A rectangle has perimeter 6m and the length of a diagonal is radical 5m. what are dimensions?

Question

we are working on solving systems of equations and i have a really hard time with this because my teacher says not to follow the book. iIalso need to fined the answer as points like a graph and using the quadratic formula 

Answer 1

If you draw a rectangle with length = L on the x-axis and width = W on the y-axis with its bottom left corner at the origin (0,0) then its bottom right corner will be at (L,0), its upper left corner will be at (0,W) and its upper right corner will be at (L,W).  The diagonal of the rectangle is the line segment from (0,0) to (L,W).  

 

The perimeter of the rectangle is L + W + L + W ==>  2*L + 2*W = 6 

 

The length of the diagonal is sqrt (L² + W²) by the distance formula or equivalently the Pythagorean Theorem 

 

Solve the 1st equation for L  ==>  2*L = 6 - 2*W  ==>  L = 3 - W 

 

Substitute this value for L into the 2nd equation 

 

sqrt(5) = sqrt ( (3-W)² + W²) 

 

Square both sides and expand the right hand side 

 

5 = 9 - 6*W + W² + W² 

0 = 2*W² - 6*W + 4

0 = W² - 3*W + 2

0 = (W - 2)*(W - 1) 

 

So W = 2  or  W = 1 

 

If W = 2 then L = 1, if W = 1 then L = 2, so the dimensions are the same in either case, the rectangle is just rotated 90 degrees 

 

So the dimensions of the rectangle are 2m by 1m 

 

Check:  Perimeter = 2 + 1 + 2 + 1 = 6m 

            Diagonal = sqrt(2² + 1²) = sqrt(5)