the chord of a circle of radius 10 cm subtends a right angle at its centre find the length of the chord
Answers
Answered by
10
Heya!!
◆As we know that the bisector of centre angle bisect the chord in two equal part at 90°
so,we join the line AD which is bisector of centre .
now,The triangular part wil appear ..and it make an angle 45° between bisector and radius .
so,sin45°=p/h=p/5 【where p=perpendicular and b =base of triangle...
now, 1/√2=p/5
p=5/√2
So,chord =2BD
chord=2*5/√2
=)10/√2
=)10*√2/√2*√2
=)5√2 Is lenght if chord .
hope it help you.☺ :)
@rajukumar☺
◆As we know that the bisector of centre angle bisect the chord in two equal part at 90°
so,we join the line AD which is bisector of centre .
now,The triangular part wil appear ..and it make an angle 45° between bisector and radius .
so,sin45°=p/h=p/5 【where p=perpendicular and b =base of triangle...
now, 1/√2=p/5
p=5/√2
So,chord =2BD
chord=2*5/√2
=)10/√2
=)10*√2/√2*√2
=)5√2 Is lenght if chord .
hope it help you.☺ :)
@rajukumar☺
Attachments:
Similar questions