+66 ![[asy] size(5cm);pair A,B,C,D;A=(0,0);B=(5,0);C=(3,3);D=(0,1);draw(A--C,0.5*red + 0.5*white);draw(A--B--C--D--A);draw(rightanglemark(B,A,D)^^rightanglemark(D,C,B));add(pathticks(B--C, 1, .5, 6, 10));add(pathticks(C--D, 1, .5, 6, 10));label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F3%2Fb%2F1%2F3b18c58f6e9ade0e7e2bdbee84c5d8ea10a824f2.png)
Find AC
I figured out that CB and DC are 5, but I can't figure out AC.
+129852 You are correct in that CB = DC = 5 ......so....let's carry this a little further....
Note that the four interior angles must sum to 360°
This means that angles D and B are supplementary (since A and C = 90° )
This also means that the cos B = -cos D (since D) is > 90°
Put another way..... -cosB = cos D
Since AC forms two triangles we can use the Law of Cosines twice and we have that
AC^2 = 1^2 + 5^2 - 2 ( 1)(5) (-cos B) (1)
AC^2 = 5^2 + 7^2 - 2 ( 5) (7) (cos B) (2)
Which implies that
1^2 + 5^2 + 10cos B = 5^2 + 7^2 - 70 cos B
26 + 10cos B = 74 - 70 cos B
80cos B = 48
cos B = 48 / 80 = 3/5
So using (2)
AC^2 = 74 - 70 (3/5)
AC^2 = 32
AC = sqrt (32) = 4sqrt (2) ≈ 5.66
