In that case you get s^2 + s^2 = 32^2 which becomes 2s^2 = 32^2 which gets you s^2 = 512 which gets you s = sqrt(512), s being the length of a side of the square. Now if you're talking about a square rather than a rectangle, then you can solve for the length of a side because length and width are now the same measure. Pick any other value of W^2 less than 32^2 and you'll get another value of L that will satisfy the equation. You have a length of sqrt(63) and a width of 31 and a diagonal of 32.ģ1^2 + sqrt(63)^2 = 32^2 becomes 961 + 63 = 1024 which becomes 1024 = 1024 which confirms that the solution of L^2 = 63 is correct. If L^2 is equal to 63, then L is equal to sqrt(63) You can pretty much pick any W^2 that is less than 32^2 and you will get a corresponding length that will satisfy the equation.Īssuming W^2 = 31^2, you will find a corresponding L^2 that is equal to 63. In algebraic terms, this looks like L^2 = (32^2 - W^2) Solve for it's length and you get it's length is equal to the square root of (it's length squared minus its width squared). Question 355440: the width, length, and diagonal of a rectangle are consecutive even integers. Diagonals are equal in length Diagonals perpendicularly bisect each other Adjacent sides make a right angle at the vertex Perimeter of a rectangle is two times the sum of its length and width Area is the multiplication of the length and width The sum of all the interior angles as well as exterior angles is 360°. Evaluating this, d 25 and therefore the diagonal is 5cm long. For example, the diagonal of a rectangle of length 4cm and width 3cm is given by d (4 + 3). So we can use a square with side length 24/4 6 feet, which then has area 62 36 square feet. Since 24 is divisible by 4, we can make the length and width equal even though the problem states that they have to be integers. The diagonal of a rectangle is equal to 32.īy the pythagorean formula, the diagonal squared of the rectangle is equal to its length squared plus its width squared. To find the diagonal of a rectangle using the Pythagorean Theorem, use the formula d (l + w), where l is the length and w is the width of the rectangle. For a rectangle of a given perimeter, the area is maximized by making the length and width as close together as possible. ![]() You don't have enough information to determine this exactly. You can put this solution on YOUR website!
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