Construct a right triangle where one side is $3.5 \, cm$ and the sum of the other side and the hypotenuse is $5.5 \, cm$.
(N/A) In $\triangle ABC$,let the base $BC = 3.5 \, cm$,the sum of the other side and the hypotenuse be $AB + AC = 5.5 \, cm$,and $\angle ABC = 90^{\circ}$.
Steps of construction:
$1.$ Draw a ray $BX$ and cut off a line segment $BC = 3.5 \, cm$ from it.
$2.$ Construct $\angle XBY = 90^{\circ}$ at point $B$.
$3.$ From ray $BY$,cut off a line segment $BD = 5.5 \, cm$.
$4.$ Join $CD$.
$5.$ Draw the perpendicular bisector of $CD$,which intersects $BD$ at point $A$.
$6.$ Join $AC$. Then,$\triangle ABC$ is the required triangle.
Steps of construction:
$1.$ Draw a ray $BX$ and cut off a line segment $BC = 3.5 \, cm$ from it.
$2.$ Construct $\angle XBY = 90^{\circ}$ at point $B$.
$3.$ From ray $BY$,cut off a line segment $BD = 5.5 \, cm$.
$4.$ Join $CD$.
$5.$ Draw the perpendicular bisector of $CD$,which intersects $BD$ at point $A$.
$6.$ Join $AC$. Then,$\triangle ABC$ is the required triangle.
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