In the figure, tangents and are drawn to a circle with centre from an external point . If is a tangent to the circle at and = 15 cm, find the perimeter of $Δ$ PCD

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Solution

We know that the lengths of the two tangents from an exterior point to a circle are equal.

CA = CE, DB = DE and PA = PB.

Now, the perimeter of $Δ$ PCD = PC + CD + DP

= PC + CE + ED + DP

= PC + CA + DB + DP

= PA + PB = 2 PA (PB = PA)

Thus, the perimeter of $Δ$ PCD = 2 $\times$ 15 = 30 cm.

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Q. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at the point E and PA = 14 cm, find the perimeter of ∆PCD.
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Q. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

From an external point P, 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 $P A = 14 c m$ , find the perimeter of $Δ P C D .$

Q. In the given figure, from an external point

P

, tangent

P X

and

P Y

are drawn to a circle with centre

O

A B

is another tangent to the circle at

C

and

P X = 14 c m

, then find the perimeter of

△ P A B

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of $Δ$ PCD.

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In the figure, tangents and are drawn to a circle with centre from an external point . If is a tangent to the circle at and = 15 cm, find the perimeter of ΔPCD

We know that the lengths of the two tangents from an exterior point to a circle are equal. CA = CE, DB = DE and PA = PB. Now, the perimeter of ΔPCD = PC + CD + DP = PC + CE + ED + DP = PC + CA + DB + DP = PA + PB = 2 PA (PB = PA) Thus, the perimeter of ΔPCD = 2×15 = 30 cm.

In the figure, tangents and are drawn to a circle with centre from an external point . If is a tangent to the circle at and = 15 cm, find the perimeter of $Δ$ PCD

We know that the lengths of the two tangents from an exterior point to a circle are equal.

CA = CE, DB = DE and PA = PB.

Now, the perimeter of $Δ$ PCD = PC + CD + DP

= PC + CE + ED + DP

= PC + CA + DB + DP

= PA + PB = 2 PA (PB = PA)

Thus, the perimeter of $Δ$ PCD = 2 $\times$ 15 = 30 cm.