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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
Answer:
A line RT is drawn parallel to SP which intersect QP produced at T.
Given: PS is the bisector of $\angle QPR \: \: of\: \: \Delta PQR$ .
$\angle QPS=\angle SPR.....................................1$
By construction,
$\angle SPR=\angle PRT.....................................2$ (as PS||TR)
$\angle QPS=\angle QTR.....................................3$ (as PS||TR)
From the above equations, we get
$\angle PRT=\angle QTR$
$\therefore PT=PR$
By construction, PS||TR
In $\triangle$ QTR, by Thales theorem,
$\frac{QS}{SR}=\frac{QP}{PT}$
$\frac{QS }{SR } = \frac{PQ }{PR }$
Hence proved.
Answer:
Join BD
Given : D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB.Also DN || BC, DM||NB
$\angle CDB=90 ^\circ$
$\Rightarrow \angle 2+\angle 3=90 ^\circ.............................1$
In $\triangle$ CDM, $\angle 1+\angle 2+\angle DMC=180 ^\circ$
$\angle 1+\angle 2=90 ^\circ.......................2$
In $\triangle$ DMB, $\angle 3+\angle 4+\angle DMB=180 ^\circ$
$\angle 3+\angle 4=90 ^\circ.......................3$
From equation 1 and 2, we get $\angle 1=\angle 3$
From equation 1 and 3, we get $\angle 2=\angle 4$
In $\triangle DCM\, \, and\, \, \triangle BDM,$
$\angle 1=\angle 3$
$\angle 2=\angle 4$
$\triangle DCM\, \, \sim \, \, \triangle BDM,$ (By AA)
$\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}$
$\Rightarrow \frac{DN}{DM}=\frac{DM}{MC}$ (BM=DN)
$\Rightarrow$ $DM^2 = DN . MC$
Hence proved
Answer:
In $\triangle$ DBN,
$\angle 5+\angle 7=90 ^\circ.......................1$
In $\triangle$ DAN,
$\angle 6+\angle 8=90 ^\circ.......................2$
BD $\perp$ AC, $\therefore \angle ADB=90 ^\circ$
$\angle 5+\angle 6=90 ^\circ.......................3$
From equation 1 and 3, we get $\angle 6=\angle 7$
From equation 2 and 3, we get $\angle 5=\angle 8$
In $\triangle DNA\, \, and\, \, \triangle BND,$
$\angle 6=\angle 7$
$\angle 5=\angle 8$
$\triangle DNA\, \, \sim \, \, \triangle BND$ (By AA)
$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$
$\Rightarrow \frac{AN}{DN}=\frac{DN}{DM}$ (NB=DM)
$\Rightarrow$ $DN^2 = AN . DM$
Hence proved.
Answer:
In $\triangle$ ADB, by Pythagoras theorem
$AB^2=AD^2+DB^2.......................1$
In $\triangle$ ACD, by Pythagoras theorem
$AC^2=AD^2+DC^2.......................2$
$AC^2=AD^2+(BD+BC)^2$
$\Rightarrow AC^2=AD^2+(BD)^2+(BC)^2+2.BD.BC$
$AC^2 = AB^2 + BC^2 + 2 BC . BD.$ (From 1)
Answer:
In $\triangle$ ADB, by Pythagoras theorem
$AB^2=AD^2+DB^2$
$AD^2=AB^2-DB^2...........................1$
In $\triangle$ ACD, by Pythagoras theorem
$AC^2=AD^2+DC^2$
$AC^2=AB^2-BD^2+DC^2$ (From 1)
$\Rightarrow AC^2=AB^2-BD^2+(BC-BD)^2$
$\Rightarrow AC^2=AB^2-BD^2+(BC)^2+(BD)^2-2.BD.BC$
$AC^2 = AB^2 + BC^2 - 2 BC . BD.$
Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that : $AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2$
Answer:
Given: AD is a median of a triangle ABC and AM $\perp$ BC.
In $\triangle$ AMD, by Pythagoras theorem
$AD^2=AM^2+MD^2.......................1$
In $\triangle$ AMC, by Pythagoras theorem
$AC^2=AM^2+MC^2$
$AC^2=AM^2+(MD+DC)^2$
$\Rightarrow AC^2=AM^2+(MD)^2+(DC)^2+2.MD.DC$
$AC^2 = AD^2 + DC^2 + 2 DC . MD.$ (From 1)
$AC^2 = AD^2 + (\frac{BC}{2})^2 + 2(\frac{BC}{2}). MD.$ (BC=2 DC)
$AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2$
Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that : $AB ^2 = AD ^2 - BC .DM + \left ( \frac{BC}{2} \right ) ^2$
Answer:
In $\triangle$ ABM, by Pythagoras theorem
$AB^2=AM^2+MB^2$
$AB^2=(AD^2-DM^2)+MB^2$
$\Rightarrow AB^2=(AD^2-DM^2)+(BD-MD)^2$
$\Rightarrow AB^2=AD^2-DM^2+(BD)^2+(MD)^2-2.BD.MD$
$\Rightarrow AB^2=AD^2+(BD)^2-2.BD.MD$
$\Rightarrow AB^2 = AD^2 + (\frac{BC}{2})^2 -2(\frac{BC}{2}). MD.=AC^2$ (BC=2 BD)
$\Rightarrow AD^2 + (\frac{BC}{2})^2 -BC. MD.=AC^2$
Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that: $AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2$
Answer:
In $\triangle$ ABM, by Pythagoras theorem
$AB^2=AM^2+MB^2.......................1$
In $\triangle$ AMC, by Pythagoras theorem
$AC^2=AM^2+MC^2$ ..................................2
Adding equation 1 and 2,
$AB^2+AC^2=2AM^2+MB^2+MC^2$
$\Rightarrow AB^2+AC^2=2AM^2+(BD-DM)^2+(MD+DC)^2$
$\Rightarrow AB^2+AC^2=2AM^2+(BD)^2+(DM)^2-2.BD.DM+(MD)^2+(DC)^2+2.MD.DC$
$\Rightarrow AB^2+AC^2=2AM^2+2.(DM)^2+BD^2+(DC)^2+2.MD.(DC-BD)$ $\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2.MD.(\frac{BC}{2}-\frac{BC}{2})$ $\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2$
$AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2$
Answer:
In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.
