I’m rohmbus ABCD, AB= 14 and AC= 26. Find the area of the rhombus to the nearest tenth.

From the pythagoras theorem AB² = AX² + BX²
14² = 13² + BX²
196 = 169 + BX²
27 = BX²----- > BX=√27
BX = 5.20
The four triangles formed by construction of the diagonals are all congruent since the diagonals are perpendicular bisectors.
Total area of the rhombus is 4 times of ABX triangle