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In the given figure, a circle with center O is given in which a diameter AB bisects the chord CD at a point E such that $CE=E D=8 cm$ and $E B=4 cm$. Find the radius of the circle.
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Given :
In the given circle, O is the center, and diameter AB bisects the chord CD at the point E.
$CE=E D=8 cm$ and $E B=4 cm$.
To do :
We have to find the radius of the circle.
Solution :
Let the radius of the circle be 'x'.
From the figure, $OB = x$
$OB = OE+EB$
$OE = OB-EB$
$OE = x-4$
We know that "the perpendicular bisector of the chord passes through the center of the circle".
So, OEC is a right-angled triangle.
$x^2 = (x-4)^2+8^2$
$x^2 = x^2 -8x + 16+64$
$x^2=x^2-8x+80$
$x^2-x^2+8x=80$
$8x = 80$
$x = \frac{80}{8}$
$x = 10$.
Therefore, the radius of the circle is 10 cm.
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