
Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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Answer:
y=-1/5 x+c
0=-1/5(5)+c
0=-1+c
1=c
y=-1/5x+1
Step-by-step explanation:
Answer:
Ab
Bd
bc
(I'm not pretty sure but I double-checked)
Narancia out~!