I’m rohmbus abcd, ab= 14 and ac= 26. find the area of the rhombus to the nearest tenth.

The complete question in the attached figure
 we know
In a rhombus ABCD , AC  = 26 , AB  = 14
the diagonals of a rhombus are perpendicular bisectors of each other
AX  = XC  = 13. triangle ABX  is a right triangle with hypotenuse 14.

From the pythagoras theorem AB² = AX² + BX²

14² = 13² + BX²

196 = 169 + BX²

27 = BX²-----  > BX=√27

BX  = 5.20

Area of ABX  triangle=(1/2)*5.2*13=33.77

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

 Area = 4* 33.77=135.10 units²

the answer is the option D) 135.1 units ²

I’m a smart person so that’s why i think it was 6