Verified answer

Answer:

Step-by-step explanation:

Given that the inner circle has a radius OB=4 cm

The outer circle has a radius OA=6 cm

The ∠OAP=90°; since OA is radius and PA is tangent, OA⊥PA

So, the ΔOAP is a right-angled triangle.

In ΔOAP OA is perpendicular, PA is the base and OP is the hypotenuse.

OA=6 cm

PA=10 cm.

By Pythagoras theorem, OP=[tex]\sqrt{OA^{2}+PA^{2} }[/tex]

∴, OP=[tex]\sqrt{6^{2}+10^{2} }[/tex]

⇒ OP=[tex]\sqrt{36+100}[/tex]

⇒ OP=[tex]\sqrt{136}[/tex]

Now, for the inner circle, OB is the radius and PB is the tangent to the circle.

∴, OB⊥PB

∴, ΔOPB is a right-angled triangle.

OB is the perpendicular, PB is the base and OP is the hypotenuse.

By Pythagoras theorem, [tex]OP^{2} =OB^{2}+PB^{2}[/tex]

Now, OP=[tex]\sqrt{136}[/tex] cm, OB=4 cm

∴, [tex]OP^{2}=OB^{2} + PB^{2}[/tex]

⇒ 136 = 4² + [tex]PB^{2}[/tex]

⇒  [tex]PB^{2}=136 - 16[/tex]

⇒  [tex]PB^{2} = 120[/tex]

⇒ [tex]PB=\sqrt{120}[/tex]

⇒ [tex]PB = 10.9[/tex]

Hence, The length of PB is 10.9 cm