### Video Transcript

The difference between the lengths of the diagonals of a rhombus is five centimeters. Given that its perimeter is 304 centimeters, find the lengths of the diagonals to the nearest centimeter.

Weβre asked to find the lengths of the diagonals of a rhombus. Letβs call our diagonals π one and π two. If we let π one be the longest of the two diagonals, then π one is greater than π two. And weβre told that the difference between the lengths of the diagonals is five centimeters, so π one minus π two is equal to five centimeters. So letβs use this now to write π one in terms of π two. If we discard our units for the moment and add π two to both sides, we have π one is equal to π two plus five. Okay, so weβll put this to one side for the moment and weβll come back to it later.

Weβre also told that the perimeter of the rhombus is 304 centimeters. Now, by definition, a rhombus is a quadrilateral whose four sides have equal length, which means that the side length for our rhombus is 304 divided by four, that is, 76 centimeters. Now, something else we know about rhombuses, or rhombi, is that their diagonals both bisect and are perpendicular to each other. Now remember, weβre trying to find π one and π two, and we can use the Pythagorean theorem to do this using one of our right angle triangles. In our case, this gives us π two over two all squared plus π one over two all squared is equal to 76 squared, where π is π two over two, π is π one over two, and π is 76.

But now we recall that we have π one in terms of π two. So now if we substitute this in as shown, our equation is in terms of π two only. We then have π two squared over four plus π two plus five all squared over four is equal to 5776. We have a common denominator of four on our left-hand side. And multiplying both sides by four and dividing through on the left-hand side by four, we have π two squared plus π two plus five all squared on the left-hand side and 5776 times four on the right-hand side. That is 23104 on the right-hand side. Distributing our parentheses on the left-hand side, we have π two squared plus π two squared plus 10π two plus 25. And collecting like terms, we have the equation shown. And now subtracting 23104 from both sides, we have two π two squared plus 10π two minus 23079 is equal to zero.

So now making some space, we have a quadratic in π two. And to solve this for π two, we can use the quadratic formula or any other method for solving quadratics. And so applying this to the coefficients in our equation, we have π two is negative 10 plus or minus the square root of 100, thatβs 10 squared, minus four times two times negative 23079 all over four. Evaluating inside the square root, that gives us negative 10 plus or minus the square root of 184732 all over four. And evaluating the square root to four decimal places gives us negative 10 plus or minus 429.8046 all over four. And of course, since length is always positive, we can ignore the negative solution.

Evaluating then to four decimal places, we have π two is equal to 104.9511. And so to the nearest centimeter, thatβs 105. And now making some space, we recall that π one is equal to π two plus five, that is, 105 plus five, which is 110 centimeters. To the nearest centimeter then, our diagonals are 110 centimeters and 105 centimeters.