The least area of a circle circumscribing any | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the legs of the right triangle be $x$ and $y$. The area of the triangle is $S = \frac{1}{2}xy$,so $xy = 2S$.The hypotenuse of the right triang

Vedclass · Accepted Answer

$\pi S$

Vedclass · Answer

(A) Let the legs of the right triangle be $x$ and $y$. The area of the triangle is $S = \frac{1}{2}xy$,so $xy = 2S$.The hypotenuse of the right triangle is $h = \sqrt{x^2 + y^2}$.$A$ circle circumscribing a right triangle has the hypotenuse as its diameter. Thus,the diameter $d = \sqrt{x^2 + y^2}$,and the radius $r = \frac{\sqrt{x^2 + y^2}}{2}$.The area of the circle is $A = \pi r^2 = \pi \left( \frac{\sqrt{x^2 + y^2}}{2} \right)^2 = \frac{\pi}{4}(x^2 + y^2)$.Since $y = \frac{2S}{x}$,we substitute this into the area formula:$A(x) = \frac{\pi}{4} \left( x^2 + \left( \frac{2S}{x} \right)^2 \right) = \frac{\pi}{4} \left( x^2 + \frac{4S^2}{x^2} \right)$.To find the minimum area,we differentiate $A(x)$ with respect to $x$ and set it to zero:$A'(x) = \frac{\pi}{4} \left( 2x - \frac{8S^2}{x^3} \right) = 0$.$2x = \frac{8S^2}{x^3} \implies x^4 = 4S^2 \implies x^2 = 2S$.Substituting $x^2 = 2S$ back into the area equation:$A = \frac{\pi}{4} \left( 2S + \frac{4S^2}{2S} \right) = \frac{\pi}{4} (2S + 2S) = \frac{\pi}{4} (4S) = \pi S$.Thus,the least area of the circle is $\pi S$.

The least area of a circle circumscribing any right triangle of area $S$ is

Similar Questions

The number $28$ is divided into two positive parts such that the sum of the cube of one part and the square of the other part is minimum. Find the absolute difference between the two parts.

The difference between the greatest and least values of the function $f(x) = -x + \sin 2x$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is

The sum of all the local minimum values of the twice differentiable function $f: R \rightarrow R$ defined by $f(x)=x^{3}-3 x^{2}-\frac{3 f^{\prime \prime}(2)}{2} x+f^{\prime \prime}(1)$ is:

The function $f(x) = x e^{-x}, \forall x \in R$ attains a maximum value at $x$ equal to:

The maximum value of $xy$ subject to $x+y=7$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module