# 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.

Result

S =  80 cm2

#### Solution:

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

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