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 XY = 11 cm (As AB+  BC + CA = 11cm)
ii) Construct an angle PXY of  60^o at point X and an angle ㄥQYZ of 45^o at point Y
iii) Bisect ㄥPXY and ㄥQYZ .These bisectors intersect each other at point A
iv) Draw perpendicular bisectors ST of XA and UV of YA.
v) Perpendicular bisector ST intersects XY at B and UV intersects XY at C Join AB, AC MBC is the required triangle.

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.

(i) Steps of Construction
1. Draw the base PQ = 8.4 cm.
2. At point P make an angle say XPQ=45°.
3. Cut the line segment PD = 2.8 ern from rayPX.
4. Join DQ and draw the perpendicular bisector of DQ.
5. Let it intersect PX at a point R. Join RQ. Then PQR is the required triangle.

(ii) By measurement, PR = 10cm
(iii) By measurement, QR = 7.2cm
(iv) PR - QR = 10-7.2 = 2.8 cm
(v) By measurement,
ㄥPQR = 54°
∴ Gaffar is correct.
∴ The value 'intelligence' is depicted by his statement.

(i) Steps of Construction
1. Draw the base BC = 5 em.
2. At point B make an angle XBC = 45°.
3. CutthelinesegmentBD=AB-AC(=2.8 cm) from the ray BX.
4. Join DC.
5. Draw the perpendicular bisector, say PQ of DC.
6. Let it intersect BX at a point A.
7. Join AC.
Then, ABC is the required triangle.

(ii) By measurement, AB = 13cm
(iii) By measurement, AC = 10.2cm
(iv) AB - AC = 13 - 10 .2 = 2.8cm
(v) Yes! Hari is true as by measurement ㄥACB = 112°.
The value 'wise' is depicted by comment of Hari.

(i) Steps of Construction
1. Draw the base AB = 5.8 ern.
2. At the point A, make an angle, say XAB=60°.
3. Cut a line segment AD equal to BC + CA = 8.4 cm from the ray BX.
4.Join DB
5.Make an angle DBY equal to ADB.
6.Let BY intersect AD at C. Then, ABC is the required triangle.

(ii) By measurement, AC = 3.4 cm
(iii) By measurement, BC = 5cm
(iv) Yes! AC + B C =3.4 + 5 = 8.4cm
(v) By measurement, ㄥACB = 84°
Meenu is right.
The value 'exactness' is depicted by Meenu's statement.

Steps of Construction
1. Draw a line segment AB = 5 cm.
2. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point P.
3. Taking P as centre and with the same radius as before, draw an arc intersecting the previously draw arc, say at a point E.
4. Draw the ray AC passing through E.
5. From ray AC, cut off AD = 6 cm. Then, \(\angle DAB\) is the required angle of 60° such that AD = 6 cm.
6. Now, taking A and D as centres and radius 1 more than \(\frac { 1 }{ 2 } \) AD, draw arcs on both sides of the line segment AD (to intersect each other).
 

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
1. Draw BC = 6 cm.
2. With B as centre and 6 cm as radius, draw an arc on one side of BC.
3. With C as centre and 6 cm as radius, draw another arc on the same side of BC to intersect the former arc at A.
4. Join AB and AC. Then, \(\Delta \) ABC is the required equilateral triangle.