Center of gravity

In the isosceles triangle ABC the lengths of AB and the height to AB is the ratio of 10:12. The arm has a length of 26 cm. If the center of gravity is T, find the area of the triangle ABT.

Correct answer:

S =  80 cm²

Step-by-step explanation:

$$\begin{gathered} r=26 \\ {r}^2=h^2+(a/2{)}^2 \\ a:h=10:12 \\ {r}^2=h^2+(10/12\cdot h/2{)}^2 \\ {r}^2=h^2+(5/12{)}^2\cdot h^2 \\ h=\sqrt{r^2/(1+(5/12{)}^2)}=\sqrt{26^2/(1+(5/12{)}^2)}=24 \\ a=10/12\cdot h=10/12\cdot 24=20 \\ {h}_2=h/3=24/3=8 \\ S=a\cdot h_2/2=20\cdot 8/2=80{\text{cm}}^2 \end{gathered}$$

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