A chord of a circle divides the circle into two parts such that the squares inscribed in two parts have areas 16 and 144 square unit .the radius of the circle is
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A chord of a circle divides the circle into two parts such that the squares inscribed in the two parts have areas 16 and 144 unit as shown in figure.
Let radius of Circle is r
And distance between centre of circle to chord = x [ see attachment]
so, distance between centre to side of big square {e.g., AB} = 12 - x
Now, ∆ABC is right angled triangle .
so, AC² = AB² + BC²
Here, AC = r
AB = 12 - x
BC = half of side length = 12/2 = 6 [ because radius is the perpendicular bisector on chord ]
r² = (12 - x)² + 6² ----------(1)
Similarly,
∆ADE is also a right angled triangle ,
so, AD² = AE² + DE²
r² = (x + 4)² + 2² ----------(2)
Now, from equations (1) and (2),
(12 - x)² + 6² = (x + 4)² + 2²
⇒144 + x² - 24x + 36 = x² + 16 + 8x + 4
⇒ 180 -24x = 20 + 8x
⇒ 160 = 32x
⇒ x = 5
Now, radius , r = √{(x +4)² + 2²}
r = √{9² + 2²} = √85
Let radius of Circle is r
And distance between centre of circle to chord = x [ see attachment]
so, distance between centre to side of big square {e.g., AB} = 12 - x
Now, ∆ABC is right angled triangle .
so, AC² = AB² + BC²
Here, AC = r
AB = 12 - x
BC = half of side length = 12/2 = 6 [ because radius is the perpendicular bisector on chord ]
r² = (12 - x)² + 6² ----------(1)
Similarly,
∆ADE is also a right angled triangle ,
so, AD² = AE² + DE²
r² = (x + 4)² + 2² ----------(2)
Now, from equations (1) and (2),
(12 - x)² + 6² = (x + 4)² + 2²
⇒144 + x² - 24x + 36 = x² + 16 + 8x + 4
⇒ 180 -24x = 20 + 8x
⇒ 160 = 32x
⇒ x = 5
Now, radius , r = √{(x +4)² + 2²}
r = √{9² + 2²} = √85
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