Question
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.
Answers
Steve
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.
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.