If the length of the beams is made three times the original length on each side, which statement is correct about the maximum area available for an exhibit to be displayed? It becomes nine times the original area. It becomes twenty-seven times the original area. It becomes three times the original area. It becomes eighty-one times the original area.
i guess.. option 1.. It becomes nine times the original area..
how do you know that?
Hey sorry, i thought it was perfect square.. But it is not, it looks like a trapezoid to me.. So for the 1st figure, A1 = h + (l+b)/2 And for the second.. A2 = 3h + (3l + 3b) / 2 = 3h + 3(l + b)/2 = 3 (h + (l+b)/2) = 3A1 So its option 3... 3 A1 I still doubt, but worth a try... |dw:1342785504724:dw|
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