Question

In The given figure , PQ, PR and BC are the
tangents to the circle BC touches the circle
at x if P Q = 7cm , then the perimeter
4 triangle
PBC
is​

Answer 1

Correct question :-

In the given figure , PQ, PR and BC are the

tangents to the circle. BC touches the circle

at X. If PQ = 7cm , then the perimeter ΔPBC.

Solution :-

We know that

The lengths of tangents drawn from an external point of a circle are equal

CR and XR are two tangents drawn from an external point C

⇒ CR = XC

BQ and XB are two tangents drawn from an external point B

⇒ BQ = XB

PR and PQ are two tangents drawn from an external point P

⇒ PR = PQ = 7 cm

Perimeter of ΔPBC = PC + PB + BC

From figure,

= PC + PB + (BX + CX)

= PC + CX + BX + PB

= (PC + CR) + (BQ + PB)

[ Because CX = CR, BQ = BX ]

= PR + PQ

= 7 + 7

[ Because PR = PQ = 7 CM ]

= 14 cm

Therefore, the perimeter of ΔPBC is 14 cm.

Answer 2

Correct Question:

In the given figure , PQ, PR and BC are the

tangents to the circle. BC touches the circle

at X. If PQ = 7cm , then the perimeter ΔPBC.

Answer:

$\large\boxed{\sf{14\;\;cm}}$

Step-by-step explanation:

°.° Tangents from an external point to a circle are equal.

.°. CR = CX ........(1)

and, BQ = BX ...........(2)

And, PR = PQ = 7 cm ...........(3)

But, PR = PC + CR

=> PR = PC + CX ........( from 1 )

Also, PQ = PB + BQ

=> PQ = PB + BX ...............( from 2 )

Now, Perimeter of ∆PBC

= PB + BC + PC

= PB + BX + CX + PC ........(°.° BC = BX + CX )

= PR + PQ

= 7 + 7 ...........( from 3 )

= 14 cm

Hence, perimeter if ∆PBC is 14 cm