Given;
XY = 30, XZ = 24, JQ = 8
From the given diagram we can see that;
XJ = JY
So,
XJ = JY = XY/2
XJ = JY = 30/2 = 15
Here, Point Q is the centroid of the ∆XYZ, which means that XQ = QZ = QY will be the radius of the circumscribed circle.
Since,
We know JQ and XJ, let us consider triangle XJQ.
By Using Pythagoras Theorem;
(XQ)² = (XJ)² + (JQ)²
(XQ)² = (15)² + (8)²
(XQ)² = 225 + 64
(XQ)² = 289 +
XQ = √289
XQ = 17
Thus, The radius of the circumscribed circle of ∆XYZ is 17
-TheUnknownScientist 72
