NCERT solutions for class 10 Maths chapter 6 Triangles (Exercise 6.5)

NCERT solutions for class 10 Maths


Chapter 6


Triangles


Exercise 6.5


Ex 6.5 Question 1.


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.

(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Solution:

(i)Sides of the triangle = 7 cm, 24 cm, and 25 cm.

Squaring these sides

= 49, 576, and 625.

49 + 576 = 625

(7)2 + (24)2 = (25)2

The above equation satisfies, Pythagoras theorem. Hence, these are sides of a right angled triangle.

∴ Its lenght = 25 cm

(ii) Sides of the triangle =3 cm, 8 cm, and 6 cm.

Squaring these sides

=9, 64, and 36.

9 + 36 ≠ 64

Or, 32 + 62 ≠ 82

The sides do not satisfies the Pythagoras theorem.Hence, these are not the sides of a right angled triangle.

(iii)Sides of triangle = 50 cm, 80 cm, and 100 cm.

Squaring these sides

=2500, 6400, and 10000.

2500 + 6400 ≠ 10000

Or, 502 + 802 ≠ 1002

The sides do not satisfies the Pythagoras theorem.Hence, these are not the sides of a right angled triangle.

(iv) Sides of triangle = 13 cm, 12 cm, and 5 cm.

Squaring these sides

=169, 144, and 25.

144 +25 = 169

122 + 52 = 132
The above equation satisfies, Pythagoras theorem. Hence, these are sides of a right angled triangle.

∴ Its lenght = 13 cm


Ex 6.5 Question 2.


PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.

Solution:

To prove = PM2 = QM × MR

In ΔPQM,

Using pythagoras theorem

PQ2 = PM2 + QM2

Or, PM2 = PQ2 – QM2 ………..(i)

In ΔPMR, by Pythagoras theorem

PR2 = PM2 + MR2

Or, PM2 = PR2 – MR2 ……………..(ii)

Adding equation, (i) and (ii),

2PM2 = (PQ2 + PM2) – (QM2 + MR2)

= QR2 – QM2 – MR2        [∴ QR2 = PQ2 + PR2]

= (QM + MR)2 – QM2 – MR2

= 2QM × MR

∴ PM2 = QM × MR


Ex 6.5 Question 3.


In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD

Solution:

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB       [Each 90°]

∠ABD = ∠CBA       [Common]

∴ ΔADB ~ ΔCAB    [AA similarity]

⇒ AB/CB = BD/AB

⇒ AB2 = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° – x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD,

∠CBA = ∠CAD   [Proved above]

∠CAB = ∠CDA    [proved above]

∠ACB = ∠DCA     [Each 90°]

∴ ΔCBA ~ ΔCAD   [AAA similarity]

⇒ AC/DC = BC/AC

⇒ AC2 =  DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB    [Each 90°]

∠CDA = ∠ADB    [common]

∴ ΔDCA ~ ΔDAB   [AA similarity]

⇒ DC/DA = DA/DA

⇒ AD2 = BD × CD


Ex 6.5 Question 4.


ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .

Solution:

In ΔACB,

AC = BC   [isosceles triangle property]

∠C = 90°

Using pythagoras theorem

AB2 = AC2 + BC2

= AC2 + AC2       [Because, AC = BC]

AB2 = 2AC2


Ex 6.5 Question 5.


ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Solution:

Given, AB2 = 2AC2

AB2 = AC+ AC2

= AC2 + BC2      [Because, AC = BC]

These sides satisfies Pythagoras theorem. Hence, the triangle is a right angled triangle.


Ex 6.5 Question 6.


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

Solution:

In ΔADB and ΔADC,

AB = AC

AD = AD

∠ADB = ∠ADC [Each 90°]

∴ ΔADB ≅ ΔADC    [by RHS congruence rule]

Hence, BD = DC        [by CPCT]

In  ΔADB,

AB2 = AD+ BD2

(2a)2 = ADa

⇒ AD2 = 4a2 – a2

⇒ AD2 = 3a2

⇒ AD = √3a


Ex 6.5 Question 7.


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

Solution:

Given that- ABCD is a rhombus. Diagonals AC and BD intersect at O.

TO prove – AB+ BC+ CD2 + AD= AC+ BD2

The diagonals of a rhombus bisect each other at right angles.

∴AO = CO and BO = DO …….(i)

In ΔAOB,

∠AOB = 90°

AB2 = AO+ BO………….. (ii) [Using Pythagoras theorem]

Similarly,

AD2 = AO+ DO………….. (iii)

DC2 = DO+ CO………….. (iv)

BC2 = CO+ BO………….. (v)

Adding equations (ii) + (iii) + (iv) + (v),

AB+ AD+ DC+ BC2 = 2(AO+ BO+ DO+ CO2)

= 4AO+ 4BO2                                [from equation (i)]

= (2AO)+ (2BO)2 = AC+ BD2

AB+ AD+ DC+ BC2 = AC+ BD2


Ex 6.5 Question 8.


