Center of gravity

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

Correct result:

S =  80 cm²

Solution:

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

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