ACT Preparation / Unit 13 ACT 1229 worked example, 60 to 90 seconds
Applying a Scale Factor
The idea
Scale-factor problems are solved in three steps. First, find the length scale factor k from a pair of corresponding lengths, new over old. Second, decide what kind of quantity is asked: a length uses k, an area k², a volume k³. Third, multiply the original quantity by that power. If an area or volume ratio is given instead, take the square or cube root to recover k.
In practice
Worked example: two similar triangles have corresponding sides of 4 and 10, so k = 10/4 = 2.5. If the smaller triangle's area is 12 square units, the larger's is 12 × 2.5² = 12 × 6.25 = 75. Second: two similar cubes have volumes of 8 and 125, a ratio of 125/8, so k = ∛(125/8) = 5/2, and their edges are in the ratio 2 to 5.
Where people slip
Applying k to an area is the slip to avoid: it would give 12 × 2.5 = 30 here, less than half the true 75.
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