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Question

Two perpendicular chords AB and CD of a circle intersect at O, (inside the circle AB is bisected at O). If AO = 4 units and OD = 6 units. Find circumradius of $$\displaystyle \Delta ACO$$.


A
13
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B
2133
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C
152
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D
2154
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Solution

The correct option is B $$\displaystyle \frac{2\sqrt{13}}{3}$$

When 2 chords intersect, the product of segment of one chord is equal to the product of segments of the other

$$\implies AO \times BO = OC \times OD$$

Given $$AO = BO = 4$$

$$\implies 4 \times 4 = OC \times 6$$

$$OC = \dfrac{8}{3}$$

Given $$AB$$ is perpendicular to $$CD$$ $$\implies \angle COA = 90$$

$$AC^2 = OC^2 + OA^2 = \dfrac{64}{9} + 16$$

$$AC = \dfrac{\sqrt{208}}{3}$$

Since $$AOC$$ is a right triangle

$$Circumradius = \dfrac{1}{2}(hypotenuse)$$

$$= \dfrac{1}{2} \times AC = \dfrac{\sqrt{208}}{6} = \dfrac{4\sqrt{13}}{6}$$

$$r = \dfrac{2\sqrt{13}}{3}$$


857210_296547_ans_40b6cb18f293486baf5152c75f8dff90.png

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