PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
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Given :
PQ is a chord of length 8 cm of a Circle
of radius 5cm .
The Tangents at P and Q intersect at a
point T.
Let OT intersects PQ at a point R.
We know that ,
OP perpendicular to PQ.
and OR bisects PQ .
PR = PQ/2 = 8/2 = 4cm
In ∆PRO ,
<PRO = 90° ,
PR = 4 cm , OR = 5 cm
By Pythagorean theorem ,
OR² + PR² = OP²
OR² = OP² - PR²
= 5² - 4²
= 25 - 16
= 9
OP = 3 cm
In ∆OPT and ∆ORP ,
<OPT = <ORP = 90°
<TOP = <TOR ( common angle )
Therefore ,
∆OPT ~ ∆ORP
TP/PR = OP/OR
[ Corresponding ratios in proportion ]
TP/4 = 5/3
=> TP = ( 5 × 4 )/3
=> TP = 20/3 cm
*****
PQ is a chord of length 8 cm of a Circle
of radius 5cm .
The Tangents at P and Q intersect at a
point T.
Let OT intersects PQ at a point R.
We know that ,
OP perpendicular to PQ.
and OR bisects PQ .
PR = PQ/2 = 8/2 = 4cm
In ∆PRO ,
<PRO = 90° ,
PR = 4 cm , OR = 5 cm
By Pythagorean theorem ,
OR² + PR² = OP²
OR² = OP² - PR²
= 5² - 4²
= 25 - 16
= 9
OP = 3 cm
In ∆OPT and ∆ORP ,
<OPT = <ORP = 90°
<TOP = <TOR ( common angle )
Therefore ,
∆OPT ~ ∆ORP
TP/PR = OP/OR
[ Corresponding ratios in proportion ]
TP/4 = 5/3
=> TP = ( 5 × 4 )/3
=> TP = 20/3 cm
*****
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