SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

Exercise 6.6 - Chapter 6 Triangles class 10 ncert solutions Maths - SaraNextGen [2024]


Question 1:

In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2866/Chapter%206_html_m2f73de40.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2866/Chapter%206_html_m47b7cdd8.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2866/Chapter%206_html_m6240dbfb.jpg

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for ΔQTR,

QSSR=QPPT

⇒QSSR=PQPR          ∵PT=PR

Question 2:

In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:

(i) DM2 = DN.MC

(ii) DN2 = DM.AN

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2874/Chapter%206_html_7c4addaa.jpg

Answer:

(i)Let us join DB.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2874/Chapter%206_html_29c83c00.jpg

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2874/Chapter%206_html_46371165.gif

⇒ DM2 = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ∼ ΔBND (AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2874/Chapter%206_html_m7153b0c1.gif

⇒ DN2 = AN × NB

⇒ DN2 = AN × DM (As NB = DM)

Question 3:

In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2881/Chapter%206_html_468bcd8a.jpg

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AB2 = AD2 + DB2 … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC2 = AD+ DC2

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB × BC

AC2 = AB2 + BC2 + 2DB × BC [Using equation (1)]

 

Question 4:

In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 − 2BC.BD.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2885/Chapter%206_html_5777f071.jpg

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD2 + DB2 = AB2

⇒ AD2 = AB2 − DB… (1)

Applying Pythagoras theorem in ΔADC, we obtain

AD2 + DC2 = AC2

AB2 − BD2 + DC2 = AC2 [Using equation (1)]

AB2 − BD2 + (BC − BD)2 = AC2

AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD

= AB2 + BC2 − 2BC × BD

Question 5:

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_m7f0e9614.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_3f9e2260.gif

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_747ce29a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_735c4b14.jpg

Answer:

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM2 + MD2 = AD… (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM2 + MC2 = AC2

AM2 + (MD + DC)2 = AC2

(AM2 + MD2) + DC2 + 2MD.DC = AC2

AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)]

Using the result, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_m3c69647f.gif, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_6330889b.gif

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB2 = AM2 + MB2

= (AD2 − DM2) + MB2

= (AD2 − DM2) + (BD − MD)2

= AD2 − DM2 + BD2 + MD2 − 2BD × MD

= AD2 + BD2 − 2BD × MD

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_1f2442f4.gif

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM2 + MB2 = AB2 … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM2 + MC2 = AC2 … (2)

Adding equations (1) and (2), we obtain

2AM2 + MB2 + MC2 = AB2 + AC2

2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2

2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB+ AC2

2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2894/Chapter%206_html_53003a67.gif

Question 6:

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2902/Chapter%206_html_m56fc8869.jpg

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC.

Applying Pythagoras theorem in ΔDEA, we obtain

DE2 + EA2 = DA2 … (i)

Applying Pythagoras theorem in ΔDEB, we obtain

DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2

(DE2 + EA2) + AB2 + 2EA × AB = DB2

DA2 + AB2 + 2EA × AB = DB2 … (ii)

Applying Pythagoras theorem in ΔADF, we obtain

AD2 = AF2 + FD2

Applying Pythagoras theorem in ΔAFC, we obtain

AC2 = AF2 + FC2

= AF2 + (DC − FD)2

= AF2 + DC+ FD2 − 2DC × FD

= (AF2 + FD2) + DC2 − 2DC × FD

AC2 = AD2 + DC2 − 2DC × FD … (iii)

Since ABCD is a parallelogram,

AB = CD … (iv)

And, BC = AD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEAD https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2902/Chapter%206_html_m3437cc6c.gif ΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

Adding equations (i) and (iii), we obtain

DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2

BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2

[Using equations (iv) and (vi)]

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Question 7:

In the given figure, two chords AB and CD intersect each other at the point P. prove that:

(i) ΔAPC ∼ ΔDPB

(ii) AP.BP = CP.DP

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2907/Chapter%206_html_m6027943e.jpg

Answer:

Let us join CB.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2907/Chapter%206_html_721c58f9.jpg

(i) In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

ΔAPC ∼ ΔDPB (By AA similarity criterion)

(ii) We have already proved that

ΔAPC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2907/Chapter%206_html_c4bf31c.gif

∴ AP. PB = PC. DP

Question 8:

In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) ΔPAC ∼ ΔPDB

(ii) PA.PB = PC.PD

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2912/Chapter%206_html_m13f1603f.jpg

Answer:

(i) In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB

(ii)We know that the corresponding sides of similar triangles are proportional.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2912/Chapter%206_html_m7db4e707.gif

∴ PA.PB = PC.PD

Question 9:

In the given figure, D is a point on side BC of ΔABC such thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2920/Chapter%206_html_692a7414.gif. Prove that AD is the bisector of ∠BAC.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2920/Chapter%206_html_1e68f68d.jpg

Answer:

Let us extend BA to P such that AP = AC. Join PC.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2920/Chapter%206_html_4cc0d008.jpg

It is given that,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2920/Chapter%206_html_7d3953c9.gif

By using the converse of basic proportionality theorem, we obtain

AD || PC

⇒ ∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2920/Chapter%206_html_m2513be96.jpg

By construction, we have

AP = AC

⇒ ∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.

Question 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, ho much string does she have out (see Fig. 6.64)? 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?

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2943/Chapter%206_html_52d3c97c.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2943/Chapter%206_html_m7ef6ff29.jpg

Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

AC2 = AB2 + BC2

AB2 = (1.8 m)2 + (2.4 m)2

AB2 = (3.24 + 5.76) m2

AB2 = 9.00 m2

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2943/Chapter%206_html_7e489692.gif

Thus, the length of the string out is 3 m.

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2943/Chapter%206_html_m2006c23b.jpg

Let the fly be 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 ΔADB,

AB2 + BD2 = AD

(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 fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

Also Read : Exercise-7.1-Chapter-7-Coordinate-Geometry-class-10-ncert-solutions-Maths

SaraNextGen