**Answer
1** :

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.

(iii) Join AB and AC. Then ABC is the triangle.

(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB1 = B1B2= B2B3.

(v) Join B3C.

(vi) Draw B’C’ parallel to B3C and C’A’ parallel to CA then ΔA’BC’ is the required triangle.

**Answer
2** :

Steps of construction :

(i) Draw a line segment BC = 7 cm.

(ii) Draw a ray BX making an angle of 50° and cut off BA = 5 cm.

(iii) Join AC. Then ABC is the triangle.

(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B1B2=B2B3=B3B4=B4Bs=B5B6=B6B7

(v) Join B7 and C

(vi) Draw B5C’ parallel to B7C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

**Answer
3** :

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) Draw a ray BX making an angle of 50° and CY making 60° with BC which intersect each other at A. Then ABC is the triangle.

(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB1 =B1B2 = B2B2

(iv) Join B3C.

(v) From B2, draw B2C’ parallel to B3C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

**Answer
4** :

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs’intersecting eachother at A.

(iii) Join AB and AC. Then ABC is the triangle,

(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB1= B1B2 = B2B3 = B3B4.

(v) Join B4 and C.

(vi) From B3C draw C3C’ parallel to B4C and from C’, draw C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

**Answer
5** :

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting eachother at A.

(iii) Join AB and AC. Then ABC is the triangle.

(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.

(v) Join B5 and C.

(vi) From B7, draw B7C’ parallel to B5C and C’A’ parallel CA. Then ΔA’BC’ is the required triangle.

**Answer
6** :

Steps of construction :

(i) Draw a line segment AB = 4.5 cm.

(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.

(iii) Join BC. Then ABC is the triangle.

(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 =A3A4 = A4A5

(v) Join A4 and B.

(vi) From 45, draw 45B’ parallel to A4B and B’C’ parallel to BC.

Then ΔAB’C’ is the required triangle.

**Answer
7** :

Steps of construction :

(i) Draw a line segment BC = 5 cm.

(ii) At B, draw perpendicular BX and cut off BA = 4 cm.

(iii )join Ac , then ABC is the triangle

(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5

(v) Join B3 and C.

(vi) From B5, draw B5C’ parallel to B3C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

**Answer
8** :

Steps of construction :

(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.

(ii) Join AB and AC. Then ABC is the triangle.

(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD1 = D1D2 =D2D3 = D3D4

(iv) Join D2

(v) Draw D3A’ parallel to D2A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.

Then ΔB’A’C’ is the required triangle.

**Answer
9** :

Steps of construction :

(i) Draw a line segment BC = 6 cm.

(ii) At B, draw a ray BX making an angle of 60° with BC and cut off BA = 5 cm.

(iii) Join AC. Then ABC is the triangle.

(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB1= B1B2 B2B3=B3B4.

(v) Join B4 and C.

(vi) From B3, draw B3C’ parallel to B4C and C’A’ parallel to CA.

Then ΔA’BC’ is the required triangle.

**Answer
10** :

Steps of construction :

(i) Draw a line segment AB = 4.6 cm.

(ii) At A, draw a ray AX making an angle of 60°.

(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.

(iv) Join BC. Then ABC is the triangle.

(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 = A3A4=A4A5.

(vi) Join A4 and B.

(vii) From A5, drawA5B’ parallel to A4B and B’C’ parallel to BC.

Then ΔC’AB’ is the required triangle.

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