Two equal chords ab and cd of a circle intersect at x, show that ax = xd and bx = cx
Answers
Intersecting Chord Theorem:
When two chords intersect each other in a circle, the product of the segments of each chord are equal.
Here, the two chord AB and CD intersect at a point X. So, the chord AB has two segments AX and XB. Similarly, the chord CD has two segments CX and XD.
AX * XB = CX * XD
Given: AX = 8 cm AB = 14 cm CX - XD = 8 cm
So, XB = AB - AX = 6 cm and CX = 8 + XD
Using the theorem,
AX * XB = CX * XD
8 * 6 = (8 + XD) * XD
48=8∗XD+X
D
2
48=8∗XD+XD2
Let XD = x
x 2 +8x−48=0
x2+8x−48=0
x2 +12x−4x−48=0
x2+12x−4x−48=0
x(x+12)−4(x+12)=0
x(x+12)−4(x+12)=0
(x+12)(x-4) =0
x = -12 or x =4
Since x represents the length of a segment, it can't be negative. Rejecting x =-12
So x = 4
XD = 4 cm
CX = 8 + XD = 8 = 4 = 12 cm
Hence, CD = CX + XD = 12 + 4 = 16 cm
I hope it helps!
Step-by-step explanation:
Intersecting Chord Theorem:
When two chords intersect each other in a circle, the product of the segments of each chord are equal.
Here, the two chord AB and CD intersect at a point X. So, the chord AB has two segments AX and XB. Similarly, the chord CD has two segments CX and XD.
AX * XB = CX * XD
Given: AX = 8 cm AB = 14 cm CX - XD = 8 cm
So, XB = AB - AX = 6 cm and CX = 8 + XD
Using the theorem,
AX * XB = CX * XD
8 * 6 = (8 + XD) * XD
48=8∗XD+X
D
2
48=8∗XD+XD2
Let XD = x
x 2 +8x−48=0
x2+8x−48=0
x2 +12x−4x−48=0
x2+12x−4x−48=0
x(x+12)−4(x+12)=0
x(x+12)−4(x+12)=0
(x+12)(x-4) =0
x = -12 or x =4
Since x represents the length of a segment, it can't be negative. Rejecting x =-12
So x = 4
XD = 4 cm
CX = 8 + XD = 8 = 4 = 12 cm
Hence, CD = CX + XD = 12 + 4 = 16 cm
I hope it helps!
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