Rectangles in Triangles
Geometry
Level
3
What is the largest possible area of rectangle inscribed in a triangle with side lengths 10,10 and 12?
25
8
24
98
48
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1 solution
I'm sure someone can code this solution better than I can - but here goes:
Divide the triangle in half, with the midpoint of the base at the origin. The line created by the right-hand leg has the equation y=(-4/3)x+8. Let the maximally sized rectangle have two vertices on the x-axis and 2 mirrored vertices along the legs at (x, y) and (-x, y).
The area of the rectangle can be defined as 2 x y; we know that y=(-4/3)x+8 and can sub that into the area formula. Thus Max area = (2)(x)(-4x/3+8) = (-8/3)x^2 + 16x The maximum value of the area will occur at -b/2a or (-16)/[(2 (-8/3)] = 3; thus the maximal area of the rectangle occurs when the "upper" vertices are at (3,4) and (-3, 4); giving a rectangle with an area of 2 3*4 = 24