Question
A piece of wire x cm long is to be cut into two pieces, each to bent to be a square. the length of a side of one square is to be 9 times the length of a side of the other . Express the sum of the areas of two squares in term of x .
Answers
bobpursley
perimeter of new squares sum=x
so one length is 9/10 x, the other x
Then
area sum= (.9x/4)^2+(.1x/4)^2
= 1/16( .81x^2+.01x^2)
=1/16( .82x^2)
so one length is 9/10 x, the other x
Then
area sum= (.9x/4)^2+(.1x/4)^2
= 1/16( .81x^2+.01x^2)
=1/16( .82x^2)
I think you read "nine times" as "nine tenths". Also, the sum of the perimeters is x. In that case, the two perimeters are 9/10 x and 1/10 x. So, the areas sum to
((9/10 x)/4)^2 + ((1/10 x)/4)^2 = 41/800 x^2
((9/10 x)/4)^2 + ((1/10 x)/4)^2 = 41/800 x^2
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