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Triangles

Long Answer Type

221. In the given Fig, 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 + CE^2,(ii) AF² + BD² + CE² = AE² + CD² + BF².

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Short Answer Type

222.

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.

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223. 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?

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Long Answer Type

224. An aeroplane leaves an airport and flies due north at 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

1 1 half hours ?

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225.

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.

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Short Answer Type

226. 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.

Given : A right triangle ABC, right angled at C. D and E are points on sides AC and BC respectively.
To Prove : AE² + BD² = AB² + DE²Const: Join AE, BD and DE.
Proof: In ∆ACE
AE² = AC² + CE² ...(i)
[Using Pythagoras theorem]
In ∆BCD, BD² = CD² + BC² ...(ii)
[Using Pythagoras theorem]
Adding (i) and (ii), we get
AE² + BD² = (AC² + BC²) + (CE² + CD²⁾⇒ AE² + BD² = AB² + DE² Hence Proved.