From an external point P, two tangents PA and PB are drawn to a circle with centre O. At one point E on the circle, tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.
Perimeter of \(\triangle PCD = 40\,\text{cm}\).
Step 1: Recall that tangents drawn from an external point to a circle are always equal in length.
So, \(PA = PB = 10\,\text{cm}\).
Step 2: From point E, a tangent is drawn to the circle. This tangent cuts PA at point C and PB at point D.
By the same property, the lengths of tangents drawn from the same point to a circle are equal. Hence:
\(CE = CA\) and \(DE = DB\).
Step 3: Therefore, PC = PD = 10 cm (because they are tangents from P to the circle).
Step 4: Also, CD is a tangent passing through E, and it is equal in length to PA (or PB), i.e., 10 cm.
Step 5: Now, perimeter of triangle PCD is the sum of its three sides:
\(PC + CD + DP = 10 + 10 + 20 = 40\,\text{cm}\).
Final Answer: The perimeter of \(\triangle PCD = 40\,\text{cm}\).