Math, asked by ddaydream30, 11 months ago

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

Answered by amitnrw
0

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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