SOLUTION: Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.

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Question 1043677: Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.
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let x = the width of the rectangle
let y = the length
:
Draw out a right triangle with the rectangle inscribed therein, x by y
the height of the right triangle = 12 and the base = 9
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We are interested in the small right triangles adjacent the inscribed rectangle
The tangent of the top angle = 9/12 which = x%2F%28%2812-y%29%29 of the small right triangle on top
Cross multiply
12x = 9(12-y)
12x = 108 - 9y
x = 9 - 9%2F12y
similarly, the tangent of the base angle = 12/9 = y%2F%28%289-x%29%29
Cross multiply
9y = 12(9-x)
9y = 108 - 12x
y = 12 - 12%2F9x
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Area = xy
replace y with (12-12%2F9x)
A = x(12-12%2F9x)
A = 12x - 12%2F9x^2
A quadratic equation, max occurs at the axis of symmetry; x=-b/(2a)
x = %28-12%29%2F%282%28-12%2F9%29%29
x = %28-12%29%2F%28-24%2F9%29
x = +4.5 cm is the width of the rectangle with max area
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Find y
y = 12 - 12%2F9*4.5
y = 12 - 6
y = 6 cm is length
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Max area: 6 * 4.5 = 27 sq/cm
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NOt sure if I explained this clearly, let me know. CK