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 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 a line segment BC = 4 cm.
ii) Bisect BCat D
iii) From B and D, draw arcs at distances 5 cm each on the same side of BC, cutting each other at A
iv) Join AB and AC.
v) Then, ΔABC is the required triangle.

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.

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 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:
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)