How To Find A Diagonal Of Rectangle

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When you're working with rectangles, y'all tin discover out a lot of information well-nigh them just by knowing a few key points of information. If you've been given the length of the diagonal and at to the lowest degree one side, you can calculate the area of the whole shape with just a few equations. Always employ a calculator to bank check your math, and don't forget to add your units onto the answer!

i
Mark the diagonal on the rectangle. If you have a visual of your rectangle, it will be a lot easier to figure out the surface area given a diagonal. Draw out a rectangle (it doesn't accept to match the proportions given, simply a rectangle shape will do) and mark out a diagonal line to create 2 triangles.^[i]
- You'll exist able to see the two triangles that make up the rectangle now.
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iii
4

Solve the equation to detect the missing side. Break downwardly every number to its simplest class one pace at a time to solve for the missing expression. When you're washed, you can assign the answer to the missing side on your rectangle.^[iv]
v

Plug the new number into $A=l\times w$ . To find the area of your entire rectangle, you need to solve for Area = length x width, or A = l ten w. Now that yous've used the Pythagorean Theorem to find the second side length, you tin can solve for the area. Don't forget your units!^[5]

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1
Mark the diagonal and side length on your rectangle. If you lot take a visual in forepart of you, information technology will exist much easier to see how you're solving your equation. Describe a simple rectangle with a ruler and create a diagonal line beyond it to brand 2 triangles.^[6]
- It's not mandatory to depict your triangle, but it can assistance you lot, specially if yous're just starting out.
ii
Enter the length and diagonal into $A=l\times {\sqrt {d^{2}-l^{two}}}$ . But plug the length and the diagonal into the formula above. Remember, the diagonal is e'er longer than the length of the rectangle. ^[7]
- For example, if the length is 8 inches and the diagonal is 10 inches, the formula would be $A=viii\times {\sqrt {10^{2}-8^{2}}}$ .
three

Solve the equation to observe the area. All you take to practice now is suspension down the equation (using PEMDAS) in guild to simplify it and get the surface area of the rectangle. For example:^[eight]

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Question

If the diagonal of a rectangle is10cm then what is its area?

You don't accept enough data to solve.
Question

A rectangle with diagonal of 20 cm has sides in the ratio two:ane. How do I find the area of rectangle?

Such a rectangle would have its sides and diagonal in the ratio of ii:i:√5 (per the Pythagorean theorem). The ratio of √v to the diagonal (20) is equal to the ratio of two to the longer side (or the ratio of 1 to the shorter side). Thus √v : twenty = ii : ten, where x is the longer side. Expressed every bit fractions, that's √v / 20 = ii / x. Cantankerous-multiplying: (√5)(ten) = (20)(2). Dividing by √v: ten = twoscore / √5 = 40 / 2.236 = 17.9 cm. The shorter side is half that, or viii.94 cm.
Question

If the side = 48 meters and the diagonal = 50 meters, how do I find the area of the rectangle?

Permit s be the shorter side. And so (by the Pythagorean Theorem) s² + 48² = 50². Then s² + 2304 = 2500. So s² = 196, and southward = 14 m. The area is 48 x xiv = 672 foursquare meters.
Question

The length of a rectangle is 2 inches more than the width. If the diagonal is the square root of twenty inches, how practise I notice the area of the rectangle?

Allow the width be w. Then the length is (w+ii). Past the Pythagorean theorem, (w²) + (w+two)² = (√20)². So w² + (due west² + 4w + 4) = 2w² + 4w + 4 = xx. Divide by 2: west² + 2w + ii = 10. Subtract 10 from both sides: w² + 2w - viii = 0. And then (due west + 4)(w - 2) = 0. So west equals either -iv or +two. A width cannot be a negative number, so w = 2. The width is 2 inches, the length is 2+2 or 4 inches, and the area of the rectangle is 2x4 or eight square inches.
Question

If the diagonal length of a rectangle is 19cm and the area is 120cm. How do I find the length and width?

Start by defining unknowns; 50 = length and westward = width. We're given ii equations; l^two + westward^2 = xix^2 = 361 and lw=120. You can solve by substitution from here -- you'll get a quartic, merely it'due south really a quadratic in fifty^2 (or w^2 if you lot substitute the other fashion) so you tin yet solve it. But it goes a bit nicer if we don't suspension symmetry just yet and calculate (l+w)^ii = fifty^2 + w^two + 2lw = 361 + 240 = 601, and (fifty-west)^2 = fifty^ii + w^2 -2lw = 361 - 240 = 121. Then 50+w = sqrt(601) and l-w = 11. (Here, I assume l > w > 0.) So l = (sqrt(601) + 11)/two and w = (sqrt(601)-11)/two.

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You lot tin can also employ an online reckoner, like Omni Calculator, to cheque your work.
To detect the area of a rectangle, you need to know the diagonal and at least one side length.
If all you know is the length of the diagonal, you can summate the area of a foursquare, just not a rectangle.

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