2025 AMC 12B Problem 21

With sides $8$ and $9$ and included angle $\theta,$ the area is $36\sin\theta,$ so equal areas come from $\theta$ and $180^\circ - \theta$

By the law of cosines the third sides satisfy $t^2 = 145 \mp 144\cos\theta,$ so $t_1^2 + t_2^2 = 290$

The area with included angle $\theta$ is $36\sin\theta,$ so two triangles of equal area use angles $\theta$ and $180^\circ - \theta,$ with cosines $\pm\cos\theta.$ By the law of cosines the third sides satisfy $t^2 = 145 \mp 144\cos\theta,$ hence $t_1^2 + t_2^2 = 290.$ Since $290\equiv2\pmod8,$ both integer sides must be odd. Checking the odd squares in the triangle-inequality range $1\lt t\lt17$ leaves only $11^2+13^2=121+169=290.$ Therefore the sum is $24.$

Thus, the correct answer is C.