[tex]\text{Area of Rhombus = }\frac{1}{2}\times\text{ diagonal 1}\times\text{ diagonal2}[/tex]
we will need to find x, using Pythagoras theorem, we have
[tex]\begin{gathered} \text{hyp}^2=opp^2+Adj^2 \\ 13^2=5^2+x^2 \\ 169=25+x^2 \\ 169-25=x^2 \\ 144\text{ =}x^2 \\ x^2\text{ = 144} \\ x\text{ = }\sqrt[]{144} \\ x\text{ = 12} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{Diagonal 1 = 10} \\ \text{Diagonal 2 = 12+ 12 = 24} \\ \text{Area =}\frac{1}{2}\times10\times24 \\ \text{Area =}\frac{240}{2}=120u^2 \end{gathered}[/tex]
