by construction of triangle
then, make an angle of 45° on point B
then, draw a ray BX of 45° and then, mark an arc of AB + BC = 5.6 cm on BX and name the point as D and then join D to C
and draw a perpendicular bisector of DC
and where the bisector intersect BX name that point as A and join AC
Hence your triangle is ready
Step-1: Draw the base line segment QR= 6 cm.
Step-2 : From Q, draw an angle of using compass, say ∠XQR.
Step-3 :Extend the line QX in the opposite direction as well to T. Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. Now join the point to R in order to get SR.
Step-4 : Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.
Thus ΔPQR is the required triangle.
then the answer is 9.45
ya me too msg