2nd length y+2=2y-2.8 y=4.8
The lengths of the kite would be 4 and 4.8
2y = 3(4)
2y = 12
y = 6
GF = 6+2 = 8
DE = 3+4 = 7
10x = 6(7)
10x = 42
x = 4.2
MN = 6+7 = 13
LJ = 4.2 + 10 = 14.2
5. y=6; DE=7; FG=8
6. x=4.2; JL=14.2; MN=13
When chords cross inside a circle, the product of the segment lengths is a constant.
5. DH×EH = FH×GH
y = GH = DH×EH/FH = 3×4/2 . . . . . divide by the coefficient of GH
y = 6
Of course, the total chord length is the sum of the lengths of its segments.
DE = DH +EH = 3+4 = 7
FG = FH +GH = 2+6 = 8
6. This problem works the same way as the previous one.
x = LK = MK×NK/JK = 6×7/10 = 4.2
JL = JK +LK = 10 +4.2 = 14.2
MN = MK +NK = 6 +7 = 13
Comment on chords and secants
You can think of the points H (problem 5) and K (problem 6) as points where the chords meet. A chord is part of a secant line, a line that intersects the circle in two points. One can also consider these points (H or K) to be the points where the respective secant lines meet.
Then, the product that is a constant is the product of the distance from the meeting point (H or K) to one circle intersection with the distance from that meeting point to the other circle intersection. For problem 5, the constant is the product HD×HE = HF×HG.
It turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.
That is, you only need to remember one rule to work both kinds of problems.
-For chords intersecting each other within a circle, the product of their segments are always equal.
-We can therefore use this relationship to calculate the lengths of missing chords and segments as:
Hence, the length of the segment HG is 6
The length of a chord is equivalent to the sum of its segments;