Question

Involves triangle centroid

Answer 1

Here, you do the same as the previous one, keeping in mind that X cuts TW in a 2:1 ratio.

$\bf TW-TX=XW\implies \cfrac{8a}{5}+\cfrac{1}{10}-\left(a +\cfrac{4}{5} \right)=XW \\\\\\ \cfrac{8a}{5}+\cfrac{1}{10}-a-\cfrac{4}{5}=XW\implies \cfrac{3a}{5}-\cfrac{7}{10}=XW\\\\ -------------------------------$

$\bf TX:XW\qquad 2:1\qquad \cfrac{TX}{XW}=\cfrac{2}{1}\implies \cfrac{a+\frac{4}{5}}{\frac{3a}{5}-\frac{7}{10}}=\cfrac{2}{1} \\\\\\ \cfrac{\frac{5a+4}{5}}{\frac{6a-7}{10}}=\cfrac{2}{1}\implies \cfrac{5a+4}{5}\cdot \cfrac{10}{6a-7}=\cfrac{2}{1}\implies \cfrac{10a+8}{6a-7}=\cfrac{2}{1} \\\\\\ 10a+8=12a-14\implies 22=2a\implies \cfrac{22}{2}=a\implies 11=a$