A homework question that I CANNOT figure out..
How do the areas of two parallelograms compare when the dimensions of one are 3 times the dimensions of the other?
2007-01-31 11:57:10 · 3 answers · asked by Roxanne-Chan 2 in Science & Mathematics ➔ Mathematics
Just imagine a paralleogram having length of l units and breadth of b units.its area would naturally be ab square units.
now if the length is increased ti 3a and the breadth to 3b,its aea becomes 3a*3b or 9ab
Hence it is evident that the area increases by 9 times.
2007-01-31 12:06:18 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
OK - how do you calculate the area of a parallelogram? base times height, right? So if the base is x and the height is y, then the area of the first one is x(y). Now in the second parallelogram if the base is 3 times x and the height is 3 times y, calculate the area and compare it with the area of the first parallelogramj.
2007-01-31 20:02:04 · answer #2 · answered by koolkat 3 · 0⤊ 0⤋
Assuming the parallelograms are congruent (meaning they have the same interior angles), a 3 times increase in proportional sides would yield a 3^2 or 9 times increase in area.
2007-01-31 20:01:25 · answer #3 · answered by disposable_hero_too 6 · 0⤊ 0⤋