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 r2=h2+(a/2)2 a:h=10:12 r2=h2+(10/12⋅ h/2)2 r2=h2+(5/12)2⋅ h2 h=r2/(1+(5/12)2)=262/(1+(5/12)2)=24 a=10/12⋅ h=10/12⋅ 24=20 h2=h/3=24/3=8 S=a⋅ h2/2=20⋅ 8/2=80 cm2r=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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