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Of any two chords of a circle ...

# Of any two chords of a circle show that one which is larger is nearer to the centre.

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Updated On: 17-4-2021

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Transcript

TimeTranscript
00:00 - 00:59any two chords of a circle then show that one which is larger is nearest to the centre so we have given in the question that is there are two chords of a circle Subah have to prove that the course which is larger is needed to the centre Subah stop all we will draw a circle let's suppose we have a circle and the centre of the circle is O and we will do our two coats of the circle sobin cord is this select support this court's is a b and we will draw another cause of the circle s cord is this let's suppose this code is CD and hear the perpendicular distance of the chord from the centre it is Centre is Osho this distance is oem is it perpendicular distance of the costs from the centre and the distance of the a b chord from the centre is this so let's suppose this
01:00 - 01:59o n this is also a perpendicular distance and here we have also given that is because that is ab is greater than the cord CD and here we have to prove in the question that is to prove that is one is less than am because we have given that is ab is the larger cause in the circle prove this we will do a sump construction that is the construction is we will meet a 28 and we will also meets e2open lexapro radius of the circle is are that is a equal to c equal to r a p talk about the triangle A and so this is a right angle triangle this is a right angle at point and so this is 90 degree Subah using the Pythagoras
02:00 - 02:59by Pythagoras S se the hypotenuse square that is a square is the importance so I put in a square is equal to the sum of the other two sides square that is a n a square and Plus and Square so when we simplify this venue is the radius of the circle is a square is equal to a n is equal to square and Plus and square and here we know the another property of the circle that is if we draw a perpendicular from the centre of the circle to the cord show the perpendicular is bisect the cord so here we can also write in place of an n is equal to MB that is equal to and be so here we can say that is in place of A and B can also write one way to ab so we will put this value here then R square equal to 1 by 2 a b
03:00 - 03:59squire Plus and Square then it will be our square is equal to a b square by 4 and + and Square so let's suppose this is equation first a bit awkward the triangle c o m angle C O M this is also a right angle triangle this is writing a letter M Subah using the Pythagoras Theorem we can say that have put in a square to the hypotenuse AC you so here is a square the sum of the other two sides called the other two side is oem in OC that is oem square and plus CM square the figure that is CM square so here so so here we will again discuss the property of the circle that is if we draw a perpendicular of the
04:00 - 04:59circle from the centre to the chords so it will bisect the codes into two equal parts so here CM CM is equal to MD and calculate the value of CM then it will be one way to work CD so in place of cmv will put one by two CD so here it is a square is equal to M squire plus in place of cmv will put one way to CD square so again when we will simplify this and it will be a square + c square by this is the value of the square leg support this is a question second question first this is the value of the square unit question second this is also the value of square when we will equip them then it will be a b square upon 4
05:00 - 05:59+ and the square equal to MC square place C D square by 4 so here in the question we have given that is ab is greater than CD ab is greater than CD and the value is equal so if ab is greater than CD so here it will be definitely then n is less than oem this is the only condition by which we can say these two values are equal so this is something we have to prove that we have two chords of the circle to the larger cause will be nearer to the centre of the circle so this is our answer thank you
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