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# Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of sideL?

The rest of it says..... if one side of the rectangle lies on the base of the triangle. I am not given any thing else.

### 2 Answers

- falzoonLv 71 decade agoFavorite Answer
Draw equilateral triangle ABC, with BC as the base.

Draw median line from A to M (midpoint of BC).

Draw rectangle PQRS, with P on AB, Q on BM, R on MC, and S on AC.

Given L = side of triangle, calculate using Pythagoras, that AM = L√3/2.

Let MR = x, so that RC = L/2 - x, and let h = SR = height of rectangle.

Triangles AMC and SRC are similar, so ratios of sides are equal, that is :

SR / RC = AM / MC

or

h / (L/2 - x) = (L√3/2) / (L/2)

Therefore, h = √3(L/2 - x)

Now, area of rectangle, A = base * height = 2xh = 2x√3(L/2 - x) = xL√3 - 2x^2√3.

Largest area is found by taking the derivative and setting it equal to zero.

dA/dx = L√3 - 4x√3 = 0, from which, x = L/4.

Thus, by substitution, h = L√3/4, and largest area, A = L^2√3/8.

Dimensions of rectangle are 2x = L/2 and h = L√3/4.

So, P and S lie on the midpoints of their respective sides

and Q and R lie on the midpoints of BM and MC respectively.