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**Q.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****Sol.** (i) 7 cm, 24 cm, 25 cm

(7)^{2} + (24)^{2} = 49 + 576

= 625 = (25)^{2} = 25

∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm

(8)^{2} = 64

(3)^{2} + (6)^{2} = 9 + 36 = 45

64 ≠ 45

The square of larger side is not equal to the sum of square of other two sides.

∴ The given triangle is not a right angled.

(iii) 50 cm, 80 cm, 100 cm

(100)^{2} = 10000

(80)^{2} + (50)^{2} = 6400 + 2500 = 8900

The square of larger side is not equal to the sum of squares of other two sides.

∴ The given triangle is not a right angled.

(iv) 13 cm, 12 cm, 5 cm

(13)^{2} = 169

(12)^{2} + (5)^{2} = 144 + 25 - 169

Sides make a right angled triangle with hypotenuse 13 cm.**Q.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.**

∠P = 90°, PM

∴ ΔPMQ ~ ΔRMP

[If

⇒

[Corresponding sides of similar triangles]

AC

AB

= 2AC

AB

AB

⇒ ∠ACB = 90°

[By converse of Pythagoras theorem]

∴ ΔABC is right angled triangle

AB = BC = AC = 2a

and AD ⊥BC

In right angled triangle ADB,

AD

⇒ AD

= (2a)

To Prove: AB

Proof: In ΔAOB, AB

AB

AB

[∵ AB = BC = CD = DA]

OA

AF

In ΔBDO, OB

BD

In ΔCEO, OC

CE

Equating (i) and (ii)

AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}**Q.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.****Sol.** Let AC be the ladder of length 10 m and AB = 8 m

In ΔABC,

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

⇒ BC^{2} = AC^{2} - AB^{2}

= (10)^{2} - (8)^{2}

BC^{2} = 100 - 64 = 36

Hence distance of foot of the ladder from base of the wall is 6 m.**Q.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?****Sol. **In Δ ABC, AC^{2} = AB^{2} + BC^{2}

⇒ (24)^{2} = (18)^{2} + BC^{2}

⇒ 576 = 324 + BC^{2}

⇒ 576 - 324 = BC^{2}

BC^{2} = 252**Q.11. An aeroplane leaves an airport and flies due north a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after **** hours ?****Sol.**

Distance travelled by the aeroplane due west in**Q.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.****Sol.** Length of poles is 6 m and 11m.

DE = DC - EC = 11 m - 6 m = 5 m

In ΔDAE, AD^{2} = AE^{2} + DE^{2} [∴ AE - BC]

= (12)^{2} + (5)^{2} = 144 + 25 = 169**Q.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}.**

AB

In ΔDCE, DE

⇒ AB

= AC

= AE

On dividing both sides by 4, we get,

7AB

AB = BC = AC

AD ⊥ BC

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

Proof: In ΔADB,

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

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

⇒ 4AD^{2} = 4AB^{2} - BC^{2}

= 4AB^{2} - AB^{2} [∴BC = AB = AC]

4AD^{2} = 3AB^{2}**Q.17. Tick the correct answer and justify: In Δ ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:****(A) 120° ****(B) 60° ****(C) 90° ****(D) 45°****Sol.** In ΔABC,

AB = 6√3 cm, AC = 12 cm and BC = 6 cm

ΔABC is right angled at B.

∴ ∠B = 90°

Correct answer is (c).

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