AB is a diameter of ⦿ (O, 10). A tangent is drawn from B to ⦿ (O,8) which touches ⦿ (O,8) at D. BĎ intersects ⦿ (O, 10) in C.Find AC
Answers
Answered by
53
It is given that ,
OA = OB = 10
Diameter = AB = 20
OD = 8
Chord BC of large circle is tangent
to smaller circle.
point of contact is D.
OD is perpendicular bisector of BC
In ∆OBD , we have
<BDO = 90°
OB² = OD² + BD²
10² = 8² + BD²
100 - 64 = BD²
BD² = 36
BD = √36
BD = 6
Therefore ,
BC = 2BD
BC = 2 × 6 = 12
ii ) In ∆ACD ,
<ACB = 90° ( angle in semicircle )
∆ACB right angled triangle.
AB² = AC² + BC²
20² = AC² + 12²
AC² = 20² - 12²
AC² = ( 20 + 12 ) ( 20 - 12 )
= 32 × 8
AC = √ 256
AC = 16
I hope this helps you.
: )
OA = OB = 10
Diameter = AB = 20
OD = 8
Chord BC of large circle is tangent
to smaller circle.
point of contact is D.
OD is perpendicular bisector of BC
In ∆OBD , we have
<BDO = 90°
OB² = OD² + BD²
10² = 8² + BD²
100 - 64 = BD²
BD² = 36
BD = √36
BD = 6
Therefore ,
BC = 2BD
BC = 2 × 6 = 12
ii ) In ∆ACD ,
<ACB = 90° ( angle in semicircle )
∆ACB right angled triangle.
AB² = AC² + BC²
20² = AC² + 12²
AC² = 20² - 12²
AC² = ( 20 + 12 ) ( 20 - 12 )
= 32 × 8
AC = √ 256
AC = 16
I hope this helps you.
: )
Attachments:
Similar questions