Construction:
Steps (1) Draw a line segment BC = 5.6 cm
Steps (2) At B, draw BE such that \(\angle CBE={ 60 }^{ 0 }\)
Steps (3) At B draw BF such that \(\angle EBF={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to BC, which intersects BF at O and BC at G.
Steps (5) With O as centre and OB as radius draw a circle
Steps (6) From B, marked an arc of 4 cm on BC at D.
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID.
Steps (8) ID produced meets the circle at A. Now joined AB and AC. Then \(\triangle\)ABC is the required triangle.

Construction:
Step (1) Draw \(\bar { AB } =5.5cm\)
Step (2) Draw \(\angle BAE={ 25 }^{ 0 }\)
Step (3) Draw \(\angle FAE={ 90 }^{ 0 }\)
Step (4) Drawn the perpendicular bisector XY to AB which intersects AF at O and AB at G.
Step (5) With O as center and OA as radius drawn a circle
Step (6) XY intersects AB at G. On XY from G marked an arc at M such that GM = 4 cm
Step (7) Drawn PQ through M which is parallel to AB.
Step (8) PQ meets the circle at C and S.
Step (9) Joined AC and BC. Now \(\triangle\)ABC is the required triangle

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.

Construction:
Step (1) Draw a line segment QR = 5 cm.
Step (2) At Q, draw QE such that \(\angle RQE\) = 40°.
Step (3) At Q, draw QF such that \(\angle EQF\) = 90^o
Step (4)Drawn a perpendicular bisector to QR, which intersects QF at 'O' and QR at G.
Step (5) With O as centre and OQ as radius, draw a circle
Step (6) From G marked arcs of radius 4.4 cm on the circle. Marked them as P and S.
Step (7) Joined QP and PR. Now \(\triangle\)PQR is the required triangle
Step (8) From P draw a line PN which is \(\bot \) to LR. LR meets PN at M.
Step (9) The length of the altitude is PM = 2.1cm

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle