In the diagram below, chords AB and CD are perpendicular, and meet at
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16 Nov 2020, 10:22
Concept: any line that is drawn from the Center of the Circle and that is Perpendicular to a Chord will Bisect that Chord
Since the Diameter is drawn from 1 end of the Circle to the other end of the Circle and passes through the Center, if CD were the Diameter, since it is Perpendicular to Chord AB, it would have Bisected AB.
However, CD does not Bisect AB.
Since AX = 3 and XB = 4, judging by how Chord CD cut Chord AB at a 90 degree angle, the actual Diameter must be underneath CD and will be Parallel to CD. The actual Diameter will Bisect AB when it is Perpendicular to Chord AB.
We can draw another Chord PQR immediately beneath Chord CXD such that it is parallel to Chord CXD and cuts the Chord AB at Point Q.
We will make Chord PQR a MIRROR REFLECTION of Chord CXD over the Actual Diameter. Thus, Chord PQR will have the Same Length of 8 and will cut the Chord AB at Point Q such that:
AQ = 4 and QB = 3 (just reverse how Chord CXD cut Chord AB when it did so at a 90 degree angle).
Since a Circle is composed of points on the Circumference that are equidistant from the center, the Actual Diameter must lie exactly between these 2 chords ——-> the Given Chord CXD and our created Chord PQR.
The Actual Diameter will be Parallel to both these chords, exactly in the middle, and will meet Chord AB at a 90 degree Angle and Bisect the Chord AB. The center of this circle will lie at the center of these 2 Chords. Therefore, any chord that cuts through this center and connects opposite ends of the Circle’s perimeter will be a Diameter.
Given Points C , D , P , and R, we can connect CP and DR to make an inscribed Rectangle.
The Length of the Rectangle will be 8 and the Width will be the Perpendicular Distance from Point X to Point Q.
Again, since our new Chord is a mirror reflection, the width from X to Q will equal 1.
(You can draw a 3rd Line in the Exact middle of our 2 Chords and have this line be the Actual Diameter that will Bisect Chord AB, 3.5 and 3.5, and see how the Distance from X to Q must equal = 1)
Finally, we can draw a Diameter they will be the Diagonal line that passes through the geometric center of this inscribed Rectangle we created, which is also the center of the circle——->connecting 2 Diagonally Opposite Vertices of the Inscribed Rectangle.
Have the Diameter connect Vertex P with Vertex D.
Then, using Pythagoras to find the Diagonal/Diameter:
(1)^2 + (8)^2 = (D)^2
65 = (D)^2
D = sqrt(65)
-E-
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