
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:
Rs 175
Step-by-step explanation:
Suppose the cost is x and at Rs150 the loss is 150-x (this should be a negative number).
At Rs200, the profit is 200-x.
So we have an equation: minus 150 minus x is equal to 200 minus x.
To solve the equation, the cost price X is Rs175.
4^2 = 16
so x must be greater than 16
answer
E. 17
F. 19