$\vartriangle ABC$ is an isosceles right triangle with area P. The radius of the circle that passes through the points A, B and C is :
(a) $\sqrt{P}$
(b) $\sqrt{\dfrac{P}{2}}$
(c) $\dfrac{\sqrt{P}}{2}$
(d) $\sqrt{2P}$

Hint: We know that the angle subtended by the diameter is a right angle. Using the converse of this theorem, we can find the centre(O) of our circle. We then need to prove that $\angle AOB=\angle AOC={{90}^{\circ }}$ , by congruence of $\vartriangle AOB\text{ and }\vartriangle AOC$ . Now, we can equate the area P with the area taking BC as base and OA as height of the triangle. Hence, we can find the radius.

Complete step by step solution:
We are given that $\vartriangle ABC$ is an isosceles right triangle. So, two of its sides will be equal.
So, we have AB = AC.
The circle passes through all of its points A, B and C. Thus, the figure is
Now, we know that the angle subtended by a diameter on the same circle is a right angle. Thus, by the converse of this theorem, we can say that if the angle subtended on the circle is a right angle, then that angle is subtended by the diameter.
Since $\angle BAC={{90}^{\circ }}$ , we can say that BC is the diameter of our circle.
Let us assume a point O, that is the midpoint of BC. Then, O will also the centre of our circle.
Let us construct a line AO, as shown in the figure below

Now, in $\vartriangle AOB\text{ and }\vartriangle AOC$ , we have
AB = AC (it is given that $\vartriangle ABC$ is an isosceles triangle)
OB = OC (both are radius of the circle)
AO = AO (common)
Hence, by SSS congruence criteria, it is clear that $\vartriangle AOB\cong \vartriangle AOC$ .
Now by corresponding angles of congruent triangles, we can say that
$\angle AOB=\angle AOC...\left( i \right)$
But BC is a straight line and the sum of angles of a straight line is ${{180}^{\circ }}$ .
Hence, $\angle AOB+\angle AOC={{180}^{\circ }}$
Using equation (i), we can write
$\angle AOB+\angle AOB={{180}^{\circ }}$
$\Rightarrow 2\angle AOB={{180}^{\circ }}$
$\Rightarrow \angle AOB=\angle AOC={{90}^{\circ }}$
Also, OA, OB and OC all are the radius of our circle. Hence, OA = OB = OC.
Let the radius of the circle be r.
Then, OA = OB = OC = r.
So, now our figure looks like this.

Now, we are given that are area of our triangle $\vartriangle ABC$ is P.
$Area=P$
$\dfrac{1}{2}\times BC\times OA=P$
In $\vartriangle ABC$ , we have BC = 2r and OA = r. Therefore,
$\dfrac{1}{2}\times 2r\times r=P$
$\Rightarrow {{r}^{2}}=P$
$\Rightarrow r=\sqrt{P}$
So, the radius of the circle is $\sqrt{P}$ .

So, the correct answer is “Option A”.

Note: Some students may directly equate the area found by taking BC as base and OA as height with P, which will be an incomplete solution. We will first have to show that the line OA is perpendicular to line BC. We should take care while writing the corresponding angles of congruent triangles, to avoid any confusion.