Two chords ab and cd of a circle with its centre o are equal. If angle aob is of 60and cd=6cm . Calculate the length of radius
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Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED have a length of 2cm , 6cm and 3 cm respectively. What is the length of diameter of a circle in cm?
some will argue that OF should be 3–3.5 as per the diagram, but d diagram is just for reference , not accurate. The actual calculations are on the right.
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We need to use two geometry properties to answer this:
(1) When two chords intersect inside a circle, the product of the two lengths on one chord = the product of the two lengths on the other chord.
So AE * EB = CE * ED
2 * 6 = 4 * ED,
12 = 4ED
ED = 3.
So now we know the lengths of all 4 of the little segments on the chords.
(2) The center of a circle is on the perpendicular bisector of any chord.
Let’s locate the chords conveniently on an x/y axis. We can put E at the origin, and the chords will fall along the x and y axes. Locate A and B on the x-axis and C and D on the y-axis with
A = (-2, 0) and B = (6, 0)
C = (0, 4) and D = (0, -3)
Now you can easily find the midpoints of the two chord by averaging the endpoints (or just looking at your graph).
AB: (2, 0) CD : (0, .5)
Since the chords are vertical and horizontal, their intersection is easy to see. The intersection must be the center of the circle.
Center = (2, .5)
The length r of the radius would be the distance from the Center to any of the four points A, B, C, or D. We can use the Pythagorean theorem.
Let’s use point C: (0, 4)
r^2 = (2–0)^2 + (4 - .5)^2
r^2 = 4 + 49/4 = ( 16 + 49)/ 4 = 65 / 4
So r =( Sqrt65)/ 2
The diameter is twice that: d = Sqrt 65
You can confirm this by doing the same calculation with A, B, or D.
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Chords AB and CD of a circle intersect at E and are perpendicular to each other. Segments AE, EB and ED have a length of 2 cm 6 cm and 3 cm, respectively. What is the length of diameter of a circle in cm?
Since the the two chords AB and CD intersect at E, AE * EB = CE * ED. So
2 * 6 = 3 * EC, or
EC = 2 * 6/3 = 4 cm.
So the chord CD = EC + ED =4+3 = 7 cm and AB = AE+EB = 2+6 = 8 cm
Let OM be a perpendicular from the center, O on AB and let ON be a perpendicular from the center, O on CD.
EM = (2+6)/2 - AE = 4–2 = 2 cm
EN = 4 -(4+3)/2 = 0.5 cm
OE = [2^2 + 0.5^2]^0.5 = 4.25^0.5 = 2.061552813 cm.
Let R be the radius of the circle. Let EO extended meet the circumference at S and T. TE = R-OE or (R - 2.061552813) and ES = R + OE = (R + 2.061552813)
TE * ES = AE * EB
(R - 2.061552813)*(R + 2.061552813) = 2*6 = 12
R^2 - 2.061552813^2 = 12, or
R^2 = 12 +2.061552813^2 = 16.25
So R = 16.25^(0.5) = 4.031128874 cm
The radius of the circle is 4.031128874 cm.