Please solve q8 and 12 8 Parallel Chords AB and CD are 3 9 cm apart and lie on - Maths - Circles

Question

Please solve q8 and 12
8   P a r a l l e l   C h o r d s   A B   a n d
  C D   a r e   3 9   c m   a p a r t
  a n d   l i e   o n   t h e   o p p o s
i t e   s i d e s   o f   c e n t r e    
  a   c i r c l e   I f   A B   =   1
4   c m   a n d   C D   =   4   c m ,
  f i n d   t h e   r a d i u s   o f   t
h e   c i r c l e 12   A n   e q u i l a t e r a l
  t r i a n g l e   o f   s i d e   9   c
m   i s   i n s c r i b e d   i n   a   c
i r c l e   F i n d   t h e   r a d i u s   o f
  t h e   c i r c l e

Brijendra Pal · Accepted Answer

Hi, 12,

âˆ†ABC is an equilateral triangle AB = BC = CA = 9 cm O is the circumcentre of âˆ†ABC ∴ OD is the perpendicular bisector of the side BC (O is the point of intersection of the perpendicular bisectors of the sides of the triangle) In âˆ†OBD and âˆ†OCD, OB = OC (Radius of the circle) BD = DC (D is the mid point of BC) OD = OD (Common) ∴ âˆ†OBD ≅ âˆ†OCD (SSS congruence criterion) ⇒ ∠BOD = ∠COD (CPCT) ∠BOC = 2 ∠BAC = 2 × 60° = 120° ( The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle) $∴ ∠ BOD = ∠ COD = \frac{∠ BOC}{2} = \frac{120 °}{2} = 60 ° BD = DC = \frac{BC}{2} = \frac{9}{2} cm$ In âˆ†BOD, Sin ∠BOD $= \sin 60 ° = \frac{BD}{OB}$ $∴ \frac{\sqrt{3}}{2} = \frac{\frac{9}{2} cm}{OB} \Rightarrow OB = \frac{9}{2} \times \frac{2}{\sqrt{3}} cm = 3 \sqrt{3} cm$ Thus, the radius of the circle is $3 \sqrt{3} cm$ Please post single query in a thread