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# just trying to make my first discussion :D and try to solve :D

Three distinct diameters are drawn on a unit circle such that chords are drawn as shown in the figure. If the length of one chord is $$\sqrt{2}$$ units and the other two chords are of equal lengths, what is the common length of these chords?

Note by Ian Mana
4 years, 4 months ago

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since radius is 1(unit circle). the angle opposite to the Root 2 side is 90 (converse of pythagoras theorem. hence its vertically opposite angle is also 90.

calling the centre as o and the diameter on which the 2 equal sides sit on as AB.one side as AC and the other equal side as BD since the radii are equal and AC=BD from sss postulate angle BOD = angle AOC Since they are on a straight line with one angle as 90 degrees angle AOC=angle bod=45

then use cosine rule to get the answer · 4 years, 4 months ago

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sqrt [ 2 - (sqrt 2) ] · 4 years, 4 months ago

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$$\sqrt (2- \sqrt 2)$$ · 4 years, 4 months ago

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root of 2-root 2

sorry didnt have time to put in latex · 4 years, 4 months ago

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nice can you explain how did you get it :D · 4 years, 4 months ago

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easy thing ans- sqrt[2-sqrt(2)] · 4 years, 4 months ago

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He said to explain. · 4 years, 4 months ago

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\sqrt (2- \sqrt 2) · 4 years, 4 months ago

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We have a isosceles triangle whit a=x b=1 and angle at A is 45 (at B and C are 67.5 ) so we can make a right triangle witch has hypotenuse 1 and a side x/2. Since we have all the angles we can see that x/2 = 1cos(67.5) so x=2cos(67.5) or x=2*sin(22.5) witch is approximately 0.765 · 4 years, 4 months ago

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I like this :] · 4 years, 4 months ago

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Sweet.... · 4 years, 4 months ago

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sqrt(2-sqrt(2)). solved using basic geometry no trigonometry. · 4 years, 4 months ago

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