^\circ+80 ^\circ+\angle F=180 ^\circ$

$\Rightarrow 150 ^\circ+\angle F=180 ^\circ$

$\Rightarrow \angle F=180 ^\circ-150 ^\circ=30 ^\circ$

In $\triangle PQR$ , we know that

$\angle P+\angle Q+\angle R=180 ^\circ$

$\Rightarrow 30 ^\circ+80 ^\circ+\angle R=180 ^\circ$

$\Rightarrow 110 ^\circ+\angle R=180 ^\circ$

$\Rightarrow \angle R=180 ^\circ-110 ^\circ=70 ^\circ$

$\angle Q=\angle P=70 ^\circ$

$\angle E=\angle Q=80 ^\circ$

$\angle F=\angle R=30 ^\circ$

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

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

1635924052755

Answer:

Given : $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 ^\circ$ and $\angle CDO = 70 ^\circ$

$\angle DOC+\angle BOC=180 ^\circ$ (DOB is a straight line)

$\Rightarrow \angle DOC+125 ^\circ=180 ^\circ$

$\Rightarrow \angle DOC=180 ^\circ-125 ^\circ$

$\Rightarrow \angle DOC=55 ^\circ$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 ^\circ$

$\Rightarrow 55 ^\circ+ 70 ^\circ+\angle DCO=180 ^\circ$

$\Rightarrow \angle DCO+125 ^\circ=180 ^\circ$

$\Rightarrow \angle DCO=180 ^\circ-125 ^\circ$

$\Rightarrow \angle DCO=55 ^\circ$

Since , $\Delta ODC \sim \Delta OBA$ , so

$\Rightarrow\angle OAB= \angle DCO=55 ^\circ$ ( 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 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: $\Delta ABC \sim \Delta AMP$

1635924254422

Answer:

To prove : $\Delta ABC \sim \Delta AMP$

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

$\angle ABC=\angle AMP$ ( Each $90 ^\circ$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( 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 $90 ^\circ$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( By AA criterion )

$\frac{CA }{PA } = \frac{BC }{MP}$ ( corresponding parts of similar triangles )

Hence proved.

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

Answer:

1635924284479

To prove :

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

Given : $\Delta ABC \sim \Delta EGF$

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

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

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

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

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

$\angle A=\angle F$ ( proved above)

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

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

Hence proved.

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

Answer:

1635924301949

To prove : $\Delta DCB \sim \Delta HGE$

Given : $\Delta ABC \sim \Delta EGF$

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

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

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

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

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

Answer:

1635924370515

To prove : $\Delta DCA \sim \Delta HGF$

Given : $\Delta ABC \sim \Delta EGF$

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

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

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

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

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

1635924369822

Answer:

To prove : $\Delta ABD \sim \Delta ECF$

Given: ABC is an isosceles triangle.

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

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

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

$\angle ADB=\angle EFC$ ( Each $90 ^\circ$ )

$\Delta ABD \sim \Delta ECF$ ( 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 $90 ^\circ$ )

$\angle DCF=\angle BAE$ (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)

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


Topics covered in Chapter 10 Triangles: Exercise 6.3

1. Understanding the relationship between sides and angles of a triangle: All triangles follow the pattern where large angles pair with extended sides, but small angles match short sides, thus creating a direct relationship between angle measures and side lengths.

2. Applying the triangle inequality property to compare sides: Understand how the rule functions, that the two connecting triangle sides should exceed the third side measure for triangle formation.

3. Determining which angle is larger based on side lengths: Strategies must be developed to understand side length ratios so that one can predict angles' relative size measurements without performing actual measurements.

4. Determining which side is longer based on angle measures: Interpretation of given angle measures allows students to determine opposite side relations, leading to improved triangle analysis abilities.

5. Logical reasoning with inequalities in triangles: Logical reasoning with inequalities in triangles: Students should apply triangle inequality properties to critical problem-solving through proofs and comparisons, and effective problem-solving methods.

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NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

JEE Main Important Mathematics Formulas

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

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

1. What is the core concept of Class 10 Maths chapter 6 exercise 6.3 ?

This exercise is about the criteria or theorems for congruence of triangles. Practice this class 10 ex 6.3 to command the concepts.


2. Name all the criteria for similarity of triangles according to NCERT solutions Class 10 Maths chapter 6 exercise 6.?

In this class 10 maths ex 6.3 following theorems are discussed in detail.

  1. Angle-angle-angle

  2. Angle-side-angle

  3. Side-side-side

  4. Angle-angle

  5. Angle-angle-side

  6. Side-angle-side

Go through this ex 6.3 class 10 to get deeper understanding of related concepts.

3. What is the total of a triangle's angels?

The sum of a triangle's angels is 180 degrees.


4. Is similarity a sign of congruence between the figures?

No, similarity simply refers to the shape of the figures; it does not imply equality.


5. If two corresponding angels of two triangles are same does this mean the triangles are similar?

Yes, because a triangle has three angles and their sum remains 180 degrees. So, if two angels are equal then the third also becomes equal. 


6. If all the three sides of two triangles are in proportion, then does this means the triangles are similar?

Yes, if all the three sides of two triangles are in proportion, then does this means the triangles are similar. By side-side-side criteria.


7. What will happen if the ratio of the sides of two similar triangles becomes equal to 1?

If the ratio becomes equal to 1 then the triangles are congruent to each other.


8. For similar triangles, what is the purpose of side ratios?

We can find the length of the corresponding side of the other triangle using the side ratio and the length of any side. Practice 10th class maths exercise 6.3 answers to command the concepts.

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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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