NCERT Solutions for Exercise 6.5 Class 10 Maths Chapter 6 - Triangles

Q1 (2) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 3 cm, 8 cm, 6 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 3 cm, 6 cm

By Pythagoras theorem,

$h^2=3^2+6^2$

$h^2=9+36$

$h^2=45$

$h=\sqrt{45}\ne 8$

Hence, it is not the right triangle.

Q1 (3) Sides of triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. 50 cm, 80 cm, 100 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 50 cm, 80 cm

By Pythagoras theorem,

$h^2={50}^2+{80}^2$

$h^2=2500+6400$

$h^2=8900$

$h=\sqrt{8900}\ne 100$

Hence, it is not a right triangle.

Q1 (4) Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm

Answer:

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 5cm, 12 cm

By Pythagoras theorem,

$h^2=5^2+{12}^2$

$h^2=25+144$

$h^2=169$

$h=13$ = given third side.

Hence, it is a right triangle with h=13 cm.

Q2 PQR is a triangle right angled at P and M is a point on QR such that $PM\perp QR$ . Show that $P{M}^2=QM.MR.$

Answer:

Let $\mathrm{\angle }MPR$ be x

In $\mathrm{△}MPR$ ,

$\mathrm{\angle }MRP={180}^{\circ }-{90}^{\circ }-x$

$\mathrm{\angle }MRP={90}^{\circ }-x$

Similarly,

In $\mathrm{△}MPQ$ ,

$\mathrm{\angle }MPQ={90}^{\circ }-\mathrm{\angle }MPR$

$\mathrm{\angle }MPQ={90}^{\circ }-x$

$\mathrm{\angle }MQP={180}^{\circ }-{90}^{\circ }-({90}^{\circ }-x)=x$

In $\mathrm{△}QMPand\mathrm{△}PMR,$

$\mathrm{\angle }MPQ=\mathrm{\angle }MRP$

$\mathrm{\angle }PMQ=\mathrm{\angle }RMP$

$\mathrm{\angle }MQP=\mathrm{\angle }MPR$

$\mathrm{△}QMP\sim \mathrm{△}PMR,$ (By AAA)

$\frac{QM}{PM}=\frac{MP}{MR}$

$\Rightarrow P{M}^2=MQ\times MR$

Hence proved.

Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that $A{B}^2=BC.BD.$

Answer:

In $\mathrm{△}ADBand\mathrm{△}ABC,$

$\mathrm{\angle }DAB=\mathrm{\angle }ACB(Each{90}^{\circ })$

$\mathrm{\angle }ABD=\mathrm{\angle }CBA$ (common )

$\mathrm{△}ADB\sim \mathrm{△}ABC$ (By AA)

$\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}$

$\Rightarrow A{B}^2=BC.BD$ , hence prooved .

Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that $A{C}^2=BC.DC.$

Answer:

Let $\mathrm{\angle }CAB$ be x

In $\mathrm{△}ABC$ ,

$\mathrm{\angle }CBA={180}^{\circ }-{90}^{\circ }-x$

$\mathrm{\angle }CBA={90}^{\circ }-x$

Similarly,

In $\mathrm{△}CAD$ ,

$\mathrm{\angle }CAD={90}^{\circ }-\mathrm{\angle }CAB$

$\mathrm{\angle }CAD={90}^{\circ }-x$

$\mathrm{\angle }CDA={180}^{\circ }-{90}^{\circ }-({90}^{\circ }-x)=x$

In $\mathrm{△}ABCand\mathrm{△}ACD,$

$\mathrm{\angle }CBA=\mathrm{\angle }CAD$

$\mathrm{\angle }CAB=\mathrm{\angle }CDA$

$\mathrm{\angle }ACB=\mathrm{\angle }DCA$ ( Each right angle)

$\mathrm{△}ABC\sim \mathrm{△},ACD$ (By AAA)

$\frac{AC}{DC}=\frac{BC}{AC}$

$\Rightarrow A{C}^2=BC\times DC$

Hence proved

Q3 (3) In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that $A{D}^2=BD.CD.$

Answer:

In $\mathrm{△}ACDand\mathrm{△}ABD,$

$\mathrm{\angle }DCA=\mathrm{\angle }DAB(Each{90}^{\circ })$

$\mathrm{\angle }CDA=\mathrm{\angle }ADB$ (common )

$\mathrm{△}ACD\sim \mathrm{△}ABD$ (By AA)

$\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}$

$\Rightarrow A{D}^2=BD\times CD$

Hence proved.

