Let's begin by listing out the information given to us:
We start out by observing that Triangles MKR & ACD are similar or proportional
[tex]\begin{gathered} MK=21;AC=\text{?} \\ MR=24;AD=28\frac{4}{5} \\ KR=CD=\text{?} \end{gathered}[/tex]
We will solve for the missing side by using the similar triangle theorem. This is shown below:~
[tex]\begin{gathered} \Delta MKR\approx\Delta ACD \\ \frac{MK}{AC}=\frac{MR}{AD} \\ \frac{21}{AC}=\frac{24}{28\frac{4}{5}} \\ \text{Cross multiply, we have:} \\ 24\cdot AC=28\frac{4}{5}\cdot21 \\ AC=\frac{28\frac{4}{5}\cdot21}{24}=25\frac{1}{5} \\ AC=25\frac{1}{5} \end{gathered}[/tex]
