What is the area of the largest isosceles triangle that can be inscribed in a circle of radius 4?

Question

a) Solve by writing the area as a function of h = height.
b) Solve by writing the area as a function of alpha,
where alpha is the angle formed by connecting the "right most side"
of the triangle with a perpendicular line drawn from the top of the triangle to the base.

c) Identify the type of triangle of maximum area.

Answer 1

coldsoup used good geometrical reasoning to decide that the largest triangle is equilateral, but he didn't figure out the area. This is a good calculus problem where you have to the shape of the triangle by maximizing its area. It's not an easy one to solve, but I'll give it a shot. Assume, for starters, that we don't know it's an equilateral triangle.

Draw the x-y axes and a circle of radius 4 centered at the origin. The top of the circle is at (0,4), and its equation is x^2 + y^2 = 16.

Now assume the vertex of the triangle is also at (0,4) and that the base angles are somewhere below the x-axis. The vertex of one of these base angles is at (a,b) on the circle in the 4th quadrant (a > 0, b < 0).

The isosceles triangle is symmetric about the y-axis, so by maximizing the right half of the triangle (for x > 0), we'll maximize the area of the whole thing.

The area of the triangle (right side only) is (1/2) a(4 - b) (because b is a negative number). We need to find a and b so that the area is maximized.

But (a,b) is on the circle, so it must satisfy x^2 + y^2 = 16. Solve this for y = -sqrt(16 - x^2) (noting that y is the square root) and substitute into the area formula:

A = (1/2) a(4 - b) = (x/2) [4 + sqrt(16 - x^2)]
A = 2x + (x/2) sqrt(16 - x^2) = 2x + (x/2)(16 - x^2)^(1/2)

Take the derivative and set it equal to zero to maximize:

dA/dx = 2 + (x/2)(1/2)(16 - x^2)^(-1/2)(-2x) + (1/2)(16 - x^2)^(1/2) = 0
dA/dx = 2 - (x^2) / [2 sqrt(16 - x^2)] + (1/2) sqrt(16 - x^2) = 0

Multiply through by 2 sqrt(16 - x^2) to get the radical out of the denominator:

4 sqrt(16 - x^2) - x^2 + (16 - x^2) = 0
4 sqrt(16 - x^2) = 2x^2 - 16
2 sqrt(16 - x^2) = x^2 - 8
4(16 - x^2) = (x^2 - 8)^2 = x^4 - 16x^2 + 64
-4x^2 = x^4 - 16x^2
x^4 = 12x^2

A double root at x = 0 falls out of this. We can ignore that because it represents a degenerate minimum (isosceles "triangle" with zero area). What's left is

x^2 = 12 ==> x = a = 2 sqrt(3)

This is the "a" value for the base vertex. We get "b" from

b = - sqrt(16 - a^2) = - sqrt(16 - 12) = -2

So the base vertex is at (2 sqrt(3), -2). The height of the triangle is 4 - b = 6.

The area of the triangle is A = a(4 - b) = 2 sqrt(3)(6) = 12 sqrt(3). Let's see if this answer is reasonable. 12 sqrt(3) = 20.78; the area of the circle is 16 pi =50.27, so the triangle fills 41% of the circle.

Now let's find the two equal sides of the isosceles triangle. Using the Pythagorean Theorem,

s = sqrt(a^2 + b^2) = sqrt(12 + 36) = 4 sqrt(3)

But the base of the triangle, 2a, also equals 4 sqrt(3), so the triangle is indeed equilateral.

Now we can answer your questions. The area of the triangle is 12 sqrt(3).

(a) The height h is 6, so A = 2h sqrt(3).

(b) Since the triangle is equilateral, alpha = 30 degrees = pi/6. sin(alpha) = 1/2 and cos(alpha) = sqrt(3)/2, so

A = 48 sin(alpha) cos(alpha)

(c) The triangle is equilateral.

This was a good problem ... fun to do, and not easy, especially the way I did it, by deriving the solution and not assuming it in advance. I'm glad you re-posted it.

Answer 2

A equilateral triangle.

Reasoning:
the perimeter of the triangle becomes less as any two sides become closer together, therefore it must necessarily get larger as all sides get farther apart, i.e: the angles between all sides are as large as possible. This state is achieved when the triangle is equilateral, as all angles in an equilateral triangle are 60 degrees, and if the size of any angle is increased any more, then the size of at least one other angle is decreased, and the sides that include it become closer together.

Answer 3

The formula for area of an isosceles triangle:

A = 1/2bh

Area equals one half the base times the height

Answer 4

3^(3/2)

Answer 5

Radius of 4 what? mm metres kilometers

Answer 6

http://mathforum.org/dr.math/faq/formulas/faq.triangle.html#isosceles