This Geometry Problem Has a Sneaky Ending

[00:00:00]Can you find d length x given o d is equal to df is equal to f h.

[00:00:17]Suppose A or D is A, DF is A and F H is A.

[00:00:29]And now if we join O G then O G S² it will be O H S A² plus G H S² and O G is a radius S².

[00:01:03]O H is A + A + A that is 3 A S² + G H is 13 S².

[00:01:18]So r a² is 9 a² + 169.

[00:01:29]And if we join O E then O E S² is O F A² plus E F² and O E is radius S² square. O F is A + A that is 2 A S² plus E F is 17 S².

[00:02:16]So r a² is 4 a² + 289.

[00:02:27]And suppose this is equation one and this is equation two.

[00:02:35]Then from equation one and equation two we can say that 9 a s² + 169 is 4 a s² + 289 and 9 a²us - 4 a s² is 289 - 169 and 5 a s² is 120.

[00:03:20]So a² is 24 and and we have r a² is 4 a² + 289.

[00:03:46]We have r a² is 4 a s² - 289.

[00:03:55]So r a² is 4 * 24 - 289 and r² is 96us 289.

[00:04:15]So, R² is 385 and Now if we join OC then O C² is O D S² + C D S² and O C is radius S² square. O D is A S²us C D is X S² and we have R² is 385 and A² is 24.

[00:05:40]So 385 is 24 + x a².

[00:05:50]So x a² is 385 - 24 and x² is 361.

[00:06:06]That means x is 19.

[00:06:14]>> [music]