### Chapter 6 Triangles – Exercise 6.5 | **Set** 1

### 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 hours?

**Solution:**

Distance covered by place left towards north = 1000 * 1.5

Distance covered by place left towards west = 1200 * 1.5 = 1800km

In right ∆ABC by Pythagoras theorem

(AC)

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

^{2 }= (1500)^{2 }+ (1800)^{2}= 250000 + 3240000

= 5490000

= √(3 * 3 * 61 * 10 * 10 * 10 * 10)

= 3 * 10 * 10√61

AC = 300√61

Distance between the two poles 300√61km

### 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:**

AB is pole of height = 11m

DC is another pole of height = 6m

BC = 12m

In fig. DE = BC = 12m

AE = AB – EB

= 11 – 6

= 5m

In right ∆AED, by Pythagoras theorem

(AD)

^{2 }= (AC)^{2 }+ (DE)^{2}(AD)

^{2 }= (5)^{2 }+ (12)^{2}(AD)

^{2 }= 25 + 144AD = √169

AD = √(13 * 13)

AD = 13

Hence, the distance between the tops of the two poles is 13m.

### 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:**

To prove:(AE)^{2 }+ (BD)^{2 }= (AB)^{2 }+ (DE)^{2}Construction: Join AE, BD and DE

Proof:In ∆ACE, by Pythagoras theorem

(AE)

^{2 }= (AC)^{2 }+ (EC)^{2}-(1)In ∆DCB, by Pythagoras theorem

(BD)

^{2 }= (DC)^{2 }+ (BC)^{2}-(2)In ∆ACB, by Pythagoras theorem

(AB)

^{2 }= (AC)^{2 }+ (CB)^{2}-(3)In ∆DCE, by Pythagoras theorem

(ED)

^{2 }= (DC)^{2 }+ (CE)^{2}Adding eq (1) and (2)

(AE)

^{2 }+ (BD)^{2 }= (AC)^{2 }+ (EC)^{2 }+ (DC)^{2 }+ (DC)^{2}= (AC

^{2 }+ BC^{2}) + (EC^{2 }+ DC^{2})= AB

^{2 }+ DE^{2}-(from 3 and 4)

### Question 14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig.). Prove that 2AB^{2 }= 2AC^{2} + BC^{2}.

**Solution:**

Given:DB = 3CD

To prove:2AB^{2 }= 2AC^{2 }+ BC^{2}

Proof:DB = 3CDBC = CD + DB

BC = CD + 3CD

BC = 4CD

BC/4 = CD -(1)

DB = 3(BC/4) -(2)

In right ∆ADB, by Pythagoras theorem

AB

^{2 }= AD^{2 }+ DB^{2}-(3)In right ∆ADC, by Pythagoras theorem

AC

^{2 }= AD^{2 }+ CD^{2}-(4)Subtract eq(3) from (4)

AB

^{2 }– AC^{2 }= AD^{2 }+ DB^{2 }– (AD^{2 }+ CD^{2})= AD

^{2 }+ DB^{2 }+ AD^{2 }– CD^{2}= DB

^{2 }– CD^{2}= (3/4BC)

^{2 }– (BC/4)^{2}= 9/6BC

^{2 }– BC/16= 8BC

^{2}/16= AB

^{2 }– AC^{2}= BC^{2}/22AB

^{2 }– 2AC^{2 }= BC^{2}2AB

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

### 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:**

Prove that 9AD

^{2}= 7AB^{2}Construction: Join AD and draw AE perpendicular BC

Let each side, AB = AC = BC = a

BD = 1/3BC = 1/3a

BC = 1/2BC = 1/2a

DE = BE – BD

= 1/2a – 1/3a

= 3a – 2a/6

= DE = a/6

In right ∆AED, by Pythagoras theorem

AD

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

^{2 }= (√3a/2)^{2 }+ (a/6)^{2}= 3a

^{2}/4+ a^{2}/36= 27a

^{2}/36 + a^{2}/36AD

^{2 }= 28a^{2}/36AD

^{2 }= 7a^{2}/99AD

^{2}= 7a^{2}9AD

^{2 }= 7AB^{2}

### 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:**

To prove:3AB^{2 }= 4AD^{2}

Proof:Let each side of one equilateral ∆ = aBD = 1/2a -[perpendicular bisects the side in on an equilateral ∆]

In right ∆ADB, by Pythagoras theorem

(AB)

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

^{2}= (AD)^{2}+ (1/2a)^{2}a

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

^{2 }– a^{2}/4 = AD^{2}3a

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

^{2 }= 4AD^{2}3AB

^{2}= 4AD^{2}

### 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:**

(AC)

^{2}= (12)^{2}= 144(AB)

^{2}= (6√3)^{2}= 6 * 6√3 = 36 * 3 = 108(BC)

^{2 }= (6)^{2 }= 36(AB)

^{2 }+ (BC)^{2 }= 108 + 36 = 144∴ It is right ∆ thus ∠B = 90°