Exercise 6.6 is the last exercise present in the NCERT Class 10 Chapter 6 of Maths, Triangles. The exercise is named an optional exercise and provides extra questions from all the topics covered in the chapter. A number of theorems are used to solve the problems given in the exercise. In order to avoid any type of confusion related to which theorem to use, the subject experts at BYJU’S ensure to solve each question by indicating the theorem used in NCERT Class 10 Maths Solutions.
The NCERT solutions for Class 10 Maths is the most efficient study material one can get for their Class 10 board exam preparation. The NCERT textbook contains different types of questions from different topics, which will help the students get enough mathematical problem-solving skills. Practising these NCERT Class 10 solutions will help the students in increasing their speed and understanding of the concepts.
NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.6
Access Other Exercise Solutions of Class 10 Maths Chapter 6 – Triangles
Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)
Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)
Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)
Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)
Access Answers to Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.6
1. In Figure, PS is the bisector of ∠QPR of ∆ PQR. Prove that QS/PQ = SR/PR
Solution:
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)
As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
∠PRT=∠QTR
Therefore,
PT=PR
In â–³QTR, by basic proportionality theorem,
QS/SR = QP/PT
Since, PT=TR
Therefore,
QS/SR = PQ/PR
Hence, proved.
2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM2 = DN . MC (ii) DN2 = DM . AN.
Solution:
- Let us join Point D and B.
Given,
BD ⊥AC, DM ⊥ BC and DN ⊥ AB
Now from the figure we have,
DN || CB, DM || AB and ∠B = 90 °
Therefore, DMBN is a rectangle.
So, DN = MB and DM = NB
The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)
In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)
In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)
From equation (i) and (ii), we get
∠1 = ∠3
From equation (i) and (iii), we get
∠2 = ∠4
In ∆DCM and ∆BDM,
∠1 = ∠3 (Already Proved)
∠2 = ∠4 (Already Proved)
∴ ∆DCM ∼ ∆BDM (AA similarity criterion)
BM/DM = DM/MC
DN/DM = DM/MC (BM = DN)
⇒ DM2 = DN × MC
Hence, proved.
(ii) In right triangle DBN,
∠5 + ∠7 = 90° ……………….. (iv)
In right triangle DAN,
∠6 + ∠8 = 90° ………………… (v)
D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)
From equation (iv) and (vi), we get,
∠6 = ∠7
From equation (v) and (vi), we get,
∠8 = ∠5
In ∆DNA and ∆BND,
∠6 = ∠7 (Already proved)
∠8 = ∠5 (Already proved)
∴ ∆DNA ∼ ∆BND (AA similarity criterion)
AN/DN = DN/NB
⇒ DN2 = AN × NB
⇒ DN2 = AN × DM (Since, NB = DM)
Hence, proved.
3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
AC2= AB2+ BC2+ 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
AB2 = AD2 + DB2 ……………………… (i)
Again, by applying Pythagoras Theorem in ∆ACD, we get,
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC) 2
AC2 = AD2 + DB2 + BC2 + 2DB × BC
From equation (i), we can write,
AC2 = AB2 + BC2 + 2DB × BC
Hence, proved.
4. In Figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that
AC2= AB2+ BC2 – 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
AB2 = AD2 + DB2
We can write it as;
⇒ AD2 = AB2 − DB2 ……………….. (i)
By applying Pythagoras Theorem in ∆ADC, we get,
AD2 + DC2 = AC2
From equation (i),
AB2 − BD2 + DC2 = AC2
AB2 − BD2 + (BC − BD) 2 = AC2
AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD
AC2 = AB2 + BC2 − 2BC × BD
Hence, proved.
5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC.DM + 2 (BC/2) 2
(ii) AB2 = AD2 – BC.DM + 2 (BC/2) 2
(iii) AC2 + AB2 = 2 AD2 + ½ BC2
Solution:
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM2 + MD2 = AD2 ………………. (i)
Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2
AM2 + (MD + DC) 2 = AC2
(AM2 + MD2 ) + DC2 + 2MD.DC = AC2
From equation(i), we get,
AD2 + DC2 + 2MD.DC = AC2
Since, DC=BC/2, thus, we get,
AD2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC2
AD2 + (BC/2) 2 + 2MD × BC = AC2
Hence, proved.
