A Very Clever Square Puzzle! | Find the Missing Length X |

[00:00:00]Hello everyone and welcome back to my channel.

[00:00:03]In today's video, we are going to solve another interesting geometric problem.

[00:00:08]In the given question, ABCD is a square.

[00:00:13]A line segment BE of length X is drawn from vertex B to a point E on the side CD such that DE is equal to 5 and angle ABE is equal to 75°.

[00:00:29]Our goal is to find the value of X.

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[00:00:51]To solve this problem, first, let's focus on the square.

[00:00:56]Since ABCD is a square, all its interior angles are 90°.

[00:01:02]And it follows that angle ABC is equal to 90° and angle BCE is equal to 90°.

[00:01:29]Given that angle ABE is equal to 75°.

[00:01:40]Since AB is parallel to CD, the alternate interior angle BEC is equal to angle ABE, which is equal to 75°.

[00:01:52]Next, let's focus on triangle BCE.

[00:01:56]In the right-angle triangle BCE with hypotenuse BE equal to X, the length of EC is equal to X times the cosine of 75°.

[00:02:12]The length of BC, side of the square, is equal to X times the sine of 75°.

[00:02:23]Because ABCD is a square, all sides are equal in length. So, BC is equal to CD.

[00:02:32]From the figure, CD is equal to DE plus EC, where DE is equal to 5.

[00:02:46]CD is equal to 5 plus EC.

[00:02:54]Since BC is equal to CD, we can say that BC is equal to 5 plus EC.

[00:03:07]But BC is equal to X times the sine of 75°.

[00:03:17]And EC is equal to X times the cosine of 75°.

[00:03:25]Substituting these values in the above expression will give us X times the sine of 75°.

[00:03:37]is equal to 5 plus X times the cosine of 75°.

[00:03:47]Rearranging this will give us X times the sine of 75 degrees minus X times the cosine of 75 degrees is equal to 5.

[00:04:05]Next, let's factor out X from both terms on the left side.

[00:04:10]Doing this will give us X multiplied by the sine of 75 degrees minus the cosine of 75 degrees is equal to 5.

[00:04:28]Let this be equation one.

[00:04:31]The sine of 75 degrees is equal to the square root of 6 plus square root of 2 all divided by 4.

[00:04:44]The cosine of 75 degrees is equal to the square root of 6 minus square root of 2 all divided by 4.

[00:04:57]The sine of 75 degrees minus the cosine of 75 degrees is equal to the square root of 6 plus square root of 2 minus the square root of 6 minus the square root of 2 all divided by 4.

[00:05:28]Expanding the bracket will give us the square root of 6 plus square root of 2 minus the square root of 6 plus the square root of 2 all divided by 4.

[00:05:43]The square root of 6 minus the square root of six will cancel out.

[00:05:49]The square root of two plus square root of two is equal to two times the square root of two.

[00:05:56]And we are left with two times the square root of two divided by four, which simplifies to the square root of two divided by two.

[00:06:18]Going further, let's substitute the sine of 75° minus the cosine of 75° with the square root of two divided by two in equation one.

[00:06:29]Doing this will give us X times the square root of two divided by two is equal to five.

[00:06:38]By cross multiplication, X times the square root of two is equal to five times two.

[00:06:49]Five times two is 10.

[00:06:52]So, X times the square root of two is equal to 10.

[00:06:58]Dividing through by the square root of two, the square root of two will cancel out.

[00:07:07]And we are left with X is equal to 10 divided by the square root of two.

[00:07:15]To rationalize the denominator, we multiply the numerator and the denominator by the square root of two.

[00:07:26]And this will give us 10 times the square root of two divided by two.

[00:07:33]10 divided by two is five.

[00:07:37]And we are left with X is equal to five times the square root of two, which is approximately 7.07.

[00:07:48]Thanks for watching.

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