NCERT Solutions for Exercise 6.6 Class 10 Maths Chapter 6 - Triangles

Triangles Class 10 Chapter 6 Exercise: 6.6

Q1 In Fig. 6.56, PS is the bisector of $\mathrm{\angle }QPRof\mathrm{\Delta }PQR$ . Prove that $\frac{QS}{SR}=\frac{PQ}{PR}$

Answer:

A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of $\mathrm{\angle }QPRof\mathrm{\Delta }PQR$ .

$\mathrm{\angle }QPS=\mathrm{\angle }SPR.....................................1$

By construction,

$\mathrm{\angle }SPR=\mathrm{\angle }PRT.....................................2$ (as PS||TR)

$\mathrm{\angle }QPS=\mathrm{\angle }QTR.....................................3$ (as PS||TR)

From the above equations, we get

$\mathrm{\angle }PRT=\mathrm{\angle }QTR$

$\therefore PT=PR$

By construction, PS||TR

In $\mathrm{△}$ QTR, by Thales theorem,

$\frac{QS}{SR}=\frac{QP}{PT}$

$\frac{QS}{SR}=\frac{PQ}{PR}$

Hence proved.

Q2(1) In Fig. 6.57, D is a point on hypotenuse AC of triangle ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB. Prove that : $D{M}^2=DN.MC$

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

$\mathrm{\angle }CDB={90}^{\circ }$

$\Rightarrow \mathrm{\angle }2+\mathrm{\angle }3={90}^{\circ }.............................1$

In $\mathrm{△}$ CDM, $\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }DMC={180}^{\circ }$

$\mathrm{\angle }1+\mathrm{\angle }2={90}^{\circ }.......................2$

In $\mathrm{△}$ DMB, $\mathrm{\angle }3+\mathrm{\angle }4+\mathrm{\angle }DMB={180}^{\circ }$

$\mathrm{\angle }3+\mathrm{\angle }4={90}^{\circ }.......................3$

From equation 1 and 2, we get $\mathrm{\angle }1=\mathrm{\angle }3$

From equation 1 and 3, we get $\mathrm{\angle }2=\mathrm{\angle }4$

In $\mathrm{△}DCMand\mathrm{△}BDM,$

$\mathrm{\angle }1=\mathrm{\angle }3$

$\mathrm{\angle }2=\mathrm{\angle }4$

$\mathrm{△}DCM\sim \mathrm{△}BDM,$ (By AA)

$\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}$

$\Rightarrow \frac{DN}{DM}=\frac{DM}{MC}$ (BM=DN)

$\Rightarrow$ $D{M}^2=DN.MC$

Hence proved

Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB. Prove that: $D{N}^2=DM.AN$

Answer:

In $\mathrm{△}$ DBN,

$\mathrm{\angle }5+\mathrm{\angle }7={90}^{\circ }.......................1$

In $\mathrm{△}$ DAN,

$\mathrm{\angle }6+\mathrm{\angle }8={90}^{\circ }.......................2$

BD $\perp$ AC, $\therefore \mathrm{\angle }ADB={90}^{\circ }$

$\mathrm{\angle }5+\mathrm{\angle }6={90}^{\circ }.......................3$

From equation 1 and 3, we get $\mathrm{\angle }6=\mathrm{\angle }7$

From equation 2 and 3, we get $\mathrm{\angle }5=\mathrm{\angle }8$

In $\mathrm{△}DNAand\mathrm{△}BND,$

$\mathrm{\angle }6=\mathrm{\angle }7$

$\mathrm{\angle }5=\mathrm{\angle }8$

$\mathrm{△}DNA\sim \mathrm{△}BND$ (By AA)

$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$

$\Rightarrow \frac{AN}{DN}=\frac{DN}{DM}$ (NB=DM)

$\Rightarrow$ $D{N}^2=AN.DM$

Hence proved.

Q3 In Fig. 6.58, ABC is a triangle in which $\mathrm{\angle }$ ABC > 90° and AD $\perp$ CB produced. Prove that $A{C}^2=A{B}^2+B{C}^2+2BC.BD.$

Answer:

In $\mathrm{△}$ ADB, by Pythagoras theorem

$A{B}^2=A{D}^2+D{B}^2.......................1$

In $\mathrm{△}$ ACD, by Pythagoras theorem

$A{C}^2=A{D}^2+D{C}^2.......................2$

$A{C}^2=A{D}^2+(BD+BC{)}^2$

$\Rightarrow A{C}^2=A{D}^2+(BD{)}^2+(BC{)}^2+2.BD.BC$

$A{C}^2=A{B}^2+B{C}^2+2BC.BD.$ (From 1)

Q4 In Fig. 6.59, ABC is a triangle in which $\mathrm{\angle }$ ABC < 90° and AD $\perp$ BC. Prove that $A{C}^2=A{B}^2+B{C}^2-2BC.BD.$

Answer:

