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NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.3 contains critical criteria or theorems for congruence such as Angle-Angle-Angle (AAA), Angle-Side-Angle (ASA), and Side-Side-Side (SSS). It contains 16 questions that cover all parts of the previously given theorems.
NCERT solutions for exercise 6.3 Class 10 Maths chapter 6 Triangles discusses the standards required to demonstrate the resemblance of triangles. 10th class Maths exercise 6.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
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Triangles Class 10 Chapter 6 Exercise: 6.3
Answer:
(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)
Answer:
Given : , and
(DOB is a straight line)
In
Since , , so
( Corresponding angles are equal in similar triangles).
Answer:
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.
Answer:
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:
Answer:
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:
Answer:
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:
Answer:
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:
Answer:
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
Answer:
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:
Answer:
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
Answer:
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:
Answer:
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:
Answer:
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:
Answer:
To prove :
Given :
In ,
( CD and GH are bisectors of equal angles)
( )
( By AA criterion )
Answer:
To prove :
Given: ABC is an isosceles triangle.
In ,
( )
( Each )
( By AA criterion)
Answer:
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
Answer:
In,
( given )
(common )
( By AA rule)
( corresponding sides of similar triangles )
Answer:
(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)
Answer:
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)
NCERT solutions Class 10 Maths chapter 6 exercise 6.3 Triangles contains well-written examples that demonstrate the key similarity theorems. Certain implications can be taken from the presented theories, such as the AA property, which states that if two angles of one triangle are equal to two angles of another triangle, then the two triangles are comparable (similar to each other) as the corresponding third angle of the triangle also becomes equal as the sum of the total angles of a triangle remains fixed to 180 degrees. Before moving on to the exercise questions, students must ensure that they have thoroughly practiced these examples. Furthermore, the chapter's highlights can be refreshed on a regular basis to keep the concepts fresh. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.
Also, See:
This exercise is about the criteria or theorems for congruence of triangles. Practice this class 10 ex 6.3 to command the concepts.
In this class 10 maths ex 6.3 following theorems are discussed in detail.
Angle-angle-angle
Angle-side-angle
Side-side-side
Angle-angle
Angle-angle-side
Side-angle-side
Go through this ex 6.3 class 10 to get deeper understanding of related concepts.
The sum of a triangle's angels is 180 degrees.
No, similarity simply refers to the shape of the figures; it does not imply equality.
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.
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.
If the ratio becomes equal to 1 then the triangles are congruent to each other.
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.
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.
Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.
Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.
There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.
You can pursue diplomas at various institutions like:
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