Find the Area of the Largest Isosceles Triangle That Can Be Inscribed in a Circle Radius 6

Notice the surface area of the largest isosceles triangle that tin be inscribed in a circle of radius vi?

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 = xc°

β = 90° - α

a = r(sin 2α/sin(90°-α))

a = r(two sinα cosα)/cosα

a = 2r sinα

a = 4r sin(α/2) cos(α/ii)

We can obtain the height h in part 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/ii) cos(α/ii)/two][cos(α/2)/sin(α/two)]

h = 2r cos²(α/two)

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

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

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

A = 4r²sin(α/2) cos³(α/ii)

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

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

On simplification,

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

Taking out mutual terms,

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

Equating the derivative to zip, nosotros get

2r²cos²(α/2)[cos^two(α/ii) - 3sin²(α/ii)] = 0

2rⁱⁱcos²(α/ii) = 0

cos²(α/2) = 0

Thus, α/2 = 90°

α = 180°

Also, cosⁱⁱ(α/2) - 3sinⁱⁱ(α/2) = 0

cos²(α/two) = 3sin^two(α/ii)

[sin^two(α/2)/cos²(α/2)] = one/3

tan^two(α/2) = one/3

tan(α/2) = one/√3

So, α/ii = xxx°

α = sixty°

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 = iv(6)² sin 30° cos³30°

= four(36)(one/ii)(√3/2)ⁱⁱⁱ

= four(36)(1/2)(3√3/8)

= 9(3√3)

= 27√3

Use √3 = one.732

= 27(1.732)

A = 46.76 square units

Therefore, the maximum expanse 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 tin be inscribed in a circle of radius half-dozen is 46.76 square units.

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Source: https://www.cuemath.com/questions/find-the-area-of-the-largest-isosceles-triangle-that-can-be-inscribed-in-a-circle-of-radius-6/#:~:text=Summary%3A-,The%20area%20of%20the%20largest%20isosceles%20triangle%20that%20can%20be,6%20is%2046.76%20square%20units.

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