In $\triangle$ DEA, by Pythagoras theorem
$DA^2=DE^2+EA^2.......................1$
In $\triangle$ DEB, by Pythagoras theorem
$DB^2=DE^2+EB^2$
$DB^2=DE^2+(EA+AB)^2$
$DB^2=DE^2+(EA)^2+(AB)^2+2.EA.AB$
$DB^2=DA^2+(AB)^2+2.EA.AB$ ....................................2
In $\triangle$ ADF, by Pythagoras theorem
$DA^2=AF^2+FD^2$
In $\triangle$ AFC, by Pythagoras theorem
$AC^2=AF^2+FC^2=AF^2+(DC-FD)^2$
$\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD$
$\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD$
$\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3$
Since ABCD is a parallelogram.
SO, AB=CD and BC=AD
In $\triangle DEA\, and\, \triangle ADF,$
$\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 ^\circ)$
$\angle DAE=\angle ADF$ (AE||DF)
AD=AD (common)
$\triangle DEA\, \cong \, \triangle ADF,$ (ASA rule)
$\Rightarrow EA=DF.......................6$
Adding 2 and, we get
$DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2$ $\Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2$
$\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2$ (From 4 and 6)
$\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2$ \
Answer:
Join BC
In $\triangle APC\, \, and\, \triangle DPB,$
$\angle APC\, \, = \angle DPB$ ( vertically opposite angle)
$\angle CAP\, \, = \angle BDP$ (Angles in the same segment)
$\triangle APC\, \, \sim \triangle DPB$ (By AA)
Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : $AP . PB = CP . DP$
Answer:
Join BC
In $\triangle APC\, \, and\, \triangle DPB,$
$\angle APC\, \, = \angle DPB$ ( vertically opposite angle)
$\angle CAP\, \, = \angle BDP$ (Angles in the same segment)
$\triangle APC\, \, \sim \triangle DPB$ (By AA)
$\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)
$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$
$\Rightarrow AP.PB=PC.DP$
Answer:
In $\Delta PAC \,and \,\Delta PDB,$
$\angle P=\angle P$ (Common)
$\angle PAC=\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, $\Delta PAC \sim \Delta PDB$ ( By AA rule)
Answer:
In $\Delta PAC \,and \,\Delta PDB,$
$\angle P=\angle P$ (Common)
$\angle PAC=\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, $\Delta PAC \sim \Delta PDB$ ( By AA rule)
24440 $\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)
$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$
$\Rightarrow AP.PB=PC.DP$
Answer:
Produce BA to P, such that AP=AC and join P to C.
$\frac{BD }{CD} = \frac{AB}{AC}$ (Given )
$\Rightarrow \frac{BD }{CD} = \frac{AP}{AC}$
Using converse of Thales theorem,
AD||PC $\Rightarrow \angle BAD=\angle APC............1$ (Corresponding angles)
$\Rightarrow \angle DAC=\angle ACP............2$ (Alternate angles)
By construction,
AP=AC
$\Rightarrow \angle APC=\angle ACP............3$
From equation 1,2,3, we get
$\Rightarrow \angle BAD=\angle APC$
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 $\triangle$ ABC, using Pythagoras theorem
$AC^2=AB^2+BC^2$
$\Rightarrow AC^2=(1.8)^2+(2.4)^2$
$\Rightarrow AC^2=3.24+5.76$
$\Rightarrow AC^2=9.00$
$\Rightarrow AC=3 m$
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.
= $12\times 5=60cm=0.6m$
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 $\triangle$ ADB,
$AB^2+BD^2=AD^2$
$\Rightarrow (1.8)^2+BD^2=(2.4)^2$
$\Rightarrow BD^2=5.76-3.24=2.52 m^2$
$\Rightarrow BD=1.587 m$
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:
Frequently Asked Questions (FAQs)
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
On Question asked by student community
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.
Hello Aspirant,
Here's how you can find it:
School ID Card: Your registration number is often printed on your school ID card.
Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.
School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.
Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.
Your school is the best place to get this information.
Hello,
It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.
The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.
For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.
Let me know if you need imformation about any exam eligibility criteria.
good wishes for your future!!
Hello Aspirant,
"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).
For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.
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