Steps of construction
(i) Draw a line segment BC = 8 cm.
(ii) At point B, Using ruler and compass, draw an angle ∠CBX = 45°.
(iii) Taking B as centre and radius 3.5 cm, mark an arc, which intersects AX at D.
(iv) Jion CD and draw a perpendicular bisector (MN) of CD, which intersects at BD produced at A.
(v) join AC.
(vi) Triangle ABC is the required triangle.
Justification
Point A lies on the perpendicular bisector of DC. So, AD = AC
Here, AB = BD – AD
⇒ AB = BD – AC [∵ AD = AC]

Steps of construction
(i) Draw a line segment AB = 11 cm.
(ii) At A, Using ruler and compass, draw an angle ∠BAX = 15° and at point B, draw an angle ∠ABX = 45°.
(iii) Draw the perpendicular bisector (MN) of AX, which intersects AB at Y.
(iv) Draw the perpendicular bisector (ST) of BX, which intersects AB at Z
(v) join X to Y and X to Z.
(vi) Triangle XYZ is the required triangle.
Justification
Point Y lies on the perpendicular bisector of AX.
So, AY = XY
Point Z lies on the perpendicular bisector of BX.
So, BZ = ZX
Here, AB = AY + YZ + ZB
⇒ AB = XY + YZ + ZX
[∵ AY = XY and BZ = XZ]
∠XYZ is the exterior angle of triangle AXY. Therefore, ∠XYZ = ∠YXA + ∠YAX = 15° + 15° = 30° Similarly, ∠XYZ is the exterior angle of triangle BXZ.
Hence, ∠XZY = ∠ZXB + ∠ZBX = 45° + 45° = 90°