The height of a rhombus, drawn from the vertex of an obtuse angle, divides its side into segments of lengths m and n. Determine the diagonals of the rhombus.
Solution
The side length of the rhombus is m+n. From triangle \triangle ABK (see the figure, assumed to be a standard diagram where B is an obtuse angle vertex, BK is the height to side AD, and K lies on AD, such that AK = m and KD = n), we find:
BK^2 = (m+n)^2 - m^2.
In triangle \triangle BKD (where BD is a diagonal), we have:
BD^2 = BK^2 + n^2 = (m+n)^2 - m^2 + n^2 = m^2 + 2mn + n^2 - m^2 + n^2 = 2mn + 2n^2 = 2n(m+n),
thus, BD = \sqrt{2n(m+n)}.
Since the sum of the squares of the diagonals of a rhombus is equal to four times the square of its side length, we have:
AC^2 + BD^2 = 4AD^2 = 4(m+n)^2.
Therefore,
AC^2 = 4(m+n)^2 - 2n(m+n).
Expanding and simplifying:
AC^2 = 4(m^2 + 2mn + n^2) - 2mn - 2n^2 = 4m^2 + 8mn + 4n^2 - 2mn - 2n^2 = 4m^2 + 6mn + 2n^2,
thus, AC = \sqrt{4m^2 + 6mn + 2n^2}.
Answer:
The lengths of the diagonals are BD = \sqrt{2n(m+n)} and AC = \sqrt{4m^2 + 6mn + 2n^2}.
