plz fast.... it's very urgent...tomorrow is my test....
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O is the centre of concentric circles.
OA=6 cm and OB= 4 cm.
PA and PB are the tangents to the outer and inner circle respectively.
PA=10 cm.
We know that ,a tangent to a circle is perpendicular to he radius through the point of contact .<OAP=<OBP=90 degree.
in right triangle OAP,
OP square = OA square + PA square
OP square=(6 cm)square +(10cm)square =(36cm)square +(100 cm)square = (136cm ) sqaure
OP=square root of 136
In right,triangle OBP,
OP square = OB square + PB sqaure
PB square=OP square-OB square
PB square = (square root of 136) square -(4 cm)square
=(136cm)square -(16 cm)square =120 cm square
PB= square root of120 cm
PB =30 cm
OA=6 cm and OB= 4 cm.
PA and PB are the tangents to the outer and inner circle respectively.
PA=10 cm.
We know that ,a tangent to a circle is perpendicular to he radius through the point of contact .<OAP=<OBP=90 degree.
in right triangle OAP,
OP square = OA square + PA square
OP square=(6 cm)square +(10cm)square =(36cm)square +(100 cm)square = (136cm ) sqaure
OP=square root of 136
In right,triangle OBP,
OP square = OB square + PB sqaure
PB square=OP square-OB square
PB square = (square root of 136) square -(4 cm)square
=(136cm)square -(16 cm)square =120 cm square
PB= square root of120 cm
PB =30 cm
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