Asked by Angy
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
Answered by
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)
Answered by
Steve
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.