Very Clever Solution

Added: 2026-06-01

[00:00:00]This is a super fun problem. We have a quadrilateral ABCD and are given that AB is 10 units and CD is 15 units.

[00:00:12]Then these two angles are right angles and then we have a circle that touches all four sides of the quadrilateral.

[00:00:20]Our goal is to determine the radius of this circle. So, can you solve it?

[00:00:25]Okay, let us start. Let us suppose the center of the circle is point O and let the radius of the circle be X. Now suppose the circle touches the top side AB at point P and touches the bottom side CD at point Q.

[00:00:43]Since OP and OQ are both radii of the same circle, their lengths are equal to X.

[00:00:50]Also, there is an important property we must remember here.

[00:00:54]In any circle, the radius drawn to a tangent is always perpendicular to the tangent line. This means OP is perpendicular to AB and similarly, OQ is perpendicular to CD. Because both these radii are perpendicular to the top and bottom horizontal sides and also both these angles are right angles, the shape APQD forms a rectangle.

[00:01:21]Now if we draw this radius of the circle, it will be equal to X. So, AP also becomes equal to X. Similarly, DQ also becomes equal to X. The total length of AB is 10 units. Therefore, the remaining part BP becomes 10 minus X.

[00:01:39]Similarly, on the bottom side, the full length CD is 15. Thus, the remaining part CQ becomes 15 minus X. Next, we will use another beautiful theorem related to tangents. From any external point, if we draw two tangents to a circle, then both tangent lengths are always equal. Suppose the slanted side touches the circle at point S. Then from point B, the tangent segments BP and BS must be equal.

[00:02:11]Since BP equals 10 minus X, we get BS also equal to 10 minus X.

[00:02:17]Similarly, from point C, the tangent segments CQ and CS must also be equal.

[00:02:24]Since CQ equals 15 minus X, we get CS also equal to 15 minus X.

[00:02:31]So the entire slanted side BC equals 10 minus X plus 15 minus X. Adding them together gives 25 minus 2X.

[00:02:41]The next step is super clever.

[00:02:44]Let us clear things up a bit.

[00:02:46]Now from point B, draw a perpendicular line to the bottom side CD.

[00:02:52]Let the foot of this perpendicular be point M.

[00:02:55]Since the top side AB is horizontal and has length 10, the horizontal distance from D to M is also 10.

[00:03:04]But the entire bottom side CD has length 15. Therefore, the remaining part CM becomes 15 minus 10, which equals 5.

[00:03:13]Then PQ equals X plus X or 2X, and thus BM also becomes 2X.

[00:03:21]So now in the right triangle BCM, we have all three side expressions, so we can easily apply the Pythagorean theorem. We get 25 minus 2X whole squared equals 2X whole squared plus 5 squared. Now, let us expand them. 25 minus 2X whole squared becomes 625 minus 100X plus 4X squared. On the right side, 2X whole squared becomes 4X squared, and 5 squared becomes squared Oh, wow. 4x squared appears on both sides, so they cancel immediately.

[00:04:01]Now move 25 to the left side. Then move 100x on the right.

[00:04:07]This gives 625 minus 25 equals 100x or 600 equals 100x.

[00:04:16]Finally, divide both sides by 100.

[00:04:19]So we get x equals 6.

[00:04:22]This means that the radius of the circle is 6 units. And that's it.

[00:04:27]Like, share, and subscribe. So good.

[00:04:35]>> [music]