How do areas of two parallelgrams compare when the dimensions of one are three times the demensions of the other?
let's take a rectangle for example. It has a width of 2 and a lengh of 3. You just take A=lw so the are is (2)(3)= 6
The one that is three times the other is 2(3) and 3(3) or 6 x 9 = 45. It is 9 times bigger. 45/6 = 9
Or you could just do the Algebra. A=lw would be A=(cm)(cm) = cm^2 or A=(m)(m) = m^2 if you had the dimensions be 2 times the first one it would be 4 times as big, since it was squared. 3 would be 9 times, 4 would be 16 times, and so on.
To compare the areas of two parallelograms when the dimensions of one are three times the dimensions of the other, you need to apply the concept of scaling.
Let's say we have two parallelograms, parallelogram A and parallelogram B. If the dimensions of parallelogram B are three times the dimensions of parallelogram A, we can represent this relationship as:
Length of parallelogram B = 3 × Length of parallelogram A
Width of parallelogram B = 3 × Width of parallelogram A
Now, the formula to calculate the area of a parallelogram is:
Area = Length × Width
Let's denote the area of parallelogram A as A_A and the area of parallelogram B as A_B.
For parallelogram A:
A_A = Length of parallelogram A × Width of parallelogram A
For parallelogram B:
A_B = Length of parallelogram B × Width of parallelogram B
= (3 × Length of parallelogram A) × (3 × Width of parallelogram A)
= 9 × (Length of parallelogram A × Width of parallelogram A)
= 9 × A_A
Therefore, the area of parallelogram B (A_B) is nine times the area of parallelogram A (A_A) when the dimensions of parallelogram B are three times the dimensions of parallelogram A.