## 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, 3^{2} + 6^{2} ≠ 8^{2}

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, 50^{2} + 80^{2} ≠ 100^{2}

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

12^{2} + 5^{2} = 13^{2}

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 PM ^{2} = QM × MR.**

**Solution:**

To prove = PM^{2} = QM × MR

In ΔPQM,

Using pythagoras theorem

PQ^{2} = PM^{2} + QM^{2}

Or, PM^{2} = PQ^{2} – QM^{2} ………..(i)

In ΔPMR, by Pythagoras theorem

PR^{2} = PM^{2} + MR^{2}

Or, PM^{2} = PR^{2} – MR^{2} ……………..(ii)

Adding equation, (i) and (ii),

2PM^{2} = (PQ^{2} + PM^{2}) – (QM^{2} + MR^{2})

= QR^{2} – QM^{2} – MR^{2 } [∴ QR^{2} = PQ^{2} + PR^{2}]

= (QM + MR)^{2} – QM^{2} – MR^{2}

= 2QM × MR

∴ PM^{2} = QM × MR

**Ex 6.5 Question 3.**

** In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB ^{2} = BC × BD
(ii) AC^{2} = BC × DC
(iii) AD^{2} = BD × CD**

**Solution:**

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB [Each 90°]

∠ABD = ∠CBA [Common]

∴ ΔADB ~ ΔCAB [AA similarity]

⇒ AB/CB = BD/AB

⇒ AB^{2} = 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

⇒ AC^{2} = DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB [Each 90°]

∠CDA = ∠ADB [common]

∴ ΔDCA ~ ΔDAB [AA similarity]

⇒ DC/DA = DA/DA

⇒ AD^{2} = BD × CD

**Ex 6.5 Question 4.**

**ABC is an isosceles triangle right angled at C. Prove that AB ^{2} = 2AC^{2} .**

**Solution:**

In ΔACB,

AC = BC [isosceles triangle property]

∠C = 90°

Using pythagoras theorem

AB^{2} = AC^{2} + BC^{2}

= AC^{2} + AC^{2 } [Because, AC = BC]

AB^{2} = 2AC^{2}

**Ex 6.5 Question 5.**

**ABC is an isosceles triangle with AC = BC. If AB ^{2} = 2AC^{2}, prove that ABC is a right triangle.**

**Solution:**

Given, AB^{2} = 2AC^{2}

AB^{2} = AC^{2 }+ AC^{2}

= AC^{2} + BC^{2 }[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,

AB^{2} = AD^{2 }+ BD^{2}

(2*a*)^{2} = AD^{2 }+ *a*^{2 }

⇒ AD^{2 =} 4*a*^{2} – *a*^{2}

⇒ AD^{2 =} 3*a*^{2}

⇒ 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^{2 }+ BC^{2 }+ CD^{2} + AD^{2 }= AC^{2 }+ BD^{2}

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

∴AO = CO and BO = DO …….(i)

In ΔAOB,

∠AOB = 90°

AB^{2} = AO^{2 }+ BO^{2 }………….. (ii) [Using Pythagoras theorem]

Similarly,

AD^{2} = AO^{2 }+ DO^{2 }………….. (iii)

DC^{2} = DO^{2 }+ CO^{2 }………….. (iv)

BC^{2} = CO^{2 }+ BO^{2 }………….. (v)

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

AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = 2(AO^{2 }+ BO^{2 }+ DO^{2 }+ CO^{2})

= 4AO^{2 }+ 4BO^{2 }[from equation (i)]

= (2AO)^{2 }+ (2BO)^{2} = AC^{2 }+ BD^{2}

AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = AC^{2 }+ BD^{2}

**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) OA ^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2} ,
(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.**

**Solution:**

(i)In ΔAOF, By Pythagoras theorem

OA^{2} = OF^{2} + AF^{2}………….. (i)

In ΔBOD, By Pythagoras theorem

OB^{2} = OD^{2} + BD^{2}………….. (ii)

In ΔCOE,By Pythagoras theorem

OC^{2} = OE^{2} + EC^{2}………….. (iii)

Adding equations, (i),(ii) and (iii)

OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2 }+ EC^{2}

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}.………….. (iv)

(ii) From equation (iv)

AF^{2} + BD^{2} + EC^{2} = (OA^{2} – OE^{2}) + (OC^{2} – OD^{2}) + (OB^{2} – OF^{2})

∴ ⇒ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

**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,

AC^{2} =^{ }AB^{2} + BC^{2}

10^{2} = 8^{2} + BC^{2}

BC^{2 }= 100 – 64

BC^{2 }= 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,

AC^{2} =^{ }AB^{2} + BC^{2}

24^{2} = 18^{2} + BC^{2}

BC^{2 }= 576 – 324

BC^{2 }= 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,

AB^{2} =^{ }AO^{2} + OB^{2}

⇒ AB^{2} =^{ }(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

AP^{2} =^{ }PC^{2} + AC^{2}

(12m)^{2} + (5m)^{2} = (AC)^{2}

AC^{2} = (144+25) m^{2} = 169 m^{2}

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 AE ^{2} + BD^{2} = AB^{2} + DE^{2}.**

**Solution:**

In ΔACE,

Using Pythagoras theorem

AC^{2} +^{ }CE^{2} = AE^{2} …………….(i)

In ΔBCD,

Using Pythagoras theorem,

BC^{2} +^{ }CD^{2} = BD^{2} ……………..(ii)

From equations (i) and (ii),

AC^{2} +^{ }CE^{2} + BC^{2} +^{ }CD^{2} = AE^{2} + BD^{2} …………..(iii)

In ΔCDE,

Using Pythagoras theorem,

DE^{2} =^{ }CD^{2} + CE^{2}…………..(iv)

In ΔABC,

Using Pythagoras theorem,

AB^{2} =^{ }AC^{2} + CB^{2}…………..(v)

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

DE^{2} + AB^{2} = AE^{2} + BD^{2}.

**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 2AB ^{2} = 2AC^{2} + BC^{2}.**

**Solution:**

In Δ ADB

By Pythagoras theorem,

AB^{2} =^{ }AD^{2} + BD^{2} …………….(i)

In ΔADC

By Pythagoras theorem,

AC^{2} =^{ }AD^{2} + DC^{2} ………..(ii)

Subtracting equation (ii) from equation (i),

AB^{2} – AC^{2} = BD^{2} – DC^{2 }

= 9CD^{2} – CD^{2 }[Given, DB = 3CD]

= 8CD^{2}

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

∴ AB^{2} – AC^{2} = BC^{2}/2

⇒ 2(AB^{2} – AC^{2}) = BC^{2}

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

∴ 2AB^{2} = 2AC^{2} + BC^{2}.

**Ex 6.5 Question 15.**

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

**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,

AD^{2} = AE^{2} + DE^{2 }

AD^{2} = (a√3/2)^{2} + (a/6)^{2}

= (3a^{2}/4) + (a^{2}/36)

= 28a^{2}/36

= 7/9AB^{2}

⇒ 9 AD^{2} = 7 AB^{2}

**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

AB^{2} = AE^{2} + BE^{2}

a^{2}= AE^{2} + (a/2)^{2}

AE^{2} = a^{2} – a^{2}/4

AE^{2} = 3a^{2}/4

4AE^{2} = 3a^{2}

⇒ 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

AB^{2} = 108

AC^{2} = 144

And, BC^{2} = 36

AB^{2} + BC^{2} = AC^{2}

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

∴ ∠B = 90°

Hence,C is the correct answer .