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# Geometry problem involving triangles

Woody Woodpecker pecked at a 17 meter wooden pole until it cracked and the upper part fell, with the top hitting the ground 10 meters from the foot of the pole. Since the upper part had not completely broken off, Wood pecked where the pole had cracked. How far was Woody above the ground?

Assuming the pole was perpendicular to the ground.  When the upper part fell it form a right triangle with the ground as one side.

The ground between the two parts = 10 m

Let the part remaining = X

The part that fell = 17 - X

102 + X2 = (17 - X)2

100 + X2 = 289 - 34X + X2

100 = 289 - 34X

34X = 189

X = 5.56 m    Woody is 5.56 m off the ground.

When the pole cracked, a right triangle formed as follows: one side is the part of the pole that stands straight (call it a), the hypotenuse (call it h) is the part that fell, the upper part, and the other side is the ground from the pole to where the broken part hit (call it b = 10 meters).

Now a + h = 17, so h = 17 - a, and according to the Pythagorean theorem: h2 = a2 + 102 = a2 + 100.

So (17-a)2 = a2 + 100

172 - 34a + a2 = a2 + 100

289 - 34a = 100

189 = 34a

a = 5.56 meters

and that's how high is Woody above the ground.