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Finding the dimensions of a rotated rectangle inside another rectangle.

  1. May 2, 2009 #1
    1. The problem statement, all variables and given/known data
    If I have a rectangle rotated at a known angle with respect to a rectangle of known dimensions that inscribes it, how can I find the dimensions of the inscribed/inner rectangle?



    If the image above is my example, I know the dimensions of ABCD and I know all the angles, such as < BPQ.

    2. Relevant equations

    3. The attempt at a solution
    I'll post if I come up with anything that looks like it's gettign anywhere =P

    Thanks for the help... let's see how my first ever post is received =)
  2. jcsd
  3. May 2, 2009 #2


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    Homework Helper

    Surely you must have tried something?

    Hint: can you find the four triangles in the figure? From there, you have the trig formulas to calculate the lengths of the sides you need...
  4. May 2, 2009 #3
    oh, I've been trying for a couple hours. But I haven't really made it anywhere =(
  5. May 2, 2009 #4


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    Insights Author

    Staff: Mentor

    So show us what you've tried.
  6. May 2, 2009 #5
    ok... I think I have something that should be able to go somewhere...

    Here's a relabelled image:

    ɵ, X, and Y are known, trying to find h and w.

    y1, y2, x1, x2, w, and h are the unknowns (6)

    I can get seven equations:

    w2 = x22+y12

    h2 = x12+y22

    Y = y1 + y2

    X = x1 + x2

    y1 = x2 tanɵ

    x1 = y2 tanɵ

    XY = x2y1 + x1y2 + hw (areas)
  7. May 2, 2009 #6
    eliminating x1,x2,y1,y2 I get...

    XY = [tex]\frac{(w^2+h^2)tan\theta}{1+tan^2\theta} + hw[/tex]

    X = [tex]\frac{htan\theta + w}{\sqrt{1+tan^2\theta}}[/tex]

    Y = [tex]\frac{wtan\theta + h}{\sqrt{1+tan^2\theta}}[/tex]

    sub some trig identities

    XY = [tex](w^2+h^2)sin\theta cos\theta + hw[/tex]

    X = [tex](htan\theta + w)cos\theta[/tex]

    Y = [tex](wtan\theta + h)cos\theta[/tex]
    Last edited: May 2, 2009
  8. May 2, 2009 #7
    Further simplifying...

    [tex]X = hsin\theta + wcos\theta[/tex]

    [tex]Y = wsin\theta + hcos\theta[/tex]

    LOL.... I could have pulled that directly off the diagram! well, at tleast I know my algebra is sound =P
  9. May 10, 2011 #8
    But with this, you find X and Y that it is supposed you already knew, what about finding h and w , huh??
  10. Jun 22, 2011 #9
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