this problem is about maxima and minima.
find the area of the largest isosceles triangle inscribed in a semi-circle radius of 10cm. the vertex of the triangle being at the center of the circle.
Answers
Area of the largest Isosceles triangle = 50 cm²
Step-by-step explanation:
isosceles triangle inscribed in a semi-circle radius of 10 cm.
and the vertex of the triangle being at the center of the circle
=> Two Equal Side of isosceles triangle = Radius = 10 cm
Let say Third side = 2a 2a < 20 cm ( diameter of circle)
as Altitude drawn will be at mid point of third side
=> Half of the Side = a
Altitude = √ 10² - a² = √(100 - a²)
Area of Triangle = (1/2) * 2a * √(100 - a²)
= a * √(100 - a²)
dA/da = √(100 - a²) + a (-2a) / (2√(100 - a²))
= √(100 - a²) - a²/√(100 - a²)
Put dA/da = 0
=> √(100 - a²) = a²/√(100 - a²)
=> 100 - a² = a²
=> 2a² = 100
=> a² = 50
=> a = 5√2
Area of Triangle = (1/2) * 2a * √(100 - a²)
= (1/2) * 2 * 5√2 * 5√2
= 50
Area of the largest Isosceles triangle = 50 cm²
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