Given a circle of radius 5cm and centre O. OM is drawn perpendicular to the chord XY. If OM = 3cm, then length of chordXY is
Question
Given a circle of radius 5cm and centre O. OM is drawn perpendicular to the chord XY. If OM = 3cm, then length of chordXY is
Solution
To solve this problem, we will use the Pythagorean theorem.
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First, we know that OM is perpendicular to the chord XY, which means that OM is a radius of the circle that bisects the chord XY. Let's call the point where OM intersects XY as point N. So, we have a right triangle OMN.
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The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. We can write this as: OM² = ON² + MN²
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We know that OM = 3cm (given) and ON = radius of the circle = 5cm.
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Substituting these values into the Pythagorean theorem gives us: 3² = 5² + MN²
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Simplifying this gives us: 9 = 25 + MN²
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Solving for MN gives us: MN² = 9 - 25 = -16, but since the square of a real number cannot be negative, there seems to be a mistake in the problem. The radius of the circle should be greater than the length of OM for the problem to make sense.
If the problem is corrected such that OM < radius, then we can solve for MN, and the length of the chord XY would be 2* MN (since MN is half the length of the chord).
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