Question

In a fig. AD=8cm AC=6cm and TB is the tangent at B to the circle with centre O find OT if BT is 4cm

Answer 1

Given:

AD= 8 cm , AC= 6 cm, BT = 4 cm

Angle CAD= 90°

[Angle in a semicircle]

In ∆ ACD

CD²= AC²+ AD²

[By Pythagoras theorem]


CD² = 6²+8²

CD²= 36 +64

CD²= 100

CD = √100= 10

CD= 10 cm

OC=OD=OB= 10/2=5 cm

[ Radius of a circle]

Angle OBT = 90°

[Angle between radius and tangent]

In ∆ OBT

OT²= OB²+BT²

OT²= 5² +4²

OT²= 25+16

OT =√41 cm


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Hope this will help you..

Answer 2

AD= 8 cm , AC= 6 cm, BT = 4 cm

Angle CAD= 90°

[Angle in a semicircle]

In ∆ ACD

CD²= AC²+ AD²

[By Pythagoras theorem]

CD² = 6²+8²

CD²= 36 +64

CD²= 100

CD = √100= 10

CD= 10 cm

OC=OD=OB= 10/2=5 cm

[ Radius of a circle]

Angle OBT = 90°

[Angle between radius and tangent]

In ∆ OBT

OT²= OB²+BT²

OT²= 5² +4²

OT²= 25+16

OT =√41 cm

HOPE THIS HELPS YOU