Answer:
5√3 units
Step-by-step explanation:
Let the center of the circle be E, the chord be AB such that A and B touch the circle, the midpoint of AB be M.
Connect point E to point B and point A.
Likewise, connect point E to point M such that EM is perpendicular to AM and BM.
Notice that this forms two right triangles.
Observe that EB and EA are the radii of the circle and the hypotenuse of the triangles and AM and BM are the bases of the two right triangles.
Thus, EB = EA = 10.
Since M is the midpoint of AB, AM = BM = 1/2(AB) = 1/2(10) = 5.
Using the Pythagorean Theorem (a² + b² = c²),
5² + EM² = 10²
25 + EM² = 100
EM² = 100 - 25
EM² = 75
EM = √75
EM = 5√3
Note that EM is the distance between the chord and the center point of the circle.
Hence, the distance of the chord to the center is 5√3 units.
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Answers & Comments
Answer:
5√3 units
Step-by-step explanation:
Let the center of the circle be E, the chord be AB such that A and B touch the circle, the midpoint of AB be M.
Connect point E to point B and point A.
Likewise, connect point E to point M such that EM is perpendicular to AM and BM.
Notice that this forms two right triangles.
Observe that EB and EA are the radii of the circle and the hypotenuse of the triangles and AM and BM are the bases of the two right triangles.
Thus, EB = EA = 10.
Since M is the midpoint of AB, AM = BM = 1/2(AB) = 1/2(10) = 5.
Using the Pythagorean Theorem (a² + b² = c²),
5² + EM² = 10²
25 + EM² = 100
EM² = 100 - 25
EM² = 75
EM = √75
EM = 5√3
Note that EM is the distance between the chord and the center point of the circle.
Hence, the distance of the chord to the center is 5√3 units.