This Simple Math Problem Will Surprise You!!

Añadido: 2026-05-25

[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 trapezoid and there is a circle inside the trapezoid.

[00:00:12]Here we have given only one base or one parallel side of the trapezoid.

[00:00:18]And here our target is to find out the area of this trapezoid.

[00:00:23]To find out the area of this trapezoid, we know that the area of trapezoid is that is simply the sum of its two bases b1 + b2 divided by two times its height h.

[00:00:41]So here we have only given one base. The second base or the second parallel side is not given and we have also not given the height.

[00:00:49]So here we have to find out these two, the height and the second base.

[00:00:53]So for that let us suppose here this is the center of the circle.

[00:00:58]So here we will join the center with these three tangent point, this point this one point and this one point. So this figure will become Now after joining the center with these three tangent points, here we know that in any circle the radius is always perpendicular to the tangent line at the point of tangency. So therefore here at this point we will have two right angles. Similarly at this point we will have two right angles and here at this point we will also have two right angles. Here in this trapezoid this angle is right angle and this is also a right angle.

[00:01:37]Look at to this one figure here, these three angles are right angles. Therefore the fourth angle will be also a right angle. So this figure looks like a square.

[00:01:46]So if this side is three, so here its all other sides will be also three. So here this radius will be three.

[00:01:54]This radius will be also three and this side will be also three.

[00:01:58]Similarly here, this angle will be a right angle. So, here all the sides will be three. This is three.

[00:02:05]This is also three.

[00:02:07]And this is also three.

[00:02:10]Now, here we will use another result.

[00:02:11]That is, if we have two tangents from an external point to a circle, then the length of that tangents will be equal.

[00:02:19]So, here from this external point, we have two tangents to this one circle.

[00:02:23]This one tangent and this one tangent.

[00:02:26]Now, this tangent has a length of four units. So, therefore here, this tangent will be also four units.

[00:02:32]Similarly, from this point, we have two tangents to this circle. This one tangent and this one tangent.

[00:02:38]But here, no one length is given. This length and this length is not given.

[00:02:43]So, let us suppose this tangent length is X units. So, this will be also X units.

[00:02:49]Now, what we will do next? Next here, we will draw a perpendicular from this point to this segment four units.

[00:02:57]So, this figure will become Now, after draw the perpendicular from that point to this segment four units, here look to this one figure. Here, the three angles are right angles. So, here the fourth angle will be also a right angle. And this figure looks like a rectangle having same opposite sides.

[00:03:15]So, this side is three plus three, which is six units. So, this side will be also six units.

[00:03:21]Here, this length is X. So, this length will be also X units. Now, what about this one length? Here, this total length is four units, and this is X. So, this will be This will be four minus X units.

[00:03:35]Now, here we have the length of this one base, which is three plus four means seven units.

[00:03:41]And we have also the value of the height, that is six units. So, we have to only find out this one length, this one base having no value of X. So, here what I did is to find out the value of X. So, for that here in this right-angled triangle, we will apply the Pythagoras theorem.

[00:03:58]So, here in this right-angled triangle, we have the value of the height or perpendicular is six units, the base is 4 - X, and the hypotenuse is X + 4 units. So, here this hypotenuse is X + 4 units.

[00:04:13]So, here we will apply the Pythagoras theorem.

[00:04:15]So, by Pythagoras theorem, the square of hypotenuse So, here our hypotenuse is X + 4.

[00:04:23]So, X + 4 whole square is equal to that is perpendicular square or height square.

[00:04:30]That is 6 square plus X base square. So, base is 4 - X whole square.

[00:04:39]So, here we will simplify this linear equation for the value of X. So, this will become Here we will expand this using a + b whole square identity. So, this will become X square + 2 * X * 4, which is 8 X plus 4 square. Now, 4 square is 16.

[00:04:59]Is equal to Here 6 square is 36.

[00:05:03]Plus and we will expand this using a - b whole square identity, which is equal to a square. So, 4 square is 16.

[00:05:11]Minus 2 * ab. So, this will become 8 X + X square.

[00:05:19]Now, look at the both sides of this equation, there is positive X square in both sides. So, therefore, we can cancel X square with X square.

[00:05:26]And there is positive 16 as well in both sides. So, we can cancel 16 with 16.

[00:05:32]So, here we will take this negative X to the left-hand side.

[00:05:36]So, this will become This is 8 X and this will become positive 8 X is equal to 30 6.

[00:05:47]So, here 8 X + 8x it is simply 16 x is equal to 36.

[00:05:56]Now, we'll divide both sides by 16.

[00:05:59]So, here 16 and 16 will be cancelled.

[00:06:02]Here, these two numbers are divisible by four.

[00:06:05]So, here 4 * 9 is 36 and 4 * 4 is 16.

[00:06:10]So, this gives the value of x as 9 by four units.

[00:06:17]So, therefore here the value of x in this figure that is 9 by four units.

[00:06:25]Here, we have the values of two bases and also the height. So, substitute these values here in the area formula.

[00:06:32]So, how So, here our b1 is simply that is 3 + 3 + 9 by four units and here our b2 is this one base that is 3 + 4 means seven units.

[00:06:46]Divided by two times height. Here, our height is six units. This will become six.

[00:06:54]So, here we simplify two with six. So, here 2 * 1 is 2, 2 * 3 is 6.

[00:06:59]So, this will become here this is 3 3 + 7 it is 10.

[00:07:04]This will become 10 + 9 by 4 times 3.

[00:07:13]And further, this will become um 10 * 4 is 40. So, it become 40 + 9 divided by 4 times 3.

[00:07:25]So, further this will become here this is 49 divided by 4 times 3.

[00:07:31]And here, 49 * 3 it is simply 147 divided by 4.

[00:07:38]So, this is here this is 147 divided by 4. Here, 147 is not exactly divisible by 4, so therefore here we leave this number as the same. So therefore, our final area will become that is 147 divided by 4 units.

[00:07:55]So therefore, the final area of this trapezoid that will be 147 divided by 4 square units. And that is our final answer and the final area of this trapezoid.