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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.
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.
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.
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.
Maths Chapter 6 Triangles class 10 Excercise: 6.1
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.
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.
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.
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.
The two different examples of a pair of similar figures are :
1. Two circles with different radii.
2. Two rectangles with different breadth and length.
Q2 (2) Give two different examples of a pair of non-similar figures.
The two different examples of a pair of non-similar figures are :
1.Rectangle and circle
2. A circle and a triangle.
Q3 State whether the following quadrilaterals are similar or not:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. 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).
(i)
Let EC be x
Given: DE || BC
By using the proportionality theorem, we get
(ii)
Let AD be x
Given: DE || BC
By using the proportionality theorem, we get
(i)
Given :
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
and
We have
Hence, EF is not parallel to QR.
(ii)
Given :
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
and
We have
Hence, EF is parallel to QR.
Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
(iii)
Given :
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
and
We have
Hence, EF is parallel to QR.
Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that
Answer:
Given : LM || CB and LN || CD
To prove :
Since , LM || CB so we have
Also, LN || CD
From equation 1 and 2, we have
Hence proved.
Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC
Given : DE || AC and DF || AE.
To prove :
Since , DE || AC so we have
Also,DF || AE
From equation 1 and 2, we have
Hence proved.
Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Given : DE || OQ and DF || OR.
To prove EF || QR.
Since DE || OQ so we have
Also, DF || OR
From equation 1 and 2, we have
Thus, EF || QR. (converse of basic proportionality theorem)
Hence proved.
Given : AB || PQ and AC || PR
To prove: BC || QR
Since, AB || PQ so we have
Also, AC || PR
From equation 1 and 2, we have
Therefore, BC || QR. (converse basic proportionality theorem)
Hence proved.
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. and
.
Using basic proportionality theorem, we have
Since
Q is the midpoint of AC.
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. and
.
we have,
From equation 1 and 2, we get
By basic proportionality theorem, we have
Draw a line EF passing through point O such that
To prove :
In , we have
So, by using basic proportionality theorem,
In , we have
So, by using basic proportionality theorem,
Using equation 1 and 2, we get
Hence proved.
Draw a line EF passing through point O such that
Given :
In , we have
So, by using basic proportionality theorem,
However, its is given that
Using equation 1 and 2 , we get
(By basic proportionality theorem)
Therefore, ABCD is a trapezium.
Class 10 Maths Chapter 6 Triangles Excercise: 6.3
(i)
(By AAA)
So ,
(ii) As corresponding sides of both triangles are proportional.
(By SSS)
(iii) Given triangles are not similar because corresponding sides are not proportional.
(iv) by SAS similarity criteria.
(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides
(vi) In , we know that
In , we know that
( By AAA)
Given : ,
and
(DOB is a straight line)
In
Since , , so
( Corresponding angles are equal in similar triangles).
In , we have
( Alternate interior angles as
)
( Alternate interior angles as
)
( Vertically opposite angles are equal)
( By AAA)
( corresponding sides are equal)
Hence proved.
Q5 S and T are points on sides PR and QR of PQR such that
P =
RTS. Show that
RPQ ~
RTS.
Given : P =
RTS
To prove RPQ ~ RTS.
In RPQ and
RTS,
P =
RTS (Given )
R =
R (common)
RPQ ~
RTS. ( By AA)
Q6 In Fig. 6.37, if ABE
ACD, show that
ADE ~
ABC.
Answer:
Given :
To prove ADE ~ ABC.
Since
( By CPCT)
(By CPCT)
In ADE and
ABC,
( Common)
and
(
and
)
Therefore, ADE ~
ABC. ( By SAS criteria)
Q7 (1) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Vertically opposite angles )
( By AA criterion)
Q7 (2) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( By AA criterion)
Q7 (3) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( By AA criterion)
Q7 (4) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( 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
To prove :
In ,
( Opposite angles of a parallelogram are equal)
( Alternate angles of AE||BC)
( 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:
To prove :
In ,
( Each
)
( common)
( By AA criterion )
Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that
To prove :
In ,
( Each
)
( common)
( By AA criterion )
( corresponding parts of similar triangles )
Hence proved.
Q10 (1) CD and GH are respectively the bisectors of and
such that D and H lie on sides AB and FE of
respectively. If
, show that:
To prove :
Given :
( CD and GH are bisectors of equal angles)
( CD and GH are bisectors of equal angles)
In
( proved above)
( proved above)
( By AA criterion)
Hence proved.
Q 10 (2) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of
respectively. If
, show that:
To prove :
Given :
In ,
( CD and GH are bisectors of equal angles)
(
)
( By AA criterion )
Q10 (3) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of
respectively. If
, show that:
To prove :
Given :
In ,
( CD and GH are bisectors of equal angles)
(
)
( By AA criterion )
To prove :
Given: ABC is an isosceles triangle.
