
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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How am i supposed to draw it ??
Answer:A) 24 waysB) 4 waysStep-by-step explanation:a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways
Step-by-step explanation:
a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways(hope this helps:)
That would be a unique triangle