In an isosceles triangle, the radius of the inscribed circle is ${ }^{2} / 7$ of the height, and the perimeter of this triangle is 56. Find its sides.
Let $CM$ be the height of the given triangle $ABC$, $AC = BC$, $O$ be the center of the inscribed circle. Then $OM$ is the radius of this circle, $AO$ is the bisector of angle $A$. Therefore, $AC: AM = CO: OM = 5: 2$. Since $AC + AM = 28$, then $AC = 20, AM = 8, AB = 16$

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Answer

$16, 20, 20$.