Question

In PQR, Q = 90, seg QM is the median. PQ² + QR² = 169. Draw a circumcircle of PQR.

Answer 1

$$\begin{gathered} \mathrm{Since}∆\mathrm{PQR}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}. \\ \mathrm{So},{\mathrm{PR}}^2={\mathrm{PQ}}^2+{\mathrm{QR}}^2(\mathrm{Phythagoras}\mathrm{theorem}) \\ {\mathrm{PR}}^2=169\left(\mathrm{Given}\right) \\ {\mathrm{PR}}^2={13}^2 \\ \mathrm{PR}=13\mathrm{units} \end{gathered}$$

As we know that if a right triangle is inscribed in a circle, then the hypotenuse will be a diameter of the circle.
So, PR is the diameter of the circumcircle.
Radius = $\frac{13}{2}=6.5\mathrm{units}$

Steps of construction:

1. Draw a circle with centre M and a radius of 6.5 units.
2. Draw the diameter PR of the circle.
3. Take a point Q on the circle.
4. Join PQ and RQ.