In Fig. 6.54, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Solution:

(i)In ΔAOF, By Pythagoras theorem

OA2 = OF2 + AF2………….. (i)

In ΔBOD, By Pythagoras theorem

OB2 = OD2 + BD2………….. (ii)

In ΔCOE,By Pythagoras theorem

OC2 = OE2 + EC2………….. (iii)

Adding  equations, (i),(ii) and (iii)

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2

⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.………….. (iv)

(ii) From equation (iv)

AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)

∴ ⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.


Ex 6.5 Question 9.


A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution:

Let

BA be the wall = 8 m

and AC be the ladder = 10 m

Using Pythagoras theorem,

AC2 = AB2 + BC2

102 = 82 + BC2

BC= 100 – 64

BC= 36

BC = 6m

Hence, the distance of the foot of the ladder from the base of the wall is 6 m


Ex 6.5 Question 10.


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?

Solution:

Let

AB be the pole = 18 m

and AC be the wire = 24 m

Udsing Pythagoras theorem,

AC2 = AB2 + BC2

242 = 182 + BC2

BC= 576 – 324

BC= 252

BC = 6√7m

Hence, the distance from the base is 6√7m.


Ex 6.5 Question 11.


An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
1½  hours?

Solution:

Speed of first aeroplane = 1000 km/hr

Distance travelled by first aeroplane (due north) in 1½ hours = 1000 x 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance travelled by first aeroplane (due west) in 1½ hours = 1200 × 3/2 km = 1800 km

Now in  ΔAOB,

Using Pythagoras Theorem,

AB2 = AO2 + OB2

⇒ AB2 = (1500)2 + (1800)2

⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, in 1½ hours the distance between two plane = 300√61 km.


Ex 6.5 Question 12.


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 tops.

Solution:

Let AB and CD be the poles of height 6m and 11 m.

Therefore, CP = 11 – 6 = 5 m and AP = 12 m

In ΔAPC,

Using Pythagoras theorem

AP2 = PC2 + AC2

(12m)2 + (5m)2 = (AC)2

AC2 = (144+25) m2 = 169 m2

AC = 13m

Therefore, the distance between top of the two pole = 13 m


Ex 6.5 Question 13.


D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Solution:

In ΔACE,

Using Pythagoras theorem

AC2 + CE2 = AE2 …………….(i)

In ΔBCD,

Using Pythagoras theorem,

BC2 + CD2 = BD2 ……………..(ii)

From equations (i) and (ii),

AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)

In ΔCDE,

Using Pythagoras theorem,

DE2 = CD2 + CE2…………..(iv)

In ΔABC,

Using Pythagoras theorem,

AB2 = AC2 + CB2…………..(v)

Putting the values  of equation (iv) & (v) in equation (iii),

DE2 + AB2 = AE2 + BD2.


Ex 6.5 Question 14.


The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.

Solution:

In Δ ADB

By Pythagoras theorem,

AB2 = AD2 + BD2 …………….(i)

In ΔADC

By Pythagoras theorem,

AC2 = AD2 + DC2 ………..(ii)

Subtracting equation (ii) from equation (i),

AB2 – AC2 = BD2 – DC

= 9CD2 – CD2        [Given, DB = 3CD]

= 8CD2

= 8(BC/4)2           [BC = DB + CD = 3CD + CD = 4CD]

∴ AB2 – AC2 = BC2/2

⇒ 2(AB2 – AC2) = BC2

⇒ 2AB2 – 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2.


Ex 6.5 Question 15.


In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.

Solution:

In ΔABC

Let the side of the equilateral triangle =  a,

and the altitude  = AE

∴ BE = EC = BC/2 = a/2

AE = a√3/2

BD = 1/3BC    [Given]

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

In ΔADE,

By Pythagoras theorem,

AD2 = AE2 + DE

AD2 =  (a√3/2)2 + (a/6)2

= (3a2/4) + (a2/36)

= 28a2/36

= 7/9AB2

⇒ 9 AD2 = 7 AB2


Ex 6.5 Question 16.


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.

Solution:

In ΔABC

Let the sides of the equilateral triangle be of length=  a,

and the altitude = AE

∴ BE = EC = BC/2 = a/2

In ΔABE,

Using Pythagoras Theorem

AB2 = AE2 + BE2

a2= AE2 + (a/2)2

AE2 = a2  – a2/4

AE2 = 3a2/4

4AE2 = 3a2

⇒ 4 × (Square of altitude) = 3 × (Square of one side)


Ex 6.5 Question 17.


Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°

(B) 60°
(C) 90° 

(D) 45°

Solution:

Given: AB = 6√3 cm, AC = 12 cm and BC = 6 cm

Squaring these sides

AB2 = 108

AC2 = 144

And, BC2 = 36

AB2 + BC2 = AC2

The above ΔABC satisfies, Pythagoras theorem. Hence, the triangle is a right angled triangle.

∴ ∠B = 90°

Hence,C is the correct answer .