# Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.5 | Set 1

### Question 1. 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.

### (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) Given:7cm, 24cm, 25cm(25)

^{2}= 25 * 25 = 625∴The are sides of a right ∆ and hypotenuse = 625

(ii) Given:3cm, 8cm, 6cm(8)

^{2}= 8 * 8 = 64They are not side of right triangle.

(iii) Given:50cm, 80cm, 100cm(100)

^{2 }= 100 * 100 = 10000(50)

^{2}_{ }+ (80)^{2 }= 2500 + 6400Therefore, this is not right triangle.

(iv) Given:13cm, 12cm, 5cm(13)

^{2 }= 13 * 13 = 169(12)

^{2 }+ (5)^{2 }= 169Therefore, this is the sides of right triangle and hypotenuse = 169

### 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:**

Given:A right ∆PQR right-angled at P and PM⊥QR

To show:PM^{2}= QM * MRPM * PM = QM * MR

In ∆PMQ and ∆PMR

∠1 =∠2 -(each 90°)

∠1 +∠2 +∠3 = 180°

90° + ∠4 + ∠5 = 180°

∠4 +∠5 = 180° – 90°

∠3 + ∠4 = 180° – 90°

∠4 +∠5 = ∠3 +∠4

∴∠5 =∠3

∴∆PMQ ~ ∆RMP

MP

^{2 }= MQ.PM

### Question 3. In Fig., 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:**

Given:A right ∆ABD, right angled at A and AC⊥ BD

To show:(i)AB

^{2}= BC × BD(ii) AC

^{2}= BC × DC(iii) AD

^{2}= BD × CD

(i)AB^{2 }= BC × BDAB * AB = BC.BD

In ∆ ABC and ∆ABD

∠B =∠B -(common)

∠BCA = ∠A -(each 90°)

∴ ∆ABC~∆DBA

AB

^{2 }= BC * DC

(ii)AC^{2 }= BC * DCAC * AC = BC * DC

In ∆ABC and ∆ACD

∠ACB = ∠ACD -(each 90°)

In ∆ ACB

∠1 + ∠2 + ∠3 = 180°

∠1 + ∠2 + 90° = 180°

∠1 + ∠2 = 90° -(1)

∠1 + ∠3 = 90° -(2)

∠1 + ∠2 =∠1 + ∠3

∴ ∠2 = ∠3

∴ ∆ABC~∆DAC

AC

^{2 }= BC * CD

(iii)AD^{2}= BD × CDIn ∆ABD and ∆ACD

∠A =∠ACD -(each 90)

∠D =∠D -(common)

∴ ∆ABD~∆CAD -(AA similarity)

AD

^{2 }= BD * CD

### Question 4. ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

**Solution:**

Given:∆ABC right angled at B

To prove:AB^{2 }= 2AC^{2}According to Pythagoras theorem,

(AB)

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

^{2 }= (AC)^{2 }+ (AC)^{2}-(BC = AC)AB

^{2 }= 2AC^{2}

### Question 5. ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is a right triangle.

**Solution:**

Given:∆ABC is an isosceles triangle in whichAC = BC

AB

^{2 }= 2AC^{2}

To prove:ABCD is right ∆AB

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

^{2 }= AC^{2}+ AC^{2}∴ By the, converse of Pythagoras theorem ∆ACB is a right ∆ at C.

### Question 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

**Solution:**

Let ABC is an equilateral triangle of each 2a units

Construction: Draw AD ⊥ BC

In right ∆ ADB

(AB)

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

^{2 }= (AD)^{2 }+ (a)^{2}(4a)

^{2 }= (AD)^{2 }+ a^{2}4a

^{2 }– a^{2 }= (AD)^{2}3a

^{2 }= (AB)^{2}√3 a

^{2 }= AD ∴ Each of its altitude= √3 a unit√3 a = AD

### Question 7. Prove that the sum of the squares of the sides of the rhombus is equal to the sum of the squares of its diagonals.

