CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
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.
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! If you are looking for the CM Shri School admission result, here is the link provided by Careers360. You can check your result directly and stay updated about the admission status. I’ll be attaching it for your reference.
https://school.careers360.com/articles/cm-shri-school-admission-test-2025
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
Hello Rajesh!!!
I have question paper of science of 9th standard . Here is your question paper link where you have click to open and wrap it up.
https://school.careers360.com/boards/dge-tamil-nadu/tn-class-9-quarterly-exam-question-paper-2025-26
Hello,
For your information, the class 7 2025 result of the CM Shri School is expected to be published on September 20, 2025. So, if you want to see and download the results, then you can visit the official website of the CM Shri School: edudel.nic.in.
I hope it will clear your query!!
Hello
Visit the official website of the Rajasthan Education Department or the Shala Darpan portal.
Look for the “Latest News” or “Examination” section. Check school notice boards or ask your class teacher for updates.
Some district education office websites may also upload the key. Avoid unofficial websites as they may provide incorrect or fake keys.
Hrllo,
If you want to view the CM Shri School 2025-26 admission test result , visit the official website, click on the "CM Shri Schools Admission Test - 2025." then select the "CM Shri Admission Test Result 2025 - Merit List" link. Here you need to log in with your credentials and view or download the merit list pdf.
I hope it will clear your query!!
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters