Question

in the given figure , PQ is tangent to outer circle and PR is tangent to inner circle . if PQ =4cm , OQ=3cm , and QR = 2 cm them find the length of PR
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Answer 1

Answer:

In ∆OQP

<OQP = 90°

and PQ = 4cm, OQ = 3cm

so, OP = √(4²+3²) = √(16+9) = √25 = 5cm

now, in ∆POR

< PRO = 90°

PO = 5cm , OR = 2cm

so, PR = √(5²-2²)

= √(25-4)

=√21cm

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Answer 2

Given:

• PQ is tangent to outer circle and PR is tangent to inner circle.
• PQ = 4cm
• OQ = 3cm
• OR = 2cm

Find:

• Length of PR

Solution:

In $\triangle$ OPQ

$\rm \boxed{\rm{H^2 = P^2 + B^2}} \qquad \qquad \big\lgroup{Pythogoras \: Theorem} \big\rgroup$

$\rm \dashrightarrow {OP}^2 = {OQ}^2 + {PQ}^2$

where,

• OQ = 3cm
• PQ = 4cm

So,

$\rm \dashrightarrow {OP}^2 = {OQ}^2 + {PQ}^2$

$\rm \dashrightarrow {OP}^2 = {3}^2 + {4}^2$

$\rm \dashrightarrow {OP}^2 =9+ 16$

$\rm \dashrightarrow {OP}^2 =25$

$\rm \dashrightarrow OP = \sqrt{25}$

$\rm \dashrightarrow OP = 5cm$

_________________________

Now, Since PR is tangent to inner circle and OR is its radius then

$\angle$ ORP = 90°

[ $\because$ The tangent is perpendicular to the radius of the Circle at the point of Contact]

Now, In $\triangle$ ORP

$\rm \boxed{\rm{H^2 = P^2 + B^2}} \qquad \qquad \big\lgroup{Pythogoras \: Theorem} \big\rgroup$

$\rm \dashrightarrow {OP}^2 = {OR}^2 + {PR}^2$

where,

• OP = 5cm
• OR = 2cm

So,

$\rm \dashrightarrow {5}^2 = {2}^2 + {PR}^2$

$\rm \dashrightarrow 25 = 4+ {PR}^2$

$\rm \dashrightarrow 25 - 4 = {PR}^2$

$\rm \dashrightarrow 21= {PR}^2$

$\rm \dashrightarrow \sqrt{21} cm= PR$

$\rm \dashrightarrow PR = \sqrt{21} cm$

$\footnotesize{ \rm \therefore PR = \sqrt{21} cm}$

Hence, PR = √(21) cm