Question

#BrainTeaser

The equation of the curve in the figure is y = (x^2)/2.
Then, find the area of semicircle in the figure.

The triangle in the figure is equilateral.

Answer 1

Answer:

agar y =length of triangle hai to

area of semi circle = πr2/2

{22/7x(1/2*X2/2)2}/2

Answer 2

Topic: Coordinate Geometry

Concepts

Equilateral triangle: A triangle in which all side lengths are equal.

→ As a characteristic of an equilateral triangle, the height of it is $\dfrac{\sqrt{3} }{2}$ times one side.

Quadratic function: A function with the highest degree of 2.

→ A quadratic function is a parabola and symmetric against $x=-\dfrac{b}{2a}$.

So let's solve our problem!

Solution

We are given that one vertice of an equilateral as a vertex of the graph. The vertex is $C(0,0)$.

Since two vertexes of an equilateral triangle lie on the graph $y=\dfrac{x^2}{2}$, the vertices are $A(x,\dfrac{x^2}{2} )$ and $B(-x,\dfrac{x^2}{2} )$.

Then according to the characteristic of the equilateral triangle,

$h=\dfrac{\sqrt{3} }{2} \times 2x=\sqrt{3} x$

We obtain two vertices as,

$A(x,\sqrt{3} x)$, $B(-x,\sqrt{3} x)$

Now we can equate the y-vertices,

$\rightarrow\dfrac{x^2}{2} =\sqrt{3} x$

$\rightarrow x^2 =2\sqrt{3} x$

$\rightarrow x^2-2\sqrt{3} x=0$

$\rightarrow x(x-2\sqrt{3} )=0$

$\rightarrow x=0\ \mathrm{or}\ x=2\sqrt{3}$

Solving the equation, we reject $x=0$ since three vertices cannot form a $\triangle ABC$. Hence there leaves $x=\sqrt{3}$ as a sole solution.

Since $\overline{AB}$ is the diameter of the semi-circle,

$\rightarrow\overline{AB}=2r$

$\rightarrow 4\sqrt{3} =2r$

$\rightarrow r=2\sqrt{3}$

Conclusion

Hence, the area of the semi-circle is,

$\dfrac{1}{2} \pi r^2=\boxed{6 \pi \ \mathrm{units^2}}$