In ,
(
)
( Each
)
( By AA criterion)
AD and PM are medians of triangles. So,
Given :
In
(SSS similarity)
( Corresponding angles of similar triangles )
In
(proved above)
Therefore, . ( SAS similarity)
Q13 D is a point on the side BC of a triangle ABC such that . Show that
In,
( given )
(common )
( By AA rule)
( corresponding sides of similar triangles )
(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
(Given )
(SSS similarity)
...................1 (Corresponding angles of similar triangles)
Similarity,
........................2
Adding equation 1 and 2,
............................3
In
( Given )
( From above equation 3)
( SAS similarity)
CD = pole
AB = tower
Shadow of pole = DF
Shadow of tower = BE
In
( Each
)
(Angle of sun at same place )
(AA similarity)
cm
Hence, the height of the tower is 42 cm.
Q16 If AD and PM are medians of triangles ABC and PQR, respectively where , prove that
Answer:
( Given )
............... ....1( corresponding sides of similar triangles )
....................................2
AD and PM are medians of triangle.So,
..........................................3
From equation 1 and 3, we have
...................................................................4
In
(From equation 2)
(From equation 4)
(SAS similarity)
Class 10 Maths Chapter 6 Triangles Excercise:6.4
Q1 Let and their areas be, respectively, 64
and 121
. If EF = 15.4 cm, find BC.
( Given )
ar(ABC) = 64 and ar(DEF)=121
.
EF = 15.4 cm (Given )
Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.
AB = 2 CD ( Given )
In
(vertically opposite angles )
(Alternate angles)
(Alternate angles)
(AAA similarity)
Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
Answer:
Let DM and AP be perpendicular on BC.
In
(Each
)
(Vertically opposite angles)
(AA similarity)
Since
Q4 If the areas of two similar triangles are equal, prove that they are congruent.
Let , therefore,
(Given )
(SSS )
D, E, and F are respectively the mid-points of sides AB, BC and CA of . ( Given )
and DE||AC
In ,
(corresponding angles )
(corresponding angles )
(By AA)
Let be x.
Similarly,
and
Let AD and PS be medians of both similar triangles.
Purring these value in 1,
In
(proved above)
(proved above)
(SAS )
Therefore,
From 1 and 4, we get
Let ABCD be a square of side units.
Therefore, diagonal =
Triangles form on the side and diagonal are ABE and
DEF, respectively.
Length of each side of triangle ABE = a units
Length of each side of triangle DEF = units
Both the triangles are equilateral triangles with each angle of .
( By AAA)
Using area theorem,
(A) 2: 1 (B) 1: 2 (C) 4 : 1 (D) 1: 4
Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
All angles of the triangle are .
ABC
BDE (By AAA)
Let AB=BC=CA = x
then EB=BD=ED=
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
Sides of two similar triangles are in the ratio 4: 9.
Let triangles be ABC and DEF.
We know that
Option D is correct.
Class 10 Maths Chapter 6 Triangles Excercise: 6.5
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,
= given third side.
Hence, it is the right triangle with h=25 cm.
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,
Hence, it is not the right triangle.
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,
Hence, it is not a right triangle.
In the case of a right triangle, the length of its hypotenuse is highest.
hypotenuse be h.
Taking, 5cm, 12 cm
By Pythagoras theorem,
= 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 . Show that
Let be x
In ,
Similarly,
In ,
In
(By AAA)
Hence proved.
Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
In
(common )
(By AA)
, hence prooved .
Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
Answer:
Let be x
In ,
Similarly,
In ,
In
( Each right angle)
(By AAA)
Hence proved
Q3 (3) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
In
(common )
(By AA)
Hence proved.
Q4 ABC is an isosceles triangle right angled at C. Prove that
Given: ABC is an isosceles triangle right angled at C.
Let AC=BC
In ABC,
By Pythagoras theorem
(AC=BC)
Hence proved.
Q5 ABC is an isosceles triangle with AC = BC. If , prove that ABC is a right triangle.
Given: ABC is an isosceles triangle with AC=BC.
In ABC,
(Given )
(AC=BC)
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.
Given: ABC is an equilateral triangle of side 2a.
AB=BC=AC=2a
AD is perpendicular to BC.
We know that the altitude of an equilateral triangle bisects the opposite side.
So, BD=CD=a
In ADB,
By Pythagoras theorem,
The length of each altitude is .
In AOB, by Pythagoras theorem,
In BOC, by Pythagoras theorem,
In COD, by Pythagoras theorem,
In AOD, by Pythagoras theorem,
Adding equation 1,2,3,4,we get
(AO=CO and BO=DO)
Hence proved .
Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE
AC and OF
AB. Show that
Join AO, BO, CO
In AOF, by Pythagoras theorem,
In BOD, by Pythagoras theorem,
In COE, by Pythagoras theorem,
Adding equation 1,2,3,we get
Hence proved
Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE
AC and OF
AB.
Join AO, BO, CO
In AOF, by Pythagoras theorem,
In BOD, by Pythagoras theorem,
In COE, by Pythagoras theorem,
Adding equation 1,2,3,we get
OA is a wall and AB is a ladder.
In AOB, by Pythagoras theorem
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
OB is a pole.
