Construct a triangle ABC with side BC = 6 cm, ∠ B = 45° , ∠ A = 105°. Then construct another triangle whose sides are times the corresponding sides of the Δ ABC.

Given,

•Side BC is 6 cm.

•∠ B is 45°.

•∠ A = 105°.

As the sum of angles in a triangle is 180°.

∠ A + ∠ B + ∠ C = 180°.

105° + 45° + ∠ C = 180°

150° + ∠ C = 180°

∠ C = 30°

Steps of construction:

1. Draw a line BC of length 6 cm.

2. With B as centre construct an angle of 45°.

a. With B as centre draw an arc of any convenient radius which cuts the line BC at D.

b. With D as centre and with the same radius as above (step a) draw an arc which cuts the previous arc at point E.

c. With E as centre and with the same radius as above (step a) draw an arc which cuts the previous arc at point F.

d. With E and F as centres and with radius more than half the length of EF, draw two arcs which intersect at G.

e. The line BG makes 90° angle with the line BC.

f. With B as centre and with some convenient radius draw an arc which cuts BG and BC at H and I.

g. With H and I as centres and with radius more than half the length of HI, draw two arcs which intersect at J.

h. Join BJ, which is the line which makes a 45° angle with line BC.

3. With C as centre construct an angle of 30°.

a. With C as centre draw an arc with some convenient radius which cuts BC at K.

b. With