Construct a right triangle whose base is $3\,cm$ and the sum of its hypotenuse and the other side is $9\,cm$. Write the steps of construction.

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(N/A) Steps of construction:
$1$. Draw a line segment $AB = 3\,cm$ as the base.
$2$. At point $A$,construct an angle of $90^{\circ}$ using a compass.
$3$. From the ray $AX$ (making $90^{\circ}$ with $AB$),cut off a line segment $AD = 9\,cm$.
$4$. Join $DB$.
$5$. Construct the perpendicular bisector of $DB$,which intersects $AD$ at point $C$.
$6$. Join $CB$. Thus,$\triangle ABC$ is the required right-angled triangle.
Verification: In $\triangle ABC$,$AB = 3\,cm$,$\angle A = 90^{\circ}$,and $AC + CB = AD = 9\,cm$ (since $C$ lies on the perpendicular bisector of $DB$,$CD = CB$).

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