An isosceles triangle has base b and equal sides of length a. Find the dimension

Question

An isosceles triangle has base b and equal sides of length a. Find the dimensions of the rectangle of maximum area that can be inscribed in the triangle if one side of the rectangle lies on the base of the triangle.

The answer provided is shown below:

Draw isocleles 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.

THE BELOW AM = L?3/2 I DO NOT UNDERSTAND
Given L = side of triangle, calculate using Pythagoras, that AM = L?3/2 -------WHY???

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.

Explanation / Answer

Since it is an isoceles triangle with AB = AC = a and BC = b

BM = b/2 since M is mid point

Now AM = sqrt(AB^2 - BM^2) = sqrt(L^2 - (L/2)^2) = sqrt(3L^2/4) = (L?3/2)

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