98% Students FAILED to Solve this Simple Math Problem?

Hinzugefügt: 2026-06-01

[00:00:00]Hello everyone, you are welcome.

[00:00:03]Today we have a very easy and a very interesting geometry math problem.

[00:00:08]Here we have given a right angle triangle which has base given.

[00:00:13]The other two sides are not given.

[00:00:16]And there is a circle inside the right angle triangle whose radius is two units.

[00:00:21]Here our target is to find out the height of this right angle triangle.

[00:00:26]So let's start our solution.

[00:00:28]To find out the height of this right angle triangle, first of all look into the circle. Here this is the center of circle.

[00:00:36]Then we will connect this center with the tangent's point here.

[00:00:40]Here the circle touches the sides of this right angle triangle. So therefore the sides of this right angle triangle are tangents to this circle.

[00:00:48]So therefore here we will connect and join the center with this one tangent point and this one tangent point.

[00:00:56]So this figure will become let's try to join these points.

[00:01:01]Also connect this point with this one tangent point.

[00:01:06]Now as the radius of this circle is two units, therefore this length this radius will be two units.

[00:01:13]And this radius will be also two units.

[00:01:16]And obviously the opposite sides this side and this side will be also two units.

[00:01:23]Look at to the height of this right angle triangle. Here this total height this is H. And this part of the height is two units. What about this one part?

[00:01:33]This one part will be height minus two means H minus two units.

[00:01:39]Also look at to the base of this right angle triangle.

[00:01:43]Here this base is total length of 10 units. But this part is two units. What about the remaining length? This remaining length will be 10 minus two units which is eight units.

[00:01:55]Now here at this step we will apply tangent tangent to a circle theorem.

[00:02:00]That is, if we have two tangents from the same external point to a circle, then the length of that tangents will be always equal. So, therefore, here from this point, we have two tangents to this circle.

[00:02:13]This one tangent and this one tangent.

[00:02:15]So, here this tangent has a length of eight units. So, therefore, this tangent will be also eight units.

[00:02:22]Similarly, here from this point, we have two tangents to this circle. This tangent and this tangent.

[00:02:28]This tangent has a length of h minus two units. So, this will be also h minus two units.

[00:02:35]Now, look into this right angle triangle. Here, we have the values of its three sides. This side is h. This side is 10 units.

[00:02:44]And this side has a sum of h minus two and eight. So, what is the value of h minus two plus eight? This is simply This is h plus six.

[00:02:58]Now, as this is a right angle triangle, so therefore, here we can apply the Pythagoras theorem.

[00:03:03]And we can find out the value of height.

[00:03:06]So, by Pythagoras theorem, the square of hypotenuse is equal to perpendicular square or height square plus base square.

[00:03:15]So, therefore, from Pythagoras theorem, here we can write This implies Here, our hypotenuse is h plus six whole square is equal to And our perpendicular is h. This become h square plus our base is 10. This become 10 square.

[00:03:35]So, let's simplify this equation for the value of h, height.

[00:03:40]So, here, we will expand the left-hand side using a plus b whole square identity, which is equal to a square plus two times ab plus B squared.

[00:03:53]So, using this algebraic identity, here our A is H and our B is 6.

[00:04:00]So, here this left-hand side will become This will become H squared plus two times H times 6 plus 6 squared is equal to and this is H squared plus 10 squared is simply 100.

[00:04:20]So, let's simplify this whole equation.

[00:04:22]So, here this will become This is H squared plus two times six is 12.

[00:04:29]This will become 12 H plus 6 squared is 36.

[00:04:34]is equal to H squared plus 100.

[00:04:39]Looking at the both sides of this equation, there is positive X squared in both sides. So, by subtracting X squared from both sides here this X squared will be cancelled.

[00:04:51]This will become 12 H plus 36 is equal to 100.

[00:05:01]Here we will take this 36 to the right-hand side. This will become This is 12 H is equal to 100 minus 36.

[00:05:14]So, what is the value of 100 minus 36?

[00:05:17]This is simply 64.

[00:05:20]So, therefore our final equation will become This is 12 H is equal to 64.

[00:05:28]To find out the value of H here, we will divide both sides by 12.

[00:05:33]So, here 12 and 12 will be cancelled and here 12 and 64 are divisible by four.

[00:05:39]So, here three times four is 12 and 16 times four it is simply 64. So, this will become 16 by 3.

[00:05:50]And by dividing 16 by 3, this gives here this is 3 * 5 is 15.

[00:05:57]So, this is 1.

[00:05:59]So, this will become this is just 5 whole 1 by 3.

[00:06:04]So, therefore, our final height will become that is 5 whole 1 by 3.

[00:06:11]And this can be written in decimals as about 3 5.

[00:06:16]33.

[00:06:18]So, this is the approximate value of the required height.

[00:06:22]So, finally, the height of this right-angle triangle that will be about 5.

[00:06:30]33 units, and that is our final answer.