anonymous
  • anonymous
I need to finish my homework ASAP! someone please help!! :( A square has sides of length 3x cm. A rectangle is 2x cm by (x+7)cm. The area of the square is twice that of the rectangle. Find the dimensions of each figure.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
area of the square is 9 , areas of the rectangle is \[2x(x+3)\] and twice that number is 9 so solve \[4x(x+3)=9\] \[4x^2+12x-9=0\]via the quadratic formula
anonymous
  • anonymous
thats it?
anonymous
  • anonymous
the area of the rectangle is (2x)(x+7) = A and the area of the square is twice that of the rectangle. Therefore area of the square = 2A . so we know (2x)(x+7) = A and (3x)^2 = 2A . so (3x)^2 = 2[ (2x)(x+7)] ====> 9x^2 = 4x^2 + 28x ====> 5x^2 = 28x ====> x = 28/5 . Now verify if my solution is correct by plugging in the value of x and finding the area of the rectangle and square

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anonymous
  • anonymous
oh i was wrong, sorry. it says side is 3x, not 3. so malcolm is right and i am wrong
anonymous
  • anonymous
\[4x(x+7)=9x^2\] \[4x^2+28x=9x^2\] \[5x^2-28x=0\] \[x(5x-28)=0\] \[x=\frac{28}{5}\]
anonymous
  • anonymous
Satelite is this how I shall write the whole thingie?

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