98% Students FAILED to Solve this Beautiful Geometry Problem

Added: 2026-05-09

[00:00:00]Hello everyone, you are welcome. Today we have a very beautiful geometry math problem.

[00:00:06]Here we have given a circle and there is a chord on the circle whose has a length of 4 units.

[00:00:13]And there is a perpendicular to this chord whose has a length of 6 units.

[00:00:17]And there is another perpendicular to the first one perpendicular which has a length of 2 units.

[00:00:23]Here our target is to find out the radius of this circle.

[00:00:27]To find out the radius of the color this circle here we will do some interesting steps. So first of all let us suppose this is a chord AB. So let us suppose this is point A.

[00:00:37]This is point B.

[00:00:39]And let us suppose this perpendicular is this is point C.

[00:00:43]And this is point D. So here the chord is AB. The first perpendicular is BC and the second is CD.

[00:00:51]Now what we will do here? Here we will draw a perpendicular from the center to this one chord.

[00:00:57]So this figure will become and here we know that any perpendicular from the center to the chord always divide the chord into two equal parts. So here this chord has a length of 4 units. So therefore this length and this will be half. So this will become 2 units and this will be also 2 units.

[00:01:18]Now what we will do next? Next here we will expand this one perpendicular and this one perpendicular. And we will also expand this radius or this perpendicular such that it join this one perpendicular.

[00:01:30]So this figure will become Now after expanding the perpendicular here, let us suppose this is point F, this is G and this is H. Now we will try to find out this one length and this one length.

[00:01:44]So here and this is a perpendicular to this side so therefore here we will have a right angle and there is also a right angle. Here all the four angles are right angles so therefore this figure is a rectangle which has same opposite sides. So, this side will be also two units.

[00:02:02]And as this this line is perpendicular to this one chord, so therefore this line will must divide this chord into two equal parts. This is because the line is passing through the center.

[00:02:13]So, here this this DG has a length of 2 + 2, which is which is four units. So, therefore this length will be also four units.

[00:02:22]And this total length CE will be 2 + 4, which is six units. So, this will be six units.

[00:02:29]Now, we'll try to find out this one length.

[00:02:31]To find out this one length, let us suppose this length is equal to X.

[00:02:36]To find out this length here, we will use chord's intersecting theorem.

[00:02:40]So, here we have two chords, chords BH and chords DF.

[00:02:45]So, by chord intersecting theorem, the product of two segments of one chord is equal to the product of two segments of the second chord.

[00:02:53]So, first we will take this one chord.

[00:02:55]So, here the product of 6 * X will become This will become 6 * X is equal to that will be the product of 2 * 6.

[00:03:07]It will become 2 * 6.

[00:03:10]Dividing both sides by 6, this will become 6 and 6 will be cancelled in both sides.

[00:03:16]And the value of X will become two units.

[00:03:20]So, therefore here this length is two units.

[00:03:25]Now, what is our next target? In our next target here, we will join this center with this chord and also this with this one point. So, this figure will become Now, after drawing the perpendicular to this one chord BH here, and here joining the center with this one point, here we got a right angle triangle. This is a right angle triangle.

[00:03:48]So, we'll try to find out the length of sides of this right angle triangle is this side is opposite to this one side and this one side so therefore this side will be two units.

[00:03:59]And this is the radius of this circle so let us suppose this is capital R which is our target to find.

[00:04:04]And what will be this one length? Here this total length BC is six units so what will be this one length? Here we will use the perpendicular bisector theorem. Here this is a chord BH and this is a perpendicular from the center so therefore this will divide this into two equal parts.

[00:04:21]What is the length of BH? This is 6 + 2 units which is 8. So therefore this half length will be just four units.

[00:04:30]So here in this right angle triangle here we have the length of three sides.

[00:04:33]This is four, this is two and this is R.

[00:04:37]So here we'll apply the Pythagoras theorem.

[00:04:39]So by Pythagoras theorem the square of hypotenuse is equal to base square plus perpendicular square.

[00:04:46]So therefore from here we can write our hypotenuse hypotenuse is R so R square is equal to Here our base is two so this will become two square plus perpendicular square which is four so this will become four square.

[00:05:01]Let's simplify this one expression for the value of R so this will become R square.

[00:05:07]Two square is simply four and four square is simply 16.

[00:05:12]And summing these two numbers this gives him R square is equal to this is 20.

[00:05:18]Here we'll take square root on both sides.

[00:05:22]So here the square and square will be cancelled so therefore the value of R will become here we can write this 20 as this is simply 4 * 5 and this is square root of 4 * square root of 5.

[00:05:39]Here square root of 4 it is simply two * square root of 5.

[00:05:44]So this is our final radius.

[00:05:48]So, finally the radius of this circle inside this interesting figure, it is simply 2 * √ of 5 units.

[00:05:59]And that is our final target.