Question

In a circle of radius 6 CM ,AB and AC are two chords such that AB is equal to AC which is equal to 8 cm. Find the length of BC. (CLASS 9TH)​

Answer 1

AB and AC are two equal chord (8cm), join OA, it intersects BC on point M.

OA is the bisector of angle BAC,

M divides BC in the ratio = 6:6= 1:1

hence M is the midpoint of BC,

so OM perpendicular on BC,

In triangle ABM,,,,,

AB² = AM² +BM² ( Pythagoras theorem),

BM² = 8² - AM².............(i ),

In right angled triangle OBM,

OB² = OM² + BM²

6² = ( 6 - AM)²+ BM²

BM² =36 - ( 6 - AM )².......( ii )

from ( i ) and ( ii) we get,

8² - AM² = 36 - (6 - AM)²

64 -AM² = 36 - 36 + 12AM + AM²

64 = 12AM

therefore AM = 64÷12

5.3

putting the value of AM in equation ( i ),

BM² = 8² - AM²

= 64 - (5•3)²

=64 - 28•9

=35•91

therefore BM= √35•91

5•99.

Hence the length of BC will be = BM × 2 = 5•99×2

11•98 cm.

length of BC will be 11•98 cm.