The radius of a circle with centre O is 15 cm find the length of the tangent drawn from an external point which is 20cm away from the centre of the circle
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PQ & PR are 2 tangents and QO & OR are 2 radius at contact point Q & R.
Angle PQO=90°
[A TANGENT TO A CIRCLE IS PERPENDICULAR TO THE RADIUS THROUGH THE POINT OF CONTACT]
By Pythagoras theorem
PQ²= OP² - OQ²
PQ² = 13²- 5² = 169- 25= 144
PQ= √ 144= 12
PQ=12cm
PQ= PR =12cm
[The Lengths of two tangents drawn from an external point to a circle are equal]
In ∆OPQ & ∆ OPR
OQ= OR (5cm) given
OP = OP ( Common)
PQ= PR( 12cm)
Hence ∆OPQ =~ ∆OPR ( by SSS congruence)
Area of ∆OPQ =Area ∆OPR
Area of quadrilateral QORP= 2×(area of ∆ OPR)
Area of quadrilateral QORP= 2× 1/2 × base × altitude
Area of quadrilateral QORP= OR× PR
Area of quadrilateral QORP=12× 5= 60 cm²
_____________________________
Area of quadrilateral QORP=60cm²
_____________________________
Hope this will help you..
PQ & PR are 2 tangents and QO & OR are 2 radius at contact point Q & R.
Angle PQO=90°
[A TANGENT TO A CIRCLE IS PERPENDICULAR TO THE RADIUS THROUGH THE POINT OF CONTACT]
By Pythagoras theorem
PQ²= OP² - OQ²
PQ² = 13²- 5² = 169- 25= 144
PQ= √ 144= 12
PQ=12cm
PQ= PR =12cm
[The Lengths of two tangents drawn from an external point to a circle are equal]
In ∆OPQ & ∆ OPR
OQ= OR (5cm) given
OP = OP ( Common)
PQ= PR( 12cm)
Hence ∆OPQ =~ ∆OPR ( by SSS congruence)
Area of ∆OPQ =Area ∆OPR
Area of quadrilateral QORP= 2×(area of ∆ OPR)
Area of quadrilateral QORP= 2× 1/2 × base × altitude
Area of quadrilateral QORP= OR× PR
Area of quadrilateral QORP=12× 5= 60 cm²
_____________________________
Area of quadrilateral QORP=60cm²
_____________________________
Hope this will help you..
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