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# 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 .

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