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NCERT Solutions for Exercise 6.6 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 6.6 help us to understand how to apply all the concepts till now we have learned through all the exercises in the chapter. The notion of triangle similarity, criteria for triangle similarity, areas of similar triangles, and Pythagoras Theorem are covered in this exercise.
10th class Maths exercise 6.6 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: 6.6
Q1 In Fig. 6.56, PS is the bisector of . Prove that
Answer:
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.
Answer:
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
Answer:
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
Answer:
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 :
Answer:
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 :
Answer:
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:
Answer:
In ABM, by Pythagoras theorem
In AMC, by Pythagoras theorem
..................................2
Adding equation 1 and 2,
Answer:
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 :
Answer:
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 :
Answer:
Join BC
In
( vertically opposite angle)
(Angles in the same segment)
(By AA)
(Corresponding sides of similar triangles are proportional)
Answer:
In
(Common)
(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, ( By AA rule)
Answer:
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.
Answer:
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
Class 10 Maths chapter 6 exercise 6: The questions in exercise 6.6 Class 10 Maths consist of many types of questions covering different types of theorems and formulas. Firstly we have a question in which we have to prove the left-hand side argument and right-hand side argument NCERT solutions for Class 10 Maths exercise 6.6 also have questions. Exercise 6.6 Class 10 Maths covers all types of questions that can be formed on the similarity of triangles. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.
Also, See:
similarity of triangle
Criteria for Similarity of Triangles
Areas of Similar Triangles
Pythagoras Theorem
If the respective angles are congruent and the corresponding sides are proportional, two triangles are said to be similar.
NCERT solutions for Class 10 Maths exercise 6.6 is different from other exercises as all other exercises are based on a single concept and theorem they are base building exercises but Class 10 Maths chapter 6 exercise 6.6 is based on all the concepts.
Yes Class 10 Maths chapter 6 exercise 6.6 because it have questions which have mix concepts
There are 13 main theorems are there which we require to solve NCERT solutions for Class 10 Maths 1 exercise 6.6
There are ten questions in exercise 6.6 Class 10 Maths question
There are two types of questions in exercise 6.6 Class 10 Maths question one type is there in which we have to proof LHS and RHS these contains uses of theorem and other is real life application of triangle
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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