### Construct a right triangle whose base is $17 \space cm$ and sum of its hypotenuse and another side is $23 \space cm$.

1.  Let $ABC$ be the required triangle such that $BC = 17 \space cm, \angle B = 90 ^ \circ$ and $AC + AB = 23 \space cm$. Draw a $BC = 17 \space cm.$ C B
2.  At $B$ construct $\angle CBX = 90^\circ.$ C B X
3.  From $B$ cut off $BD = 23 \space cm$. C B X D
4.  Join $CD$ and draw perpendicular bisector of $CD$ intersecting $BD$ at $A$. C B X D A
5.  Join $AC$ to get the required triangle $ABC$. C B X D A