i) Draw a line XY
ii) Construct perpendicular PD at any point D on the line XY
iii) From point D, cut OH line segment AD = 6 cm.
iv) Construct ㄥBAD =ㄥCAD = 30^o Then ABC is required triangle
Justification of Construction:
As A = ㄥBAD +ㄥCAD = 30^o + 30^o = 60^o and AD⊥ BC.Therefore ΔABC is an equilateral triangle with altitude AD = 6 cm.

Steps of construction:
i) Draw a line segment BC = 4.7 cm. and at point B construct an angle of 45^o i.e ㄥXBC=45^o
ii) Cut the line segment BD = 2 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. Jon AC ΔABC is the required triangle.

(i) Steps of Construction
1. Draw the base BC = 5 "7
2. At point B make XBC = 30°.
3. Cut the line segment =3 cm iron the line BX extended II opposite side of line segment Be.
4. Join DC and draw the perpendicular bisector, say PQ of De.
5. Let PQ intersect BX at A. Join AC.Then, ABC is the required triangle.

(ii) By measurement, AC = 4.5cm
(iii) By measurement, AB = 1.5cm
(iv) AC - AB =4.5 -1.5 = 3 cm
(v) The value 'ponderance' is depicted by her ponderation.

Steps of Construction
1. Draw an angle AOB = 40° with a protractor.
2. Taking O as centre and some radius, draw an arc of a circle, which intersects OA at P and OB at Q.
 
3. Taking P as centre and radius QP, draw an arc of a circle, which intersects the arc drawn in step 2, say at a point R.
4. Join OR and produce to form a ray Oc. Then, \(\angle COB=80°\)

Steps of Construction
1. Draw a line segment MN = 5 cm.
2. With M as centre and 5 cm as radius, draw an arc on one side of MN.
3. With N as centre and 5 cm as radius, draw another arc on the same side of MN to intersect the former arc at L.
4. Join LM and LN. Then, \(\Delta \) LMN is the required equilateral triangle.

5. Taking M as centre and any radius, draw an arc to intersect the line segments MN and ML at P and Q respectively.
6. Next, taking P and Q as centres and with 1 the radius more than \(\frac { 1 }{ 2 } \) PQ, draw arcs to intersect each other, say at R.
7. Draw the ray MR. This ray MR is the required bisector of the \(\angle M\).

Steps of Construction:
i) Draw any line segment PQ = 5.5. cm
ii) With P as centre and radius 5.5 cm draw an arc
iii) With Q as centre and radius 5.5 cm draw an arc to cut the previous arc at R
iv) Join PR and QR, then PQR is the required triangle.

Steps of construction:
i) Draw ㄥDEF=72^o ,using protractor
ii) Bisect it. Let the bisected angle be ㄥDEK
iii) Again bisect ㄥDEK
iv) Now ㄥGEF = \(\frac { 3 }{ 4 } \)ㄥDEF

Steps of construction:
i) Construct a triangle ABC.
ii) Mark an exterior angle outside the triangle ABC,and name the point as E.

iii) Now, ACE is the exterior angle.
iv) Draw a bisecter of ㄥACE

 We know that each angle of equilateral triangle is 60^o,So have to draw an angle 60^oand bisect it.
Construction:
i) Draw any line OP.
ii) With 0 as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius draw an arc to meet the previous arc at S.
iv) Join OS and Produce it to Q, then ㄥPOQ=60^o
v) With R as centre and any suitable radius (not necessarily)equal to radius of step 1 (but >\(\frac { 1 }{ 2 } \)RS),draw an arc. Also, with 5 as centre and radius draw another arc to meet the previous arc at Y
vi) Join OY and produced it, the OY is the required bisector of ㄥPOQ(i.e,ㄥPOY=30^o)

Steps of construction:
i) Draw a line OL using a ruler.
ii) Keeping O as center, with any radius draw an arc cutting the ray at point M using compass.
iii)T aking M as centre, draw arc to meet at previous arc at P.
iv) With P and M as centres and equal radius draw arcs intersecting at R. Join OR and extend to Q.
v) ㄥLOQ=30^o
vi) With Rand M as centre, draw arcs with same radius or more than half of RM, meeting at point S
vii)J oin OS, which makes an angle, ㄥLOS = 15^o

Steps of Construction:
i) Draw any line OP.
ii) With O as centre and any suitable radius, draw an arc to meet OP at R.
iii) With R as centre and same radius (as in step 2).draw an arc to meet the previous arc at S.
iv) Join OS and Produce it to Q,then ㄥPOQ=60^o
v) With R as centre and any suitable radius (not necessarily) equal to radius of step 1 (but > \(\frac { 1 }{ 2 } \)RS),draw an arc. Also, with 5 as centre and radius draw another arc to meet the previous arc at Y.
vi) Join OY and produced it, the OY is the required bisector of ㄥPOQ (i.e ㄥPOY=30^o)

 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 BC = 7 cm. At point B draw ㄥXBC=  75^o
ii) Cut a line segment BD = 13 cm (i.e., equal to AB + AC) from BX
iii) Join DC and make an angle DCY equal to ㄥBDC
iv) Line CY intersects BX at A. ΔABC is the required triangle.

Steps of construction:
i) Draw a ray BX and cut off line segment BC = 4cm
ii) Construct ㄥXBY = 90^o
iii) From BY cut off line segment BD = 8 cm,
iv) Join CD.
v) Draw the ⊥ bisector of CD, intersecting BD at A
vi) Join AC, ABC is the required triangle