Question

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA=12 cm,QC=QD= 3 cm, then find PC + PD.

Answer 1

Answer:

PC + PD = 18 cm

Explanation:

Given :

PA = 12 cm  &  QC = QD = 3 cm

From property of tangent we have :

PA = PB  [ Where P is external point ]

From point Q we have :

BD = QD & AC = QC

We can write :

PB as :

PB = PD + BD ... ( i )  similarly for PA

PA = PC + AC ... ( ii )

Adding both ( i )  and  ( ii ) :

PB + PA = PD + BD + PC + AC

We have value :

PA = PB = 12 cm & BD = QD = AC = QC = 3 cm

12 + 12 = 3 + 3 + PC + PD

PC + PD = 24 - 6

PC + PD = 18 cm

Hence , we get PC + PD = 18 cm .

Answer 2

Given:

• PA and PB are tangents to the circle from external point P.
• CD is another tangent that touches the circle at Q.
• PA = 12 cm
• QC = QD = 3 cm

To Find:

• The measure of PC + PD.

Solution:

• We already know that the lengths of any tangents drawn externally on any point of a circle are equal.
• From the figure, we get to know that, PA = PB = 12 cm
• AC and QC are two tangents drawn externally from point C.
• So, AC = QC = 3 cm
• The same way, QD = BD = 3 cm
• From the given figure, PC = PA - AC
• PC = 12 - 3 = 9 cm
• PD = PB-BD
• PD = 12 - 3 = 9 cm
• ∴ PC+PD = 9+9 = 18 cm.

∴ PC+PD = 18 cm.