Steps of construction:
i) Draw a line segment AB = 11 cm(As XY + YZ + ZX = 11cm)
ii) Construct an angle ㄥPAB of 30^o at point A and an angle ㄥQAB = 90^o at point B.

iii) Bisect ㄥPAB of 30^o and ㄥQBA.These bisectors intersect each other at point X.
iv) Draw perpendicular bisectors ST of AX and UV of BX
v) ⊥ bisector ST intersects AB at Y and UV intersects AB at Z. Join XY,XZ. ΔXYZ is the required triangle.

 aㄥA= \(\frac { 3 }{ 12 } \times { 180 }^{ o }\)=45^o
ㄥB=\(\frac { 4 }{ 12 } \times { 180 }^{ o }\) =60^o
ㄥC= \(\frac { 5 }{ 12 } \times { 180 }^{ o }\) =75^o

Steps of construction:
i) Draw a line PQ = 12.5 cm.
ii) At P, construct ㄥSPQ=60^o and at Q , construct ㄥRQP=75^o
iii) Draw the bisectors of ㄥSPQ and ㄥRQP ,intersecting at A.
iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at Band C respectively.
v) Join A to B and A to C.
ABC is the required triangle.

Steps of construction:
i) Draw a line segment QR = 6cm.At point Q construct an angle = 60^o i.e ㄥXQR = 60^o 

ii) Cut a line segment QS = 2 cm from the line segment QT extended on opposite side of line segment XQ.
(As PR > PQ and PR - PQ = 2cm) join SR
iii) Draw perpendicular bisector AB of line segment SR,which intersects QX at point P.Join PQ,PR. ΔPQR is the required triangle.

Steps of construction:
i) Draw the line segment BC = 8 cm and at point B construct an angle of 45o .i.e XBC = 45^o and AB - AC = 3.5 cm
ii) Cut the line segment BD = 3.5 cm(equal to AB-AC)on ray BX
iii) Join DC and draw the perpendicular bisector PQ of DC
iv) The perpendicular bisector intersects BXat point A.Join AC MBC is the required triangle.

Steps of construction:
i) Draw line segment BC = 6 cm
ii) Draw the perpendicular bisector of BC which intersect BCat D.
iii) Now with D as the centre and radius = 4.5 cm.
iv) With Bas the centre and radius = 3 cm, draw an are cutting BE produced A.
v) Join A to C. ABC is the required Δ

Steps of construction:
i) Draw a ray with end points A and B using ruler.
ii) With A as centre and any radius, draw an arc cutting the ray at point C using compass.
iii) With C as centre and same radius draw an arc cutting the arc drawn at D.
iv) With D as centre and the same radius, draw an arc intersecting the previously drawn arc at E.
v) Now, take any radius and draw two arcs with D and E as centres. Let these two arcs intersect at a point F.
vi) Join AF, ㄥFAB obtainted is the angle of measure of 90^o
vii) The line AF intersect the arc at the point named as G.
viii) With G and E as centre, draw an arc with same radius, the arc intersect at point P.
ix) Join the point P and A. ㄥPAB obtained is the angle of measurement 105^o with reference to line AB.
x) Now bisecting this angle will give angle  \(\frac { { 105 }^{ o } }{ 2 } =52\frac { 1^{ o } }{ 2 } \) angle.

Given: Side (say 4 cm) of an equilateral triangle.
Required: To construct the equilateral triangle and justify the construction.
Steps of Construction:
1. Take a ray AX with initial point A. From AX, cut off AB = 4 cm.
 
2. Taking A as centre and radius (= 4 em), draw an arc of a circle, which intersects AX, say at a point B.
3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
4. Draw the ray AE passing through C.
5. Draw the ray BF passing through e. Then \(\Delta \) ABC is the required triangle with given side 4 cm.