The side lengths of a square are tripled.

Question

a). What is the length's multiplier? What is the percentage increase of the length?

b). From the new (tripled) length, what percentage decrease would be needed to return to the original length? What is the associated multiplier?

c). What effect does the tripled length have on the square's Perimeter? It's Area?

Answer 1

Hello, Yuliya,

Let's set the length of a square's side to x. If the side lengths, x, of a square are tripled:

a. The length multioplier is 3 [3x for each side]. Percentage increase is 200% [((final-initial)/initial)*100%)] ((3x-1x)/1x)*100 = (2x/1x)*100 = 200%

b. Percentage decrease to return to original is %Decrease = -100(Final-Initial)/Initial [%Decrease = -100(1x-3x/3x), -100(-2x/3x), = 66.667% [I'm not sure what the "associated multiplier" is for this. Perhaps -92/3)??]

c. Perimeter is 4*x for the original. For the larger square it would be 4(3x) or 12 x. The original area is x². The tripled length area would be (3x)² or 9x².