Question

ABC is a right angled triangle of given area S. find the sides of the triangle for which the area of the circumscribed circle is least​

Answer 1

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Answer 2

Answer:

The sides of a triangle ABC i.e., hypotenuse, height, and base are 2√S, √2S, and √2S respectively.

Step-by-step explanation:

Let us assume the base and height of the right-angled triangle ABC circumscribed in a circle be 'b' and 'h' respectively.

The radius of the circle is 'r'.

So, the Hypotenus of a triangle ABC, c = 2r

Area of a triangle ABC, S = (1/2)×b×h  

Area of a circle, A = πr²

Therefore,

(2r)² = b² + h²

⇒ r² = (b² + h²)/4

as, h = 2S/b    -(i)

⇒ r²  = $\frac{1}{4}$×(b² + 4S²/b²)    -(ii)

A =  $\frac{1}{4}$×π(b² + 4S²/b²)     -(iii)

Differentiating A wrt b, we get

$\frac{dA}{db} = \frac{\pi}{4}(2b + 4\times \frac{-2S^2}{b^3} )$

dA/db = 0, we get

⇒ 2b - 8S²/b³ = 0

⇒ 2b = 8S²/b³

⇒ b = 4S²

⇒ b² = 2S

⇒ b = √2S  

Now, the value of 'h' by putting the value of 'b' in equation (i), we get

h = 2S/√2S

⇒ h = √2S

So, the radius 'r' by putting the value of 'b' and 'h' in equation (ii), we get

r² =  $\frac{1}{4}$((√2S)² + 4S²/(√2S)²)  

r² =  $\frac{1}{4}$(2S + 4S²/2S )

r² =  $\frac{1}{4}$(2S + 2S)

r² =  $\frac{1}{4}$(4S)

r² = S

r = √S

So, c = 2r = 2√S

Hence, the sides of a triangle ABC i.e., hypotenuse, height, and base are 2√S, √2S, and √2S respectively.

To learn more about triangles, click on the link below:

https://brainly.in/question/17424774

To learn more about hypotenuse, click on the link below:

https://brainly.in/question/1317668

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