Construct a right-angled triangle whose base is $12 \, cm$ and the sum of its hypotenuse and the other side is $18 \, cm$.
(N/A) $I.$ Draw a line segment $\overline{BC} = 12 \, cm$.
$II.$ Construct $\angle CBY = 90^{\circ}$ at point $B$.
$III.$ From the ray $\overrightarrow{BY}$,cut off a line segment $BX = 18 \, cm$.
$IV.$ Join $CX$.
$V.$ Construct the perpendicular bisector of $CX$,and let it intersect $BX$ at point $A$.
$VI.$ Join $AC$.
Thus,$\triangle ABC$ is the required right-angled triangle.
$II.$ Construct $\angle CBY = 90^{\circ}$ at point $B$.
$III.$ From the ray $\overrightarrow{BY}$,cut off a line segment $BX = 18 \, cm$.
$IV.$ Join $CX$.
$V.$ Construct the perpendicular bisector of $CX$,and let it intersect $BX$ at point $A$.
$VI.$ Join $AC$.
Thus,$\triangle ABC$ is the required right-angled triangle.
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