**1) Draw a line segment of length 7.6 cm and divide it in 5: 8 ration. In addition, find the measure of two parts.**

**Solution:**

Procedure for construction:

- Draw any ray AX, making an acute angle with AB.
- Locate 13(= 5 + 8) points A
_{1}, A_{2}, A_{3}……… A13 on AX so that AA_{1}= A_{1}A_{2}A_{l2}A_{13}. - Join BA
_{13}. - Through the point A5(m = 5), draw a line parallel to BA13 (by making an angle equal to L AA13 B at A5 intersecting AB at C. Then AC: CB = 5 : 8)

**2) Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a similar triangle to it whose sides are 2/3 of the corresponding sides of the first one.**

**Solution:**

Procedure for construction:

- Draw a line segment BC with length 5 cm.
- With B as centre and radius of 4 cm draw an arc.
- With C as centre and radius of 6 cm draw an arc.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle ∠CBX
- Along BX, mark up three points B1, B2, B3 such that BB1 = B
_{1}B_{2}= B_{2}B_{3} - Join B
_{3}C - From B2, draw B
_{2}C’llB_{3}c, meeting BC at C’ - From C’ draw C’ All CA, meeting BA at A’
- Then ∆A’BC’ is the required triangle, each of whose sides is two-third of the corresponding sides of ∆ABC.

**3) Construct a triangle with side lengths 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.**

**Solution:**

Procedure for construction :

- Draw a line segment BC with length 6 cm.
- With B as centre and keeping radius as 5 cm, draw an arc.
- With C as centre and keeping radius as 7 cm, draw another arc, intersecting the previously drawn arc at Point A.
- Join AB and AC. Then, ∆ABC is the required triangle.
- Below BC, make an acute angle∠CBX.
- Along BX, mark up seven points B
_{1}, B_{2}, B_{3}….. B_{7}such that BB_{1}= B_{1},B_{2}, B_{6}B_{7}. - Join B
_{5}to C (5 being smaller of 5 and 7 in7/5) and draw a line through B_{7}parallel to B_{5}C, intersecting the extended line segment BC at C’. - Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Then A’BC’ is the required triangle.

**4) Draw a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of ∆ABC.**

**Solution:**

**Procedure for construction:**

(i) Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii) Locate 4(the greater of 3 and 4 in ¾) points B1, B2, B3, B4 on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

(iv) Join B_{4}C and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in ¾) parallel to B_{4}C to intersect BC at C’.

(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then ∆A’BC’ is the required triangle.

Justification of construction

∆ABC ~ ∆A’BC’ , Therefore,

But,

So,

**5) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are4/3 times the corresponding sides of ∆ABC.**

**Solution:**

**Procedure for construction : **

(i) Draw a triangle ABC with BC = 7cm, ∠B = 45° and ∠A = 105°.

(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex X.

(iii) Locate 4(the greater of 3 and 4 in 4/3) points B_{1}, B_{2}, B_{3}, B_{4} on BX so that 3 BB_{1} = B_{1} B_{2} = B_{2}B_{3} = B_{3}B_{4}.

(iv) Join NC’ and draw a line through B3(the 3rd point, 3 being smaller of 3 and 4 in 1) parallel to NC’ to intersect BC’ at C. 3

(v) Draw a line through C’ parallel to the line CA to intersect BA at A’. Then A NBC’ is the required triangle.

**6) Construct a triangle of isosceles type, whose base is 8 cm and height 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.**

**Solution:**

**Given**:An isosceles triangle whose base is 8 cm and height 4 cm. Scale factor: 1 =

**Required:** To construct a similar triangle to above whose sides are 1.5 times the above triangle.

**Procedure for construction:**

(i) Draw a line segment BC = 8 cm.

(ii) Draw a perpendicular bisector AD of BC.

(iii) Join AB and AC we get a isosceles ∆ABC.

(iv) Construct an acute angle∠CBX downwards.

(v) On BX make 3 equal parts.

(vi) Join C to B_{2} and draw a line through B_{3} parallel to B_{2}C intersecting the extended line segment BC at C’.

(vii) Again draw a parallel line C’A’ to AC cutting BP at A’.

(viii) ∆A’BC’ is the required triangle.