If tangent PQ at a point P of a circle of a radius 5cm meets a line through the centre O at a point so that OQ = 12cm .Fine length of PQ.
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HEY DEAR ... ✌
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JUST CHECK THE PIC .
HOPE , IT HELPS ... ✌
____________________________
JUST CHECK THE PIC .
HOPE , IT HELPS ... ✌
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![](https://hi-static.z-dn.net/files/de5/343e8080d6248586643c9b0240ddab46.jpg)
Answered by
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aloha user^^
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given: an external point Q, tangent PQ and a circle with radius 5 cm.
to find: the length PQ
please see the attachment for help.
solution:
we know that a tangent makes 90° angle with the radius of the circle.
so, angle OPQ = 90°
in order to find the length of PQ we are going to use the Pythagoras theorem.
which is:
![{h}^{2} = {b}^{2} + {p}^{2} {h}^{2} = {b}^{2} + {p}^{2}](https://tex.z-dn.net/?f=+%7Bh%7D%5E%7B2%7D+%3D+%7Bb%7D%5E%7B2%7D+%2B+%7Bp%7D%5E%7B2%7D+)
![{12}^{2} = {5}^{2} + {pq}^{2} {12}^{2} = {5}^{2} + {pq}^{2}](https://tex.z-dn.net/?f=+%7B12%7D%5E%7B2%7D+%3D+%7B5%7D%5E%7B2%7D+%2B+%7Bpq%7D%5E%7B2%7D+)
144 - 25 = pq^2
![119 = {pq}^{2} 119 = {pq}^{2}](https://tex.z-dn.net/?f=119+%3D+%7Bpq%7D%5E%7B2%7D+)
PQ =
![\sqrt{119} cm \sqrt{119} cm](https://tex.z-dn.net/?f=+%5Csqrt%7B119%7D+cm)
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___________
given: an external point Q, tangent PQ and a circle with radius 5 cm.
to find: the length PQ
please see the attachment for help.
solution:
we know that a tangent makes 90° angle with the radius of the circle.
so, angle OPQ = 90°
in order to find the length of PQ we are going to use the Pythagoras theorem.
which is:
144 - 25 = pq^2
PQ =
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Attachments:
![](https://hi-static.z-dn.net/files/d89/23b803b0d81e49d8cafc73329e94cfcc.jpg)
locomaniac:
ty :)
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