Two Most Favorite Shapes Together

Since, the perimeter is $p$ , then the side of the equilateral triangle triangle is $\frac{p}{3}$

now the altitude of the equilateral triangle is $\frac{\sqrt{3}}{2}$ * $\frac{p}{3}$

the radius of the circle is $\frac{2}{3}$ of the altitude of the equilateral triangle

i.e. $\frac{2}{3}$ . $\frac{\sqrt{3}}{2}$ . $\frac{p}{3}$ = $\frac{p}{3\sqrt{3}}$

area of the circle is $\ πr^{2}$ = $\frac{πp^{2}}{27}$

Perimeter of the equilateral triangle = $p$ .
So, its side = $\Large \frac {p}{3}$
Now we know that radius of the circumcircle of an equilateral triangle = $\Large \frac {a}{\sqrt3}$ where $a$ is side of the equilateral triangle.
So, for the given triangle, radius = $\Large \frac {p}{3×\sqrt3}$
Now, area of a circle is $\pi × r^2$ where $r$ is the radius.
So, area of the required circle = $\pi$ × $\Large [{\frac {p}{3×\sqrt3}}]^2$
= $\Large \frac {\pi p^2}{27}$ . $_\square$

let M be the midpoint of XY and O be the centre of the circle.

OMX is a right angled triangle, whose hypotenuse is the radius, whose angle XOM is 60 degrees and whose longer short side is $\frac{P}{6}$ .

Easy trigonometry then gives the radius as

$\frac{P}{3\sqrt{3}}$

and so the area of the circle is $\frac{\pi P^2}{27}$

If r is the radius of circle, then height h of triangle is 3r/2 as center of circle would divide height of triangle in 2:1 ratio. This can be shown by using the properties of similar triangle. If O is center and M is mid point of side YZ, then triangles OMY and XMY are similar. Therefore, OM/OY=YM/XY=1/2. OY=OX=r and h=XO+OM=3r/2. Pythagoras theorem on half triangle XMY, with sides p/3, p/6 and h, gives h^2=p^2/9 - p^2/36. As h=3r/2, we get 9r^2/4=p^2/12 or r ^2=p^2/27. Area of circle=πp^2/27.