What is the largest rectangle that can be inscribed in an equilateral triangle with sides of 12?

Question

Ella Armstrong · Accepted Answer

#(3, 0), (9, 0), (9, 3 sqrt 3), (3, 3 sqrt 3)# #Delta VAB; P, Q in AB ; R in VA; S in VB# #A = (0, 0), B = (12, 0), V = (6, 6 sqrt 3)# #P = (p, 0), Q = (q, 0), 0 < p < q < 12# #VA: y = x sqrt 3 Rightarrow R = (p, p sqrt 3), 0 < p < 6# #VB: y = (12 - x) sqrt 3 Rightarrow S = (q, (12 - q) sqrt 3), 6 < q < 12# #y_R = y_S Rightarrow p sqrt 3 = (12 - q) sqrt 3 Rightarrow q = 12 - p# #z(p) = #Area of #PQSR = (q - p) p sqrt 3 = 12p sqrt 3 - 2p^2 sqrt 3# This is a parabola, and we want the Vertex #W#. #z(p) = a p^2 + bp + c Rightarrow W = ((-b)/(2a), z(-b/(2a)))# #x_W = (-12 sqrt 3)/(-4 sqrt 3) = 3# #z(3) = 36 sqrt 3 - 18 sqrt 3#

Maverick Smith · Answer

undefined The largest rectangle that can be inscribed in an equilateral triangle with sides of 12 has a height equal to the height of the equilateral triangle and a width equal to twice the height. Therefore, the height of the rectangle is $ \frac{\sqrt{3}}{2} \times 12 $, and the width is $ 2 \times \frac{\sqrt{3}}{2} \times 12 $. Simplifying, the height is $ 6\sqrt{3} $, and the width is $ 12\sqrt{3} $. Thus, the largest rectangle that can be inscribed has dimensions $ 6\sqrt{3} $ by $ 12\sqrt{3} $.