From an external point P tangents PA and PB are drawn to a circle with center O, If CD is tangent to circle at the point E and PB=16 cms, AC=7cms, CD=12 cms then the perimeter of PCD is

The correct answer is A, 32 cm. The question rests on the equal-tangent theorem: two tangents drawn to a circle from the same external point are equal in length. So PA = PB = 16 cm, and C lies on PA while D lies on PB, with CD a third tangent touching the circle at E.

Apply the same theorem at C and D. From C, the tangents are CA and CE, so CE = CA = 7 cm. From D, the tangents are DB and DE, so DE = DB. Since CD = CE + ED = 12 cm, ED = 5 cm, which gives DB = 5 cm and PD = 16 - 5 = 11 cm. Also PC = PA - AC = 9 cm.

Perimeter of PCD = PC + CD + PD = 9 + 12 + 11 = 32 cm. In general, the perimeter of a triangle formed by a third tangent between two tangents from P always equals PA + PB, which is a standard shortcut you can use in the exam.

Option B, 31 cm, and option C, 35 cm, do not follow from the tangent lengths. Option D is wrong because 32 cm is listed.

Trap: Aspirants often try to compute PC and PD using the wrong segments, or they assume CD is measured on a different path. Once you see that CE = CA and DE = DB, the perimeter collapses directly to PA + PB, with no need for the circle's radius.