In a circle of radius 5 cm, AB and AC are two equal chords such that AB = AC = 7 cm. What is the length of the chord BC? (Rounded off to 3 decimal places.)
9.998 cm
Let O be the centre with OA = OB = OC = 5 cm, and let M be the midpoint of BC. Since AB = AC, point A lies on the perpendicular bisector of BC, and so does the centre O. Hence A, O and M are collinear and AM ⟂ BC.
Let OM = d. Then \(BM^{2} = OB^{2} - OM^{2} = 25 - d^{2}\), and AM = \(5 - d\) (with M lying between A and the far side of the circle).
Apply the Pythagorean theorem in △ABM: \(AB^{2} = AM^{2} + BM^{2}\).
\(49 = (5 - d)^{2} + (25 - d^{2}) = 25 - 10d + d^{2} + 25 - d^{2} = 50 - 10d\).
So \(10d = 1\), giving \(d = 0.1\) cm.
\(BM = \sqrt{25 - 0.01} = \sqrt{24.99} \approx 4.999\) cm, so \(BC = 2 \times BM \approx 9.998\) cm.
Hence, the length of chord BC is 9.998 cm.
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on their respective axes. What is the radius of the other wheel ?- The length of a tangent drawn to a circle of radius 4 cm from a point 5 cm away from the center of the circle is:
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