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(i) PM = MQ ( Since PQ is bisected by the diameter AB)
also , OM is perpendicular to PQ ( as the theorem says , Line going through center bisect any side then it is perpendicular to that side)
Now, Join PO and QO
PQ = 15cm , so PM = 15/2 cm
In ∆OMP
by Pythagoras Theorem,
OP² = OM² + PM²
15² = 9² + PM² { since radius is given , 15cm)
PM = √15² - 9²
PM = √225-81
PM = √144 = 12cm
(ii) Join AP and AQ,
AM = radius+OM
AM = 15+9
AM = 24cm
In ∆AMP
by Pythagoras Theorem,
AP = √AM² + PM²
AP = √576+144
AP = √720 = 26.83cm
(iii)
Join PB and BQ
BM = radius - OM
BM = 15-9
BM = 6cm
In ∆BMP
by Pythagoras Theorem,
BC² = BM² + PM²
BC = √6² + 12²
BC = √36+144
BC = √180 = 13.41cm
also , OM is perpendicular to PQ ( as the theorem says , Line going through center bisect any side then it is perpendicular to that side)
Now, Join PO and QO
PQ = 15cm , so PM = 15/2 cm
In ∆OMP
by Pythagoras Theorem,
OP² = OM² + PM²
15² = 9² + PM² { since radius is given , 15cm)
PM = √15² - 9²
PM = √225-81
PM = √144 = 12cm
(ii) Join AP and AQ,
AM = radius+OM
AM = 15+9
AM = 24cm
In ∆AMP
by Pythagoras Theorem,
AP = √AM² + PM²
AP = √576+144
AP = √720 = 26.83cm
(iii)
Join PB and BQ
BM = radius - OM
BM = 15-9
BM = 6cm
In ∆BMP
by Pythagoras Theorem,
BC² = BM² + PM²
BC = √6² + 12²
BC = √36+144
BC = √180 = 13.41cm
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