How to Find the Area of a Square Using the Length of its Diagonal
Six Methods:The FormulaProofExampleAlternative Methodn.b. The Square's Diagonal and Neutral OperationsHelpful guidance
The formula for the area of a square is simple  it's side squared, or s^2. But if you've ever wondered whether you could find a square's area by using its diagonal(s), the answer is yes. And it's simple, like the saying, "The shortest distance between any two points is the hypotenuse!"
Steps
Method 1 The Formula
 1Consider a square whose diagonal measures d units.
 2The area of this square can be calculated using the formula:
Area = d^{2}/2  3Consider that, if you are given the diagonal and you know that the diagonal of a square is side * sqrt(2), then you may divide both sides by sqrt(2) and obtain from d=s*sqrt(2) that d/sqrt(2) = s. Since s^2 is the area, then (d/sqrt(2))^2 = s^2. Given d = 10 cm for example, you may get it the easy way that (10/sqrt(2))^2 = 100/2 = 50, very quickly. If you know that sqrt(2)=1.414 and are given the diagonal, say 10 cm, then you may divide 10/1.414 to get 7.072, which then squared = 50. Thus 50 sq. cm is the Area and answer. The first way is generally more accurate because sqrt(2) is irrational and transcendental. The way you approach it may depend on whether you are also asked to answer what the side's measurement is. Done.
Method 2 Proof
 1Let us consider a square whose side measures a units.
 2We know that the area of square is equal to (Side)^{2}.
The side being a here, we have
Area = a^{2}..........(Equation 1)  3Join any two opposite vertices to mark a diagonal. Let the measure of this diagonal be d units. This diagonal divides the square into two righttriangles.
 4Apply Pythagorean Theorem to any one of these triangles.
a^{2}+a^{2}=d^{2}
=> 2 a^{2} = d^{2}
=> a^{2} = d^{2}/2  5Plug in the value of a^{2} obtained in the preceding step into the Equation (1). Thus we see,
Area = d^{2}/2
Method 3 Example
Method 4 Alternative Method
 1Find the area for a square whose side measures s units by using the diagonal:
 2Because 1^2 has the diagonal sqrt(2) by the Pythagorean Theorem, i.e.
1^2 + 1^2 = (side*sqrt(2))^2, all squares have the diagonal, side * sqrt(2).
 Thus, per the above equations, the area of the square is the diagonal squared divided by 2. For side = s, [(s*sqrt(2))^2]/2 = the Area of the Square. Notice that ((sqrt(2))^2)/2 = 1, so just s^2 = A remains.
 This is useful in the more common case where the side is given in even integers, rather than the diagonal.
 Besides, if one is given that the diagonal = 10 and is asked to find both the area and the length of the side, without this alternative method, it is difficult, because one needs to divide 10 by the square root of 2 or 1.414 (1.4142135623731 to be more exact) and obtain the side's length of 7.071, which when squared = 50. 10^2 /2 = 50 also. But if the hypotenuse squared = 100, then the sides a1 and a2 each = 7.071 and square to be 50. Therefore the diagonal squared, over 2, all squared also equals the area.
 Otherwise, quickly find the root of the area by a method like the NewtonRaphson method (please see related wikiHow article below).
 In the above formula pictured, i.e. 1^2 + 1^2 = (side * sqrt(2))^2, is only correct if side = 1. But. s^2 + s^2 = 2s^2 = (side*sqrt(2))^2. Which means that sqrt(2)*s=s*sqrt(2), or d=d. That doesn't really lead anywhere. But if you take the diagonal, d=s*sqrt(2), and square it, and then divide by sqrt(2), and square that too, the result will be the Area. That is, (s*sqrt(2))^2 = 2s^2 and that divided by 2 gives s^2, the Area. So A = (s*sqrt(2))^2 / (sqrt(2))^2 = (d/sqrt(2))^2. This is another way to find the Area, given only the length of the sides, but asked to use the diagonal to determine the Area. The formula works out to the same as the one above, but the logic is a bit different. Here, we have directly reasoned that the Area of the square is half of the diagonal squared, which tells one that if one makes a square on the diagonal, exactly 1/2 of that = the Area of the originating square for the diagonal, and it does not matter how one creates that half.