In $\mathrm{△}$ ADB, by Pythagoras theorem

$A{B}^2=A{D}^2+D{B}^2$

$A{D}^2=A{B}^2-D{B}^2...........................1$

In $\mathrm{△}$ ACD, by Pythagoras theorem

$A{C}^2=A{D}^2+D{C}^2$

$A{C}^2=A{B}^2-B{D}^2+D{C}^2$ (From 1)

$\Rightarrow A{C}^2=A{B}^2-B{D}^2+(BC-BD{)}^2$

$\Rightarrow A{C}^2=A{B}^2-B{D}^2+(BC{)}^2+(BD{)}^2-2.BD.BC$

$A{C}^2=A{B}^2+B{C}^2-2BC.BD.$

Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that : $A{C}^2=A{D}^2+BCDM+{\left(\frac{BC}{2}\right)}^2$

Answer:

Given: AD is a median of a triangle ABC and AM $\perp$ BC.

In $\mathrm{△}$ AMD, by Pythagoras theorem

$A{D}^2=A{M}^2+M{D}^2.......................1$

In $\mathrm{△}$ AMC, by Pythagoras theorem

$A{C}^2=A{M}^2+M{C}^2$

$A{C}^2=A{M}^2+(MD+DC{)}^2$

$\Rightarrow A{C}^2=A{M}^2+(MD{)}^2+(DC{)}^2+2.MD.DC$

$A{C}^2=A{D}^2+D{C}^2+2DC.MD.$ (From 1)

$A{C}^2=A{D}^2+(\frac{BC}{2}{)}^2+2(\frac{BC}{2}).MD.$ (BC=2 DC)

$A{C}^2=A{D}^2+BCDM+{\left(\frac{BC}{2}\right)}^2$

Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that : $A{B}^2=A{D}^2-BC.DM+{\left(\frac{BC}{2}\right)}^2$

Answer:

In $\mathrm{△}$ ABM, by Pythagoras theorem

$A{B}^2=A{M}^2+M{B}^2$

$A{B}^2=(A{D}^2-D{M}^2)+M{B}^2$

$\Rightarrow A{B}^2=(A{D}^2-D{M}^2)+(BD-MD{)}^2$

$\Rightarrow A{B}^2=A{D}^2-D{M}^2+(BD{)}^2+(MD{)}^2-2.BD.MD$

$\Rightarrow A{B}^2=A{D}^2+(BD{)}^2-2.BD.MD$

$\Rightarrow A{B}^2=A{D}^2+(\frac{BC}{2}{)}^2-2(\frac{BC}{2}).MD.=A{C}^2$ (BC=2 BD)

$\Rightarrow A{D}^2+(\frac{BC}{2}{)}^2-BC.MD.=A{C}^2$

Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM $\perp$ BC. Prove that: $A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$

Answer:

In $\mathrm{△}$ ABM, by Pythagoras theorem

$A{B}^2=A{M}^2+M{B}^2.......................1$

In $\mathrm{△}$ AMC, by Pythagoras theorem

$A{C}^2=A{M}^2+M{C}^2$ ..................................2

Adding equation 1 and 2,

$A{B}^2+A{C}^2=2A{M}^2+M{B}^2+M{C}^2$

$\Rightarrow A{B}^2+A{C}^2=2A{M}^2+(BD-DM{)}^2+(MD+DC{)}^2$

$\Rightarrow A{B}^2+A{C}^2=2A{M}^2+(BD{)}^2+(DM{)}^2-2.BD.DM+(MD{)}^2+(DC{)}^2+2.MD.DC$

$\Rightarrow A{B}^2+A{C}^2=2A{M}^2+2.(DM{)}^2+B{D}^2+(DC{)}^2+2.MD.(DC-BD)$ $\Rightarrow A{B}^2+A{C}^2=2(A{M}^2+(DM{)}^2)+(\frac{BC}{2}{)}^2+(\frac{BC}{2}{)}^2+2.MD.(\frac{BC}{2}-\frac{BC}{2})$ $\Rightarrow A{B}^2+A{C}^2=2(A{M}^2+(DM{)}^2)+(\frac{BC}{2}{)}^2+(\frac{BC}{2}{)}^2$

$A{C}^2+A{B}^2=2A{D}^2+\frac{1}{2}B{C}^2$

Q6 Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In $\mathrm{△}$ DEA, by Pythagoras theorem

$D{A}^2=D{E}^2+E{A}^2.......................1$

In $\mathrm{△}$ DEB, by Pythagoras theorem

$D{B}^2=D{E}^2+E{B}^2$

$D{B}^2=D{E}^2+(EA+AB{)}^2$

$D{B}^2=D{E}^2+(EA{)}^2+(AB{)}^2+2.EA.AB$

$D{B}^2=D{A}^2+(AB{)}^2+2.EA.AB$ ....................................2

In $\mathrm{△}$ ADF, by Pythagoras theorem

$D{A}^2=A{F}^2+F{D}^2$

In $\mathrm{△}$ AFC, by Pythagoras theorem

$A{C}^2=A{F}^2+F{C}^2=A{F}^2+(DC-FD{)}^2$

$\Rightarrow A{C}^2=A{F}^2+(DC{)}^2+(FD{)}^2-2.DC.FD$

$\Rightarrow A{C}^2=(A{F}^2+F{D}^2)+(DC{)}^2-2.DC.FD$

$\Rightarrow A{C}^2=A{D}^2+(DC{)}^2-2.DC.FD.......................3$

Since ABCD is a parallelogram.

