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**Parallel Planes Practice Problems**

**PROBLEM 1**

Suppose you are standing inside a large warehouse. The floor is flat and level, and the ceiling is flat and is at a uniform height of 5.455 meters above the floor. You have a flashlight with a narrow beam, and hold it so its bulb is 1.025 meters above the floor. You shine the beam upward toward the ceiling. The center of the beam strikes the ceiling 9.577 meters from the point on the ceiling directly above the bulb. How long is the shaft of light representing the center of the beam? Round your answer off to two decimal places.

**SOLUTION 1**

Let’s call the flashlight bulb point *A* , the point at which the center of the light beam strikes the ceiling point *B* , and the point directly over the flashlight bulb point *C* as shown in Fig. 7-18. Call the lengths of the sides opposite these points *a, b* , and *c* . Then *ΔABC* is a right triangle because line segment *AC* (whose length is *b* ) is normal to the ceiling at point *C* , and therefore is perpendicular to line segment *BC* . The right angle is *∠ACB* . From this, we know that the lengths of the sides are related according to the Pythagorean formula:

*a* ^{2} + *b* ^{2} = *c* ^{2}

We want to know the length of side *c;* therefore:

*c* = ( *a* ^{2} + *b* ^{2} ) ^{1/2}

Calling meters “units” so we don’t have to write the word “meters” over and over, the length of side *a* is given as 9.577. The length of side *b* is equal to the height of the ceiling above the floor, minus the height of the bulb above the floor:

*b* = 5.455 − 1.025 = 4.430

Therefore:

*c* = (9.577 ^{2} + 4.430 ^{2} ) ^{1/2}

= (91.719 + 19.625) ^{1/2}

= 111.344 ^{1/2}

= 10.55

The center of the beam is 10.55 meters long.

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