This MATH Question Will Blow Your Mind!!

Añadido: 2026-05-31

[00:00:00]Hello everyone. You are welcome. Today we have a very interesting geometry math problem.

[00:00:07]Here we have given a semicircle where there are two perpendiculars over the semicircle which has a height and length of 5 minutes and 10 minutes.

[00:00:16]And the distance between these two perpendiculars is 15 minutes.

[00:00:21]Here our target is to find out the area of this semicircle.

[00:00:25]To find out the area of this semicircle here we need the radius of this semicircle. This is because here we know that the area of semicircle that is half of the area of circle. So that is half times pi r squared.

[00:00:43]So here we need the value of the radius.

[00:00:45]Now to find out the radius of this semicircle, let us suppose this is the center of this semicircle.

[00:00:51]So here we will connect the center with the top of this height, this perpendicular.

[00:00:56]But before this look into this one line segment which is 15 minutes. Here this total length is 15 minutes. But we have not given and specified this one length, this one length and this one length.

[00:01:08]So let us suppose this smaller length is x units.

[00:01:12]So what will be the remaining length?

[00:01:14]This will be that will be 15 minus x units.

[00:01:19]Now let's try to connect the center with this one point. So this figure will become Now after connecting the center with the top of the perpendicular, look into the figure here we have a right its side 10 units, base is x units and this hypotenuse is the radius of this semicircle. So let us suppose this is small r.

[00:01:40]So here in this right angle triangle we will apply the Pythagoras theorem and we will find out the radius of this semicircle.

[00:01:47]So by Pythagoras theorem here we know that hypotenuse squared is equal to base squared plus perpendicular squared or height square.

[00:01:59]So, here this 10 square is simply r squared is equal to x squared plus 10 squared is simply 100.

[00:02:10]Now, let's suppose this is our equation number first because here we cannot find out the value of our radius before without finding the value of x. Now, we will try to find out another equation like this.

[00:02:23]So, for that here we will join the center with the top of this second perpendicular.

[00:02:28]So, this figure will become Now, after joining the center with the top of the second perpendicular here this is the radius. So, let us suppose this is small r.

[00:02:38]Here in this right angle triangle the perpendicular is 5 units and the base is 15 minus x.

[00:02:44]So, again here we will apply the Pythagoras theorem in this right angle triangle.

[00:02:48]So, therefore by Pythagoras theorem this will become here r squared hypotenuse squared is equal to base squared.

[00:02:57]Here our base is 15 minus x whole squared plus perpendicular squared that is 5 squared.

[00:03:05]Let's simplify this one equations. This is r squared is equal to the we will expand this using a minus b whole squared identity. So, this will become 15 squared.

[00:03:15]And 15 squared is 225 minus 2 times 15 times x which is 30 x plus x squared plus 5 squared is simply 25.

[00:03:29]Let's simplify this one equations. So, here this is r squared is equal to this is minus 30 minus 30 x and here this is x squared.

[00:03:42]So, this will become plus x squared and this is 225 plus 25 it is simply 250.

[00:03:54]Now, let us rearrange the terms in the right-hand side. This should become r squared is equal to x squared minus 30x plus 250.

[00:04:07]Let us suppose this is equation number six.

[00:04:10]Look at equation number first and equation number second. Both has the left-hand side the same.

[00:04:16]So, therefore, here we can compare and take equal their right-hand sides.

[00:04:20]So, therefore, from equation number first and equation number second, here we can write left-hand side of equation number first, it is simply x squared plus 100 is equal to and the right-hand side of the second equation is x squared minus 30x plus 250.

[00:04:48]And there is x squared to x squared in both sides, so therefore, we can cancel them with each other.

[00:04:54]Here we will take this negative 30x to the left-hand side and this number to the right-hand side.

[00:04:59]So, this will become positive 30x.

[00:05:03]This will become 250 minus 100.

[00:05:09]And this finally become This is 30x is equal to 150 minus 100 is simply 150.

[00:05:18]And dividing both sides by 30, this gives him Here 30 and 30 will be cancelled. Zero will be cancelled with zero. Three time one is three. Three time five is 15.

[00:05:29]So, therefore, the value of x will become This is five by one, which is simply five.

[00:05:35]Now, this is the value of x. Now, we will substitute this value of x in any equation, equation number second or equation number first to find out the value of r squared.

[00:05:46]So, therefore, here we will substitute this value in equation number first.

[00:05:50]So, equation number first implies r squared is equal to x squared plus 5 squared.

[00:05:58]So, therefore, the value of r will become r squared is x squared here simply.

[00:06:03]Sorry, this is 100.

[00:06:07]So, this will become x is 5. So, this will become 5 squared plus 100.

[00:06:13]r squared is equal to 5 squared is simply 25 plus 100.

[00:06:20]And the final value of r squared will become it is simply 125.

[00:06:26]And here we will substitute this value of x in the formula, that is the formula of the area of the semicircle.

[00:06:35]So, therefore, this area will become this is half times here r squared. It is simply 125 pi.

[00:06:46]And here if we substitute the value of pi, so, from here we will get this is half times 125 times the value of pi. It is about 3.1415 up to four decimal places.

[00:07:05]And multiplying these three numbers, it gives them about 196.

[00:07:11]34 [snorts] square units.

[00:07:17]And that is the final area of the semicircle.