SO, AB=CD and BC=AD

In $\mathrm{△}DEAand\mathrm{△}ADF,$

$\mathrm{\angle }DEA=\mathrm{\angle }AFD(each{90}^{\circ })$

$\mathrm{\angle }DAE=\mathrm{\angle }ADF$ (AE||DF)

AD=AD (common)

$\mathrm{△}DEA\cong \mathrm{△}ADF,$ (ASA rule)

$\Rightarrow EA=DF.......................6$

Adding 2 and, we get

$D{A}^2+A{B}^2+2.EA.AB+A{D}^2+D{C}^2-2.DC.FD=D{B}^2+A{C}^2$ $\Rightarrow D{A}^2+A{B}^2+A{D}^2+D{C}^2+2.EA.AB-2.DC.FD=D{B}^2+A{C}^2$

$\Rightarrow B{C}^2+A{B}^2+A{D}^2+2.EA.AB-2.AB.EA=D{B}^2+A{C}^2$ (From 4 and 6)

$\Rightarrow B{C}^2+A{B}^2+C{D}^2=D{B}^2+A{C}^2$ \

Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that : $\mathrm{\Delta }APC\sim \mathrm{\Delta }DPB$

Answer:

Join BC

In $\mathrm{△}APCand\mathrm{△}DPB,$

$\mathrm{\angle }APC=\mathrm{\angle }DPB$ ( vertically opposite angle)

$\mathrm{\angle }CAP=\mathrm{\angle }BDP$ (Angles in the same segment)

$\mathrm{△}APC\sim \mathrm{△}DPB$ (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 $\mathrm{△}APCand\mathrm{△}DPB,$

$\mathrm{\angle }APC=\mathrm{\angle }DPB$ ( vertically opposite angle)

$\mathrm{\angle }CAP=\mathrm{\angle }BDP$ (Angles in the same segment)

$\mathrm{△}APC\sim \mathrm{△}DPB$ (By AA)

$\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

Q8 (1) 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 $\mathrm{\Delta }PAC\sim \mathrm{\Delta }PDB$

Answer:

In $\mathrm{\Delta }PACand\mathrm{\Delta }PDB,$

$\mathrm{\angle }P=\mathrm{\angle }P$ (Common)

$\mathrm{\angle }PAC=\mathrm{\angle }PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, $\mathrm{\Delta }PAC\sim \mathrm{\Delta }PDB$ ( By AA rule)

Q8 (2) 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 PA. PB = PC. PD

Answer:

In $\mathrm{\Delta }PACand\mathrm{\Delta }PDB,$

$\mathrm{\angle }P=\mathrm{\angle }P$ (Common)

$\mathrm{\angle }PAC=\mathrm{\angle }PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So, $\mathrm{\Delta }PAC\sim \mathrm{\Delta }PDB$ ( By AA rule)

24440 $\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

Q9 In Fig. 6.63, D is a point on side BC of D ABC such that $\frac{BD}{CD}=\frac{AB}{AC}$ Prove that AD is the bisector of $\mathrm{\angle }$ BAC.

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 $\Rightarrow \mathrm{\angle }BAD=\mathrm{\angle }APC............1$ (Corresponding angles)

$\Rightarrow \mathrm{\angle }DAC=\mathrm{\angle }ACP............2$ (Alternate angles)

By construction,

AP=AC

$\Rightarrow \mathrm{\angle }APC=\mathrm{\angle }ACP............3$

From equation 1,2,3, we get

$\Rightarrow \mathrm{\angle }BAD=\mathrm{\angle }APC$

Thus, AD bisects angle BAC.

Q10 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 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?

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 $\mathrm{△}$ ABC, using Pythagoras theorem

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow A{C}^2=(1.8{)}^2+(2.4{)}^2$

$\Rightarrow A{C}^2=3.24+5.76$

$\Rightarrow A{C}^2=9.00$

$\Rightarrow AC=3m$

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 $\mathrm{△}$ ADB,

$A{B}^2+B{D}^2=A{D}^2$

$\Rightarrow (1.8{)}^2+B{D}^2=(2.4{)}^2$

$\Rightarrow B{D}^2=5.76-3.24=2.52m^2$

$\Rightarrow BD=1.587m$

Horizontal distance travelled by fly = BD+1.2 m

=1.587+1.2=2.787 m

= 2.79 m