+4609 Question from my book: Find the length of the side of the rhombus which has area 40 and diagonals with length 2x and 3x-2.
My answer: square root of 41...Is this correct?
Well, first the area for solving the area of the rhombus is: \(d_1d_2/2\), so (2x)(3x-2)=80.
Then, take 80 to the other side, solve the quadratic and get the roots x=4, x=-10/3. Since x has to be positive, so the diagonals have length 2(4)=8 and 3(4)-2=10. We are looking for the side length, so is it square root of 4^2+5^2, since the diagonals are 8 and 10...Therefore, the answer is \(\sqrt{41}\) .
Sorry for my bad latex. Is it correct? Thanks!
