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In the given figure, AP, AQ and BC are tangents to the circle with centre O. If AB = 6 cm, AC = 7 cm and BC = 5 cm, then what is the length of AP?
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Solution: (a) $AP = AB + BP = AB + BR$, $AQ = AC + CQ = AC + CR$ (1/2 Mark)
Adding corresponding sides, $AP + AQ = AB + (BR + CR) + AC$ (1/2 Mark)
Since $AP = AQ$,
$\therefore 2AP = AB + BC + AC = 18$ (1/2 Mark)
$\Rightarrow AP = 9$ cm (1/2 Mark)
Adding corresponding sides, $AP + AQ = AB + (BR + CR) + AC$ (1/2 Mark)
Since $AP = AQ$,
$\therefore 2AP = AB + BC + AC = 18$ (1/2 Mark)
$\Rightarrow AP = 9$ cm (1/2 Mark)