Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6?

Solution:

Given, an isosceles triangle ABC is inscribed in a circle with center D and radius r.

We can obtain the side a in function of r and α by applying Law of Sines to triangle BCD,

a/sin(2α) = r/sin β

Since, 2α + β + β = 180°

2α + 2β = 180°

α + β = 90°

β = 90° - α

a = r(sin 2α/sin(90°-α))

a = r(2 sinα cosα)/cosα

a = 2r sinα

a = 4r sin(α/2) cos(α/2)

We can obtain the height h in function of r and α,

tan(α/2) = (a/2)/h

h = a/2tan(α/2)

Replacing with the value of a,

h = [4r sin(α/2) cos(α/2)/2][cos(α/2)/sin(α/2)]

h = 2r cos²(α/2)

Now, find the area of the triangle in function of a and r,

Area, A = (base)(height)/2

A = [4r sin(α/2) cos(α/2)][2r cos²(α/2)]/2

A = 4r²sin(α/2) cos³(α/2)

Now, take the derivative to find the maximum or minimum of the area of the triangle.

dA/dα = [4r²cos(α/2)(1/2)cos³(α/2)] + [4r²sin(α/2) 3cos²(α/2)(-sin(α/2))(1/2)]

On simplification,

= [2r²cos⁴(α/2)] - 6r²sin²(α/2) cos²(α/2)

Taking out common terms,

= 2r²cos²(α/2)[cos²(α/2) - 3sin²(α/2)]

Equating the derivative to zero, we get

2r²cos²(α/2)[cos²(α/2) - 3sin²(α/2)] = 0

2r²cos²(α/2) = 0

cos²(α/2) = 0

Thus, α/2 = 90°

α = 180°

Also, cos²(α/2) - 3sin²(α/2) = 0

cos²(α/2) = 3sin²(α/2)

[sin²(α/2)/cos²(α/2)] = 1/3

tan²(α/2) = 1/3 

tan(α/2) = 1/√3

So, α/2 = 30°

α = 60°

This implies that ∠B = ∠C = 60°

Thus, the isosceles triangle of the maximum area is also an equilateral triangle.

Given, radius, r = 6 units.

Maximum area of the triangle = 4(6)² sin 30° cos³30°

= 4(36)(1/2)(√3/2)³

= 4(36)(1/2)(3√3/8)

= 9(3√3)

= 27√3

Use √3 = 1.732

= 27(1.732)

A = 46.76 square units

Therefore, the maximum area of the triangle is 46.76 square units.

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Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6?

Summary:

The area of the largest isosceles triangle that can be inscribed in a circle of radius 6 is 46.76 square units.