(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD) 2
= AD2 − DM2 + BD2 + MD2 − 2BD × MD
= AD2 + BD2 − 2BD × MD
= AD2 + (BC/2)2 – 2(BC/2) MD
= AD2 + (BC/2)2 – BC MD
Hence, proved.
(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM2 + MB2 = AB2 ………………….… (i)
By applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2 …………………..… (ii)
Adding both the equations (i) and (ii), we get,
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD − DM) 2 + (MD + DC) 2 = AB2 + AC2
2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2
2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB2 + AC2
2AD2 + BC2/2 = AB2 + AC2
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.
By applying Pythagoras Theorem in ∆DEA, we get,
DE2 + EA2 = DA2 ……………….… (i)
By applying Pythagoras Theorem in ∆DEB, we get,
DE2 + EB2 = DB2
DE2 + (EA + AB) 2 = DB2
(DE2 + EA2) + AB2 + 2EA × AB = DB2
DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)
By applying Pythagoras Theorem in ∆ADF, we get,
AD2 = AF2 + FD2
Again, applying Pythagoras theorem in ∆AFC, we get,
AC2 = AF2 + FC2 = AF2 + (DC − FD) 2
= AF2 + DC2 + FD2 − 2DC × FD
= (AF2 + FD2) + DC2 − 2DC × FD AC2
AC2= AD2 + DC2 − 2DC × FD ………………… (iii)
Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)
In ∆DEA and ∆ADF,
∠DEA = ∠AFD (Each 90°)
∠EAD = ∠ADF (EA || DF)
AD = AD (Common Angles)
∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)
⇒ EA = DF ……………… (vi)
Adding equations (i) and (iii), we get,
DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2
From equation (iv) and (vi),
BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP
Solution:
Firstly, let us join CB, in the given figure.
(i) In ∆APC and ∆DPB,
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment for chord CB)
Therefore,
∆APC ∼ ∆DPB (AA similarity criterion)
(ii) In the above, we have proved that ∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
∴ AP/DP = PC/PB = CA/BD
⇒AP/DP = PC/PB
∴AP. PB = PC. DP
Hence, proved.
8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD.
Solution:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.
∠PAC = ∠PDB
Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)
(ii) We have already proved above,
∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
Therefore,
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
∴ AP. PB = PC. DP
9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.
Solution:
In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.
Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
AD || PC
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
AP = AC
⇒ ∠APC = ∠ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
∠BAD = ∠APC
Therefore, AD is the bisector of the angle BAC.
Hence, proved.
10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Solution:
Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.
To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;
AC2 = AB2+ BC2
AB2 = (1.8 m) 2 + (2.4 m) 2
AB2 = (3.24 + 5.76) m2
AB2 = 9.00 m2
⟹ AB = √9 m = 3m
Thus, the length of the string out is 3 m.
As its given, she pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m
Let us say now, the fly is at point D after 12 seconds.
Length of string out after 12 seconds is AD.
AD = AC − String pulled by Nazima in 12 seconds
= (3.00 − 0.6) m
= 2.4 m
In ∆ADB, by Pythagoras Theorem,
AB2 + BD2 = AD2
(1.8 m) 2 + BD2 = (2.4 m) 2
BD2 = (5.76 − 3.24) m2 = 2.52 m2
BD = 1.587 m
Horizontal distance of the fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
= 2.79 m
The last exercise of Class 10 Maths Chapter 6, Triangles is the sixth exercise. Exercise 6.6 contains 10 Questions that include all the topics that are covered in the chapter, from beginning to end. Out of the 10 questions in the exercise, 5 questions are short answer type questions, and 5 questions are long answer type questions. There are a number of theorems and concepts covered in this exercise. The name of these concepts and theorems or rules are given below.
- Similar Triangles.
- Thales’ Theorem
- AA Similarity
- AAA Congruence Rule.
- SSS Congruence Rule.
- SAS Congruence Rule.
- Theorems Related to Areas of Similar Triangles
- Pythagoras’ Theorem
- Converse of Pythagoras Theorem
- CPCT -‘Corresponding parts of congruent triangles are equal’.
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