In AOB, by Pythagoras theorem
Hence, the distance of the stack from the base of the pole is m.
Distance travelled by the first aeroplane due north in hours.
Distance travelled by second aeroplane due west in hours.
OA and OB are the distance travelled.
By Pythagoras theorem,
Thus, the distance between the two planes is .
Let AB and CD be poles of heights 6 m and 11 m respectively.
CP=11-6=5 m and AP= 12 m
In APC,
By Pythagoras theorem,
Hence, the distance between the tops of two poles is 13 m.
In ACE, by Pythagoras theorem,
In BCD, by Pythagoras theorem,
From 1 and 2, we get
In CDE, by Pythagoras theorem,
In ABC, by Pythagoras theorem,
From 3,4,5 we get
In ACD, by Pythagoras theorem,
In ABD, by Pythagoras theorem,
From 1 and 2, we get
Given : 3DC=DB, so
From 3 and 4, we get
Hence proved.
Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that
Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.
To prove :
Let AB=BC=CA=a
Draw an altitude AE on BC.
So,
In AEB, by Pythagoras theorem
Given : BD = 1/3 BC.
In ADE, by Pythagoras theorem,
Given: An equilateral triangle ABC.
Let AB=BC=CA=a
Draw an altitude AE on BC.
So,
In AEB, by Pythagoras theorem
Q17 Tick the correct answer and justify : In AB =
cm, AC = 12 cm and BC = 6 cm.
The angle B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°
In AB =
cm, AC = 12 cm and BC = 6 cm.
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 . Prove that
A line RT is drawn parallel to SP which intersect QP produced at T.
Given: PS is the bisector of .
By construction,
(as PS||TR)
(as PS||TR)
From the above equations, we get
By construction, PS||TR
In QTR, by Thales theorem,
Hence proved.
Join BD
Given : D is a point on hypotenuse AC of D ABC, such that BD AC, DM
BC and DN
AB.Also DN || BC, DM||NB
In CDM,
In DMB,
From equation 1 and 2, we get
From equation 1 and 3, we get
In
(By AA)
(BM=DN)
Hence proved
Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD AC, DM
BC and DN
AB. Prove that:
Answer:
In DBN,
In DAN,
BD AC,
From equation 1 and 3, we get
From equation 2 and 3, we get
In
(By AA)
(NB=DM)
Hence proved.
Q3 In Fig. 6.58, ABC is a triangle in which ABC > 90° and AD
CB produced. Prove that
In ADB, by Pythagoras theorem
In ACD, by Pythagoras theorem
(From 1)
Q4 In Fig. 6.59, ABC is a triangle in which ABC < 90° and AD
BC. Prove that
In ADB, by Pythagoras theorem
In ACD, by Pythagoras theorem
(From 1)
Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :
Given: AD is a median of a triangle ABC and AM BC.
In AMD, by Pythagoras theorem
In AMC, by Pythagoras theorem
(From 1)
(BC=2 DC)
Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :
In ABM, by Pythagoras theorem
(BC=2 BD)
Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that:
In ABM, by Pythagoras theorem
In AMC, by Pythagoras theorem
..................................2
Adding equation 1 and 2,
In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.
In DEA, by Pythagoras theorem
In DEB, by Pythagoras theorem
....................................2
In ADF, by Pythagoras theorem
In AFC, by Pythagoras theorem
Since ABCD is a parallelogram.
SO, AB=CD and BC=AD
In
(AE||DF)
AD=AD (common)
(ASA rule)
Adding 2 and, we get
(From 4 and 6)
\
Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :
Join BC
In
( vertically opposite angle)
(Angles in the same segment)
(By AA)
Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :
Join BC
In
( vertically opposite angle)
(Angles in the same segment)
(By AA)
(Corresponding sides of similar triangles are proportional)
In
(Common)
(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, ( By AA rule)
In
(Common)
(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, ( By AA rule)
24440 (Corresponding sides of similar triangles are proportional)
Q9 In Fig. 6.63, D is a point on side BC of D ABC such that Prove that AD is the bisector of
BAC.
Answer:
Produce BA to P, such that AP=AC and join P to C.
(Given )
Using converse of Thales theorem,
AD||PC (Corresponding angles)
(Alternate angles)
By construction,
AP=AC
From equation 1,2,3, we get
Thus, AD bisects angle BAC.
Let AB = 1.8 m
BC is a horizontal distance between fly to the tip of the rod.
Then, the length of the string is AC.
In ABC, using Pythagoras theorem
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.
=
Let D be the position of fly after 12 seconds.
Hence, AD is the length of the string that is out after 12 seconds.
Length of string pulled in by nazim=AD=AC-12
=3-0.6=2.4 m
In ADB,
Horizontal distance travelled by fly = BD+1.2 m
=1.587+1.2=2.787 m
= 2.79 m
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-
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.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | Triangles |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
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!
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.
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.
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.
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.
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.
Exam Date:01 January,2025 - 14 February,2025
Exam Date:01 January,2025 - 14 February,2025
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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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