Please help me I need this

Question

In rectangle ABCD, the angle bisector of ∠A intersects side 

DC

  at point M and the angle bisector of ∠C intersects side 

AB

  at point N. The length of 

CM

 is equal to the length of 

AM

 and it is 6 in longer than the length of 

DM

 . Find the perimeter of quadrilateral ANCM. 

Answer 1

We know the lengths of AD, DM, BC, and BN are equal, we know the lengths of AM, AN, CM, and CN are equal, and we know that AM, AN, CM, and CN are 6 units longer than the other segments.

ADM and BCN form right isosceles triangles in which the hypotenuse is 6 units longer than the legs. So by the Pythagorean theorem, n^2 + n^2 = (n+6)^2. 2n^2 = n^2 + 12n + 36. 0 = -n^2 + 12n + 36.

The short segments = 6 + 6 sqrt 2, and the long segments = 12 + 6 sqrt 2.

Perimeter = 4 short segments + 2 long segments = 48 + 36 sqrt 2 = approx. 98.91.

Answer 2

Did this question have an original perimeter for the rectangle? Otherwise it has to be written in terms of a variable, 4x, where x is the length of AM and CM. This is due to the fact that an angle bisector of a rectangle is 45 degrees, which forms equilateral triangles once it hits the opposite side. Draw it and you should see the relationships. You will get an exact value with a bit more information.