Answer:
9 cm
Explanation:
Given two beetles on a roof with a first face having a slope of -2/3 and the second face perpendicular to it. The first beetle travels down the first face twice as fast as the second beetle travels down the second face. You want to know their vertical separation when their horizontal separation is 72 cm.
Separation
In a given unit of time, the first beetle will travel 2×3 = 6 cm horizontally, and 2×(-2) = -4 cm vertically. The second beetle will travel 2 cm horizontally and -3 cm vertically in the same time, a perpendicular distance half as far.
After that one unit of time, the separation between the beetles is ...
horizontally: 6 +2 = 8 cm
vertically: -3 -(-4) = 1 cm
Scale
We notice the vertical separation is (1 cm)/(8 cm) = 1/8 of the horizontal separation. When the horizontal separation is 72 cm, the vertical separation will be ...
(1/8)(72 cm) = 9 cm
The second beetle is 9 cm above the first when their horizontal separation is 72 cm.
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Additional comment
The rise/run for the first face is given as (-2 cm)/(3 cm) = -2/3. The second face is perpendicular, so will have a rise/run = -1/(-2/3) = 3/2. That is, for a run of -2 cm, the rise will be -3 cm.
We note that the right triangles with legs "rise" and "run" are congruent for the two beetles. Traveling at the same speed, each beetle will cover a hypotenuse length in a given unit of time, hence the lengths "rise" and "run" on their respective roof faces. Since beetle 1 travels twice as fast, its distance will be 2·rise and 2·run in each time unit.
The attached diagram shows the geometry of the problem. The measures show beetle 1 has traveled twice as far as beetle 2 when they are separated horizontally by 72 cm. The grid is set so you can see the slopes are -2/3 and 3/2.
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