Final answer:
The length of AC, to the nearest tenth, in a rhombus with AB = 18 and BD = 20 is 30.5.
Step-by-step explanation:
Given that ABCD is a rhombus, we know that the opposite sides are congruent. Therefore, AB = CD and BC = AD. In this case, AB = 18, so CD = 18 as well. We also know that BD = 20, which is the diagonal of the rhombus. Since the diagonals of a rhombus bisect each other at right angles, we can use the Pythagorean theorem to find the length of AC.
Let's label the midpoint of BD as E. Now we have two right triangles, ABE and CDE. The length of one leg in both triangles is 9 (half of AB and CD), and the length of the other leg in both triangles is 10 (half of BD).
Using the Pythagorean theorem in triangle ABE, we have:
AB^2 = AE^2 + BE^2
18^2 = 9^2 + BE^2
324 = 81 + BE^2
BE^2 = 243
BE = √243
BE = 15.6
Now, using the Pythagorean theorem in triangle CDE, we have:
CD^2 = CE^2 + DE^2
18^2 = CE^2 + 10^2
324 = CE^2 + 100
CE^2 = 224
CE = √224
CE = 14.9
Since AC is the sum of AE and CE, we have:
AC = AE + CE
AC = 15.6 + 14.9
AC = 30.5
Therefore, the length of AC, to the nearest tenth, is 30.5.