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,

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

$h^2=9+36$

$h^2=45$

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

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,

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

$h^2=2500+6400$

$h^2=8900$

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

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 $PM ^2 = QM . MR .$

Answer:

Let $\angle MPR$ be x

In $\triangle MPR$ ,

$\angle MRP=180 ^\circ-90 ^\circ-x$

$\angle MRP=90 ^\circ-x$

Similarly,

In $\triangle MPQ$ ,

$\angle MPQ=90 ^\circ-\angle MPR$

$\angle MPQ=90 ^\circ-x$

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

In $\triangle QMP\, and\, \triangle PMR,$

$\angle MPQ\, =\angle MRP$

$\angle PMQ\, =\angle RMP$

$\angle MQP\, =\angle MPR$

$\triangle QMP\, \sim \triangle PMR,$ (By AAA)

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

$\Rightarrow PM^2=MQ\times MR$

Hence proved.

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

Answer:

In $\triangle ADB\, and\, \triangle ABC,$

$\angle DAB\, =\angle ACB \, \, \, \, \, \, \, \, (Each 90 ^\circ)$

$\angle ABD\, =\angle CBA$ (common )

$\triangle ADB\, \sim \triangle ABC$ (By AA)

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

$\Rightarrow AB^2=BC.BD$ , hence prooved .

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

Answer:

Let $\angle CAB$ be x

In $\triangle ABC$ ,

$\angle CBA=180 ^\circ-90 ^\circ-x$

$\angle CBA=90 ^\circ-x$

Similarly,

In $\triangle CAD$ ,

$\angle CAD=90 ^\circ-\angle CAB$

$\angle CAD=90 ^\circ-x$

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

In $\triangle ABC\, and\, \triangle ACD,$

$\angle CBA\, =\angle CAD$

$\angle CAB\, =\angle CDA$

$\angle ACB\, =\angle DCA$ ( Each right angle)

$\triangle ABC\, \sim \triangle ,ACD$ (By AAA)

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

$\Rightarrow AC^2=BC\times DC$

Hence proved

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

Answer:

In $\triangle ACD\, and\, \triangle ABD,$

$\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 ^\circ)$

$\angle CDA\, =\angle ADB$ (common )

$\triangle ACD\, \sim \triangle ABD$ (By AA)

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

$\Rightarrow AD^2=BD\times CD$

Hence proved.

Q4 ABC is an isosceles triangle right angled at C. Prove that $AB^2 = 2AC ^2$

Answer:

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

Let AC=BC

In $\triangle$ ABC,

By Pythagoras theorem

$AB^2=AC^2+BC^2$

$AB^2=AC^2+AC^2$ (AC=BC)

$AB^2=2.AC^2$

Hence proved.

Q5 ABC is an isosceles triangle with AC = BC. If $AB ^ 2 = 2 AC ^ 2$ , prove that ABC is a right triangle.

Answer:

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

In $\triangle$ ABC,

$AB^2=2.AC^2$ (Given )

$AB^2=AC^2+AC^2$ (AC=BC)

$AB^2=AC^2+BC^2$

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 $\triangle$ ADB,

By Pythagoras theorem,

$AB^2=AD^2+BD^2$

$\Rightarrow (2a)^2=AD^2+a^2$

$\Rightarrow 4a^2=AD^2+a^2$

$\Rightarrow 4a^2-a^2=AD^2$

$\Rightarrow 3a^2=AD^2$

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

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 $\triangle$ AOB, by Pythagoras theorem,

$AB^2=AO^2+BO^2..................1$

In $\triangle$ BOC, by Pythagoras theorem,

$BC^2=BO^2+CO^2..................2$

In $\triangle$ COD, by Pythagoras theorem,

$CD^2=CO^2+DO^2..................3$

In $\triangle$ AOD, by Pythagoras theorem,

$AD^2=AO^2+DO^2..................4$

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

$AB^2+BC^2+CD^2+AD^2=AO^2+BO^2+BO^2+CO^2+CO^2+DO^2+AO^2+DO^2$

$AB^2+BC^2+CD^2+AD^2=2(AO^2+BO^2+CO^2+DO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=2(2.AO^2+2.BO^2)$ (AO=CO and BO=DO)