This simple fact is important in crystallography and chemistry, very much so. And it is also important to artists, choreographers, machinists, et al. It is also a fact that the hypotenuse, or diagonal, is the shortest distance between two points. Thus, if you know how far you've traveled between any two points, since d=sqrt(2)* side s and Area = d^2/2, you may quickly form the grid around your location and be aware of the area you are surveying. Why? Since if you can see your destination, than at the middle of the diagonal you can see 1/2 the length of the diagonal to the destination to either side. However, if you could not originally see your destination, at least when you're halfway there, you can see half of it to either side. This fact is useful in surveying, photography and perspective in art. By the way, the shortest distance is the hypotenuse between two points, but given vectors to the lower destination and straight downwards, the brachistochrone curve is the shortest distance in terms of time between two points. See https://en.wikipedia.org/wiki/Brachistochrone_curve for a review of this important curve, especially given curved Einsteinian space. The curve is the inverted cycloid, or path traced out by one point on a moving circle in a straight line, only upside down. It has been said and shown that light moves according to this curve, to take the least possible time to arrive at a destination. It is also the fundaments of the modern infinitesimal calculus. There is no imparted force to a ball or bead descending this curve though (it moves per gravity) and friction causes a different resulting curve as the answer. What is the imparted force when light is created, for example by an electron changing orbital shells? What is the force of light? Perhaps, since light is massless, you will be the genius to figure it out.
Method 5 n.b. The Square's Diagonal and Neutral Operations
 1
 Let the diagonal be held neutral as between Addition and Multiplication. s = side:
 2Express this formulaically as

 (s+sqrt(2)^2)/2 = (s*sqrt(2)^2)/2 = c, then, taking the square root of both sides, we have
 (s+sqrt(2))/sqrt(2) = (s*sqrt(2))/sqrt(2), then multiplying both sides by sqrt(2) gives us
 (s+sqrt(2)) = (s*sqrt(2)), and subtracting s from both sides, we then have
 sqrt(2) = s*(sqrt(2)1) ... having simplified the left and factored on the right.
 sqrt(2)/(sqrt(2)1) = s when we divide both sides by (sqrt(2)1) and simplify.
 3.414213562 = s, as we solve the left side of the equation.
 Test that against the original hypothesis, that (s+sqrt(2)^2)/2 = (s*sqrt(2)^2)/2 = c;
 (3.414213562+sqrt(2)^2)/2 = (3.414213562*sqrt(2)^2)/2 = c = 11.65685425;
 Which is amazing enough, but consider also that 3.414213562^2 = 11.65685425!!
 Lastly, be aware that 2 + sqrt(2) = 3.414213562 !! That is, geometrically, two sides of 1 and the diagonal as the third leg  3.414213562 is the sum of those three numbers! That figure squared is the same as saying that Addition and Multiplication are neutral for the diagonal! And we square it the figure by swinging it 90 degrees to the z axis, which movement resembles the thumb, index and middle fingers swinging to grab an object, with the wrist at {0,0}. This is of perhaps of some interest and use to musicians, artists and sports professionals, et al.

Method 6 Helpful guidance
 1Make use of helper articles when proceeding through this tutorial:
 See the article How to Create a Spirallic Spin Particle Path or Necklace Form or Spherical Border for a list of articles related to Excel, Geometric and/or Trigonometric Art, Charting/Diagramming and Algebraic Formulation.
 For more art charts and graphs, you might also want to click on Category:Microsoft Excel Imagery, Category:Mathematics, Category:Spreadsheets or Category:Graphics to view many Excel worksheets and charts where Trigonometry, Geometry and Calculus have been turned into Art, or simply click on the category as appears in the upper right white portion of this page, or at the bottom left of the page.
Sources and Citations
 http://www.sciencedaily.com/releases/2012/04/120419102042.htm on green energy and metal oxides ....
 http://www.clarku.edu/~djoyce/trig/area.html on the area of a triangle ...
Article Info
Categories: Calculating Volume and Area  Mathematics