Right triangle ABC has side lengths AB=7 , BC=24 , and AC=25 . A second right triangle, A′B′C′ , has side lengths of 50, 14, and 48. Find the ratio of the side opposite ∠A to the hypotenuse of triangle ABC . Use this ratio to identify the location of point A′ in the second right triangle.
In triangle ABC, by the Pythagorean theorem, we have $AB^2+BC^2=AC^2$, so $7^2+24^2=25^2$ and triangle ABC is a right triangle. The angle opposite side $AB$ is $\angle A$, so the desired ratio is $\boxed{\frac{7}{25}}$.
Also in triangle ABC, notice that the ratio of $AB$ to $AC$ is the same as the ratio of $24$ to the hypotenuse $AC$. That means triangle ABC is similar to triangle AA'C, where A' is on line segment BC such that BA' has length 24. The ratio of the sides is preserved, so from triangle AA'C, the hypotenuse $A'C$ must have length $24\cdot \frac{25}{7}=\boxed{86\frac{2}{7}}$.
[asy] pair A,B,C,Ap; C=(0,0); B=(0,24); A=(7,0); draw(A--B--C--cycle); draw(rightanglemark(A,C,B,30)); Ap=(0,24*25/7); draw(Ap--C); draw(rightanglemark(A,Ap,C,30)); dot(A^^B^^C^^Ap); label("$A$",A,S); label("$A'$",Ap,NE); label("$B$",B,NW); label("$C$",C,S); [/asy]