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4(AO^2+BO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC}{2})^2+(\frac{BD}{2})^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC^2}{4})+(\frac{BD^2}{4}))$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=AC^2+BD^2$

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 $OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2,$

Answer:

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

Adding equation 1,2,3,we get

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$ $\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

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. $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.$

Answer:

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

Adding equation 1,2,3,we get

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$ $\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

$\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2$ $\Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2$

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 $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 24^2=18^2+AO^2$

$\Rightarrow 576=324+AO^2$

$\Rightarrow 576-324=AO^2$

$\Rightarrow 252=AO^2$

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

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.

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

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

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

OA and OB are the distance travelled.

By Pythagoras theorem,

$AB^2=OA^2+OB^2$

$\Rightarrow AB^2=1500^2+1800^2$

$\Rightarrow AB^2=2250000+3240000$

$\Rightarrow AB^2=5490000$

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

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 $\triangle$ APC,

By Pythagoras theorem,

$AP^2+PC^2=AC^2$

$\Rightarrow 12^2+5^2=AC^2$

$\Rightarrow 144+25=AC^2$

$\Rightarrow 169=AC^2$

$\Rightarrow AC=13m$

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 $AE^2 + BD^2 = AB^2 + DE^2.$

Answer:

In $\triangle$ ACE, by Pythagoras theorem,

$AE^2=AC^2+CE^2..................1$

In $\triangle$ BCD, by Pythagoras theorem,

$DB^2=BC^2+CD^2..................2$

From 1 and 2, we get

$AC^2+CE^2+BC^2+CD^2=AE^2+DB^2..................3$

In $\triangle$ CDE, by Pythagoras theorem,

$DE^2=CD^2+CE^2..................4$

In $\triangle$ ABC, by Pythagoras theorem,

$AB^2=AC^2+CB^2..................5$

From 3,4,5 we get

$DE^2+AB^2=AE^2+DB^2$

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

Answer:

In $\triangle$ ACD, by Pythagoras theorem,

$AC^2=AD^2+DC^2$

$AC^2-DC^2=AD^2..................1$

In $\triangle$ ABD, by Pythagoras theorem,

$AB^2=AD^2+BD^2$

$AB^2-BD^2=AD^2.................2$

From 1 and 2, we get

$AC^2-CD^2=AB^2-DB^2..................3$

Given : 3DC=DB, so

$CD=\frac{BC}{4}\, \, and\, \, BD=\frac{3BC}{4}........................4$

From 3 and 4, we get

$AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2$

$AC^2-(\frac{BC^2}{16})=AB^2-(\frac{9BC^2}{16})$

$16AC^2-BC^2=16AB^2- 9BC^2$

$16AC^2=16AB^2- 8BC^2$

$\Rightarrow 2AC^2=2AB^2- BC^2$

$2 AB^2 = 2 AC^2 + BC^2.$

Hence proved.

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

Answer:

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

To prove : $9 AD^2 = 7 AB^2$

Let AB=BC=CA=a

Draw an altitude AE on BC.

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

In $\triangle$ AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

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

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

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

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

Given : BD = 1/3 BC.

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

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

In $\triangle$ ADE, by Pythagoras theorem,

$AD^2=AE^2+DE^2$

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

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

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

$\Rightarrow AD^2=(\frac{7AB^2}{9})$

$\Rightarrow 9AD^2=7AB^2$

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 $\triangle$ AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

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

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

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

$\Rightarrow 3a^2=4AE^2$

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

Q17 Tick the correct answer and justify : In $\Delta ABC$ 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 $\Delta ABC$ AB = $6 \sqrt 3$ cm, AC = 12 cm and BC = 6 cm.

$AB^2+BC^2=108+36$

$=144$

$=12^2$

$=AC^2$

It satisfies the Pythagoras theorem.

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

Option C is correct.