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Answer 1

Answer:

hey here is your answer

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Step-by-step explanation:

so here for an adjoining circle

QM and QN are radius

thus QM=QN=9 cm (radii of same circle)

Now PM is a tangent to given circle at point of contact M

so by Tangent Theorem

We get

Angle QMP=90

So considering ∆QMP

As Angle QMP=90

Applying Pythagoras Theorem

we get

QP²=QM²+PM²

(41)²=(9)²+PM²

PM²=(41)²-(9)²

=1681-81

=1681

Taking square roots on both sides we get

PM=40. cm

so as here PM and PN are tangent segments drawn from external point P to adjoining circle

By Tangent Segment Theorem

We get

PM=PN= 40. cm

So now

Perimter of □QMPN=QM+MP+PN+QN

=40+40+9+9

=80+18

=98 cm

so hence

Perimter of □QMPN=98.cm

Option (C) is ryt