Q4 ABC is an isosceles triangle right angled at C. Prove that $A{B}^2=2A{C}^2$

Answer:

Given: ABC is an isosceles triangle right angled at C.

Let AC=BC

In $\mathrm{△}$ ABC,

By Pythagoras theorem

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

$A{B}^2=A{C}^2+A{C}^2$ (AC=BC)

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

Hence proved.

Q5 ABC is an isosceles triangle with AC = BC. If $A{B}^2=2A{C}^2$ , prove that ABC is a right triangle.

Answer:

Given: ABC is an isosceles triangle with AC=BC.

In $\mathrm{△}$ ABC,

$A{B}^2=2.A{C}^2$ (Given )

$A{B}^2=A{C}^2+A{C}^2$ (AC=BC)

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

These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.

Hence proved.

Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:

Given: ABC is an equilateral triangle of side 2a.

AB=BC=AC=2a

AD is perpendicular to BC.

We know that the altitude of an equilateral triangle bisects the opposite side.

So, BD=CD=a

In $\mathrm{△}$ ADB,

By Pythagoras theorem,

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

$\Rightarrow (2a{)}^2=A{D}^2+a^2$

$\Rightarrow 4a^2=A{D}^2+a^2$

$\Rightarrow 4a^2-a^2=A{D}^2$

$\Rightarrow 3a^2=A{D}^2$

$\Rightarrow AD=\sqrt{3}a$

The length of each altitude is $\sqrt{3}a$ .

Q7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

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

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

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

$B{C}^2=B{O}^2+C{O}^2..................2$

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

$C{D}^2=C{O}^2+D{O}^2..................3$

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

$A{D}^2=A{O}^2+D{O}^2..................4$

Adding equation 1,2,3,4,we get

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

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2(A{O}^2+B{O}^2+C{O}^2+D{O}^2)$

$\Rightarrow A{B}^2+B{C}^2+C{D}^2+A{D}^2=2(2.A{O}^2+2.B{O}^2)$ (AO=CO and BO=DO)

$\Rightarrow A{B}^2+B{C}^2+C{D}^2+A{D}^2=4(A{O}^2+B{O}^2)$

$\Rightarrow A{B}^2+B{C}^2+C{D}^2+A{D}^2=4((\frac{AC}{2}{)}^2+(\frac{BD}{2}{)}^2)$

$\Rightarrow A{B}^2+B{C}^2+C{D}^2+A{D}^2=4((\frac{A{C}^2}{4})+(\frac{B{D}^2}{4}))$

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

Hence proved .

Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD $\perp$ BC, OE $\perp$ AC and OF $\perp$ AB. Show that $O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+C{E}^2,$

Answer:

Join AO, BO, CO

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

$O{A}^2=O{F}^2+A{F}^2..................1$

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

$O{B}^2=O{D}^2+B{D}^2..................2$

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

$O{C}^2=O{E}^2+E{C}^2..................3$

Adding equation 1,2,3,we get

$O{A}^2+O{B}^2+O{C}^2=O{F}^2+A{F}^2+O{D}^2+B{D}^2+O{E}^2+E{C}^2$ $\Rightarrow O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+E{C}^2....................4$

Hence proved

Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD $\perp$ BC, OE $\perp$ AC and OF $\perp$ AB. $A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2.$

Answer:

Join AO, BO, CO

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

$O{A}^2=O{F}^2+A{F}^2..................1$

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

$O{B}^2=O{D}^2+B{D}^2..................2$

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

$O{C}^2=O{E}^2+E{C}^2..................3$

Adding equation 1,2,3,we get

$O{A}^2+O{B}^2+O{C}^2=O{F}^2+A{F}^2+O{D}^2+B{D}^2+O{E}^2+E{C}^2$ $\Rightarrow O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2=A{F}^2+B{D}^2+E{C}^2....................4$

$\Rightarrow (O{A}^2-O{E}^2)+(O{C}^2-O{D}^2)+(O{B}^2-O{F}^2)=A{F}^2+B{D}^2+E{C}^2$ $\Rightarrow A{E}^2+C{D}^2+B{F}^2=A{F}^2+B{D}^2+E{C}^2$

Q10 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer:

OB is a pole.

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

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

$\Rightarrow {24}^2={18}^2+A{O}^2$

$\Rightarrow 576=324+A{O}^2$

$\Rightarrow 576-324=A{O}^2$

$\Rightarrow 252=A{O}^2$

$\Rightarrow AO=6\sqrt{7}m$

Hence, the distance of the stack from the base of the pole is $6\sqrt{7}$ m.

Q11 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 1/2 hours?

Answer:

Distance travelled by the first aeroplane due north in $1\frac{1}{2}$ hours.

$=1000\times \frac{3}{2}=1500km$

Distance travelled by second aeroplane due west in $1\frac{1}{2}$ hours.

$=1200\times \frac{3}{2}=1800km$

OA and OB are the distance travelled.

By Pythagoras theorem,

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

$\Rightarrow A{B}^2={1500}^2+{1800}^2$

$\Rightarrow A{B}^2=2250000+3240000$

$\Rightarrow A{B}^2=5490000$

$\Rightarrow A{B}^2=300\sqrt{61}km$

Thus, the distance between the two planes is $300\sqrt{61}km$ .

Q12 Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their top

Answer:

Let AB and CD be poles of heights 6 m and 11 m respectively.

CP=11-6=5 m and AP= 12 m

In $\mathrm{△}$ APC,

By Pythagoras theorem,

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

$\Rightarrow {12}^2+5^2=A{C}^2$

$\Rightarrow 144+25=A{C}^2$

$\Rightarrow 169=A{C}^2$

$\Rightarrow AC=13m$

Hence, the distance between the tops of two poles is 13 m.

Q13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $A{E}^2+B{D}^2=A{B}^2+D{E}^2.$

Answer:

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

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

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

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

From 1 and 2, we get

$A{C}^2+C{E}^2+B{C}^2+C{D}^2=A{E}^2+D{B}^2..................3$

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

$D{E}^2=C{D}^2+C{E}^2..................4$

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

$A{B}^2=A{C}^2+C{B}^2..................5$

From 3,4,5 we get

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

Q14 The perpendicular from A on side BC of a $\mathrm{\Delta }$ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that $2A{B}^2=2A{C}^2+B{C}^2.$

Answer:

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

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

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

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

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

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

From 1 and 2, we get

$A{C}^2-C{D}^2=A{B}^2-D{B}^2..................3$

Given : 3DC=DB, so

$CD=\frac{BC}{4}andBD=\frac{3BC}{4}........................4$

From 3 and 4, we get

$A{C}^2-(\frac{BC}{4}{)}^2=A{B}^2-(\frac{3BC}{4}{)}^2$

$A{C}^2-(\frac{B{C}^2}{16})=A{B}^2-(\frac{9B{C}^2}{16})$

$16A{C}^2-B{C}^2=16A{B}^2-9B{C}^2$

$16A{C}^2=16A{B}^2-8B{C}^2$

$\Rightarrow 2A{C}^2=2A{B}^2-B{C}^2$

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

Hence proved.

Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that $9A{D}^2=7A{B}^2$

Answer:

Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove : $9A{D}^2=7A{B}^2$

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

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

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

$a^2=A{E}^2+(\frac{a}{2}{)}^2$

$\Rightarrow a^2-(\frac{a^2}{4})=A{E}^2$

$\Rightarrow (\frac{3a^2}{4})=A{E}^2$

$\Rightarrow AE=(\frac{\sqrt{3}a}{2})$

Given : BD = 1/3 BC.

$BD=\frac{a}{3}$

$DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$

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

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

$\Rightarrow A{D}^2=(\frac{\sqrt{3}a}{2}{)}^2+(\frac{a}{6}{)}^2$

$\Rightarrow A{D}^2=(\frac{3a^2}{4})+(\frac{a^2}{36})$

$\Rightarrow A{D}^2=(\frac{7a^2}{9})$

$\Rightarrow A{D}^2=(\frac{7A{B}^2}{9})$

$\Rightarrow 9A{D}^2=7A{B}^2$

Q16 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:

Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

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

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

$a^2=A{E}^2+(\frac{a}{2}{)}^2$

$\Rightarrow a^2-(\frac{a^2}{4})=A{E}^2$

$\Rightarrow (\frac{3a^2}{4})=A{E}^2$

$\Rightarrow 3a^2=4A{E}^2$

$\Rightarrow 4.(altitude{)}^2=3.(side{)}^2$

Q17 Tick the correct answer and justify : In $\mathrm{\Delta }ABC$ AB = $6\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm.
The angle B is :
(A) 120°

(B) 60°

(C) 90°

(D) 45°

Answer:

In $\mathrm{\Delta }ABC$ AB = $6\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm.

$A{B}^2+B{C}^2=108+36$

$=144$

$={12}^2$

$=A{C}^2$

It satisfies the Pythagoras theorem.

Hence, ABC is a right-angled triangle and right-angled at B.

Option C is correct.