please solve this question this is urgent I will walk he or she has brainliest
Attachments:
Answers
Answered by
1
Distance of chord from centre = perpendicular pm.
Now, perpendicular from centre bisect the chord.
So, in rt triangle apm:
Ap = 13 cm (radius)
Am = 12cm (1/2 ab)
So pm^2 = ap^2 - am^2(Pythagoras th)
Pm ^2= 169 - 144
pm ^2= 25
Pm = 5 cm.
Distance of chord from centre is 5 cm.
Answered by
0
The distance from the centre of the circle to the chord AB is PM
It means PM is perpendicular to AB.
∆APM congruent to ∆BPM by R.H.S
AM=BM (by c.p.c.t)
AB=24
AM+BM=24
2AM=24
AM=12
By the phytha gor.thrm.
In∆APM
AM^2+PM^2=AP^2
12^2+PM^2=13^2
PM^2=169-144
PM^2=25
PM=5
So the distance between the centre of circle to the chord AB is 5cm
It means PM is perpendicular to AB.
∆APM congruent to ∆BPM by R.H.S
AM=BM (by c.p.c.t)
AB=24
AM+BM=24
2AM=24
AM=12
By the phytha gor.thrm.
In∆APM
AM^2+PM^2=AP^2
12^2+PM^2=13^2
PM^2=169-144
PM^2=25
PM=5
So the distance between the centre of circle to the chord AB is 5cm
Similar questions