**Solution:**

Given:ABCD is a rhombus

To prove:AB^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2}

Proof:ABCD is rhombus and let diagonals AC and BD bisect each other at O.∴ ∠AOB =∠BOC =∠COD =∠DOA = 90°

In ∆AOB

AB

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

^{2 }= (1/2AC^{2}) + (1/2BD^{2})AB

^{2 }= 1/4(AC^{2 }+ BD^{2})4AB

^{2 }= AC^{2 }+ BD^{2}-(1)Similarly, in ∆BOC

4BC

^{2 }= AC^{2 }+ BD^{2}-(2)4CD

^{2 }= AC^{2 }+ BD^{2}-(3)4AD

^{2 }= AC^{2 }+ BD^{2}-(4)Adding 1, 2, 3, and 4

4AB

^{2 }+ 4BC^{2 }+ 4CD^{2 }+ 4DA^{2 }= 4(AC^{2 }+ BD^{2})4(AB

^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2}) = 4(AC^{2 }+ BD^{2})∴ AB

^{2 }+ BC^{2 }+ CD^{2 }+ DA^{2 }= AC^{2 }+ BD^{2 }

### Question 8. In Fig. 6.54, O is a point in the interior of a triangle diagonals.

### (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 right ∆AFO, by Pythagoras theoremOA

^{2 }= AP^{2 }+ OF^{2}-(1)^{ }In right ∆ODB, by Pythagoras theorem

OB

^{2 }= BD^{2 }+ OD^{2}-(2)^{ }In right ∆OEC, by Pythagoras theorem

OC

^{2 }= CE^{2 }+ OE^{2}-(3)Adding 1, 2, and 3

OA

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

^{2}+ OB^{2 }+ OC^{2 }– OE^{2 }– OD^{2 }– OE^{2}= AF^{2 }+ BD^{2 }+ CE^{2}

(ii)AF^{2 }+ BD^{2 }+ CE^{2 }= AE^{2 }+ CD^{2 }+ BF^{2}OB

^{2 }= BD^{2 }+ OD^{2}-(1)^{ }OC

^{2 }= DC^{2 }+ OD^{2}-(2)^{ }Subtracting (1) from (2) we get,

OB

^{2 }– OC^{2 }= BD^{2 }+ OD^{2 }– (DC^{2 }+ OD^{2})= BD

^{2 }+ OD^{2 }– DC^{2 }– OD^{2}= BD

^{2 }– DC^{2}-(3)^{ }Similarly,

OC

^{2 }– AD^{2 }= CE^{2 }– EA^{2}-(4)^{ }AO

^{2 }= BO^{2 }= AF^{2 }– FB^{2}-(5)^{ }Adding 3, 4, and 5, we get,

OB

^{2 }– OC^{2 }+ OC^{2 }– AO^{2 }+ AO^{2 }– BO^{2 }= BD^{2 }– DC^{2 }+ CE^{2 }– EA^{2 }+ AF^{2 }– FB^{2}0 + DC

^{2 }+ EA^{2 }+ FB^{2 }= BD^{2 }+ CD^{2 }+ AF^{2}

### 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:**

AB is wall = 8m

AC is ladder = 10m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)

^{2 }= (AB)^{2 }+ (BC)^{2}(10)

^{2 }= (8)^{2 }+ (BC)^{2}100 = 64 + (BC)

^{2}100 – 64 = (BC)

^{2}36 = (BC)

^{2}√36 = BC

√(6 * 6) = BC

6 = BC

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

### 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:**

In fig.

AB is pole = 18m

AC is wire = 24m

BC = ?

In right ∆ABC, by Pythagoras theorem

(AC)

^{2 }= (AB)^{2 }+ (BC)^{2}(24)

^{2 }= (18)^{2 }+ (BC)^{2}576 = 324 + (BC)

^{2}576 – 324 = (BC)

^{2}252 = (BC)

^{2}√252 = BC

√(2 * 2 * 3 * 3 * 7) = BC

2 * 3√7 = BC

6√7 = BC

∴ Hence, the stake may be placed at distance of 6√7 from the base of the pole.