I suspect it will be 4x8, but let's see what the calculations show.
If triangle ABC has vertices at (0,0),(16,0) and (h,8) then the sloping sides are lines with the equations
y = 8/h x
y = 8(x-16)/(h-16)
Now, suppose the inscribed rectangle has height k. The the line y=k intersects the sides where x is
hk/8 and k(h-16)/8 + 16
The length of the rectangle is thus 16-2k
The area of the rectangle is
a = k(16-2k) = 16k-2k^2
The vertex of this parabola is where k=4.
Thus the inscribed rectangle of maximal area is half the height of the triangle, making its base half the base of the triangle. The rectangle is thus 4x8.
A triangle has a base of 16 inches and an altitude of 8 inches. Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle.
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