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.

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 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

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.

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 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.

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.

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

 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 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)