March 29, 2017

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Find the dimensions and area of the largest rectangle that can be inscribed in a right triangle whose sides are 9cm, 12cm, 15cm.

Use quadratic functions to solve.

  • Algebra 2 - ,

    Put the points of the triangle at (0,0), (0,9) and (12,0)
    Let the insribed rectangle have corners at:
    (0,0), (x,0), (x,y'), and (0,y')
    y' is on the hypotenuse of the triangle, so that
    y = 12 - (4/3)x
    So y' = 12 - (4/3)x
    The rectangle area is
    A(x) = x*y' = 12 x - (4/3)x^2
    There is a maximum when dA/dx = 0
    12 = (8/3) x
    x = (3/8)*12 = 4.5
    y' = 12 - (4/3)(9/2) = 5
    The maximum rectangle area is xy'= 22 and the side lengths are 4.5 and 5.

  • Algebra 2 - ,

    This is normally a challenging calculus problem, however it can be worked with similar triangles, or algebra.

    First, note the triangle is a right triangle. Lay it out so the 12 dimension is on the x axis, and the 9 is on the y axis.

    Note the equation of the hypotenuse is
    y=-9/12 x + 9
    let one side of the rectangle going upward as h, and the side along the x axis as W. So the intersection with the hypotenuse is w,-9w/12 + 9

    So the drill is to find the max area.
    h= -9/12 w + 9
    Area= hw= -9/12 w^2+9w
    so when is Area max?

  • Algebra 2 (correction) - ,

    I calculated the point on the hypotenuse incorrectly. It is
    y' = 12 - (4/3)(9/2) = 6
    The rectangle side lengths are 4.5 and 6, and the area is 27. That equals half the area of the triangle, (1/2)*9*12 = 54. BobPursley tells me that Euclid proved this in many different ways, without calculus.

    Thanks to Bob for pointing this out

  • Algebra 2 - ,

    A package with square ends has a combined length and girth (girth is the perimeter of a cross section) of 120 in. The surface area of the entire package is 3600 sq. in.
    Determine the dimensions of thr package. s in.*s in.*l in.
    one solution is: 11.08in*11.08in*75.68in.
    find the other solution
    hint: if 4S + length=120, then the length=120-4S
    Round your answer to 2 decimals
    Enter the 3 dimensions separated by comas

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