Question 14 of the first survey and test in 2011 Nantong senior high school Given that the length of the middle line on the waist of an isosceles triangle is 3 under the root sign, what is the maximum area of the triangle?

Question 14 of the first survey and test in 2011 Nantong senior high school Given that the length of the middle line on the waist of an isosceles triangle is 3 under the root sign, what is the maximum area of the triangle?

Given that the length of the middle line on the waist of an isosceles triangle is 3 under the root sign, what is the maximum area of the triangle?
Analysis: let the waist length of the triangle be x and the vertex angle be θ
∵ the median line above waist is √ 3
The cosine theorem 3 = 5 / 4x ^ 2-x ^ 2cos θ = = > x ^ 2 = 12 / (5-4cos θ)
Area of triangle = 1 / 2x ^ 2Sin θ = 6sin θ / (5-4cos θ)
Let f (θ) = 6sin θ / (5-4cos θ)
Let f '(θ) = (30cos θ - 24) / (5-4cos θ) ^ 2 = 0 = = > cos θ = 4 / 5
F (θ) takes the maximum value when θ = arccos 4 / 5
When the vertex angle of the triangle is arccos 4 / 5, the maximum area is 2

What's 6 and 2 in 25?

152/25

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