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) 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 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 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.