In the given circle with centre ‘O’ ACB, AE and BD are the tangents. If AB=12cm, AE=3cm the length of BD is A. 6 cm B. 3 cm C. 8 cm D. 9 cm
Answers
Answer:
20 cm.
solution
Step-by-step explanation:
OP=OQ=5cm
OT=13cm
OP and PT are radius and tangent respectively at contact point P.
∴∠OPT=90
∘
So, by pythagoras theorem, in right angled ΔOPT,
PT
2
=OT
2
−OP
2
=13
2
−5
2
=169−25=144
⇒PT=12cm
AP and AE are two tangents from an external point A to a circle.
∴AP=AE
AEB is tangent and OE is radius at contact point E.
So, AB⊥OT ___(i)
So, by Pythagoras theorem, in right angled. ΔAET.
AE
2
=AT
2
−ET
2
⇒AE
2
=(PT−PA)
2
−[TO−OE]
2
=(12−AE)
2
−(13−5)
2
⇒AE
2
=(12)
2
+(AE)
2
−2(12)(AE)−(8)
2
⇒AE
2
−AE
2
+24AE=144−64
⇒24AE=80
⇒AE=
24
80
cm
⇒AE=
3
10
cm
REF. IMAGE
In ΔTPO and ΔTQO,
OT=OT [common]
PT=QT [Tangents from T]
OP=OQ [Radii of same circle]
∴ΔTPO≅ΔTQO [By SSS criterion of congruence]
⇒∠1=∠2 ___(ii) [CPCT]
In ΔETA and ΔETB,
ET=ET [Common]
∠TEA=∠TEB=90
∘
[From (i)]
∠1=∠2 [CPCT] [From (ii)]
∴ΔETA≅ΔETB [By ASA criterion of congruence]
⇒AE=BE [CPCT]
⇒AB=2AE=2×
3
10
⇒AB=
3
20
cm
Hence, the required length is
3
20
cm.
solutionOP=OQ=5cm
OT=13cm
OP and PT are radius and tangent respectively at contact point P.
∴∠OPT=90
∘
So, by pythagoras theorem, in right angled ΔOPT,
PT
2
=OT
2
−OP
2
=13
2
−5
2
=169−25=144
⇒PT=12cm
AP and AE are two tangents from an external point A to a circle.
∴AP=AE
AEB is tangent and OE is radius at contact point E.
So, AB⊥OT ___(i)
So, by Pythagoras theorem, in right angled. ΔAET.
AE
2
=AT
2
−ET
2
⇒AE
2
=(PT−PA)
2
−[TO−OE]
2
=(12−AE)
2
−(13−5)
2
⇒AE
2
=(12)
2
+(AE)
2
−2(12)(AE)−(8)
2
⇒AE
2
−AE
2
+24AE=144−64
⇒24AE=80
⇒AE=
24
80
cm
⇒AE=
3
10
cm
REF. IMAGE
In ΔTPO and ΔTQO,
OT=OT [common]
PT=QT [Tangents from T]
OP=OQ [Radii of same circle]
∴ΔTPO≅ΔTQO [By SSS criterion of congruence]
⇒∠1=∠2 ___(ii) [CPCT]
In ΔETA and ΔETB,
ET=ET [Common]
∠TEA=∠TEB=90
∘
[From (i)]
∠1=∠2 [CPCT] [From (ii)]
∴ΔETA≅ΔETB [By ASA criterion of congruence]
⇒AE=BE [CPCT]
⇒AB=2AE=2×
3
10
⇒AB=
3
20
cm
Hence, the required length is
3
20
cm.
solution
Answer:
Answer is 9 cm
The tangents drawn from external point to a circle are equal.
so AE=AC= 3 cm
so, BC=AB -AC
= 12- 3
9 cm
Here,the tangents BC and BD are tangents from common point B.
so, BC = BD= 9 cm
Step-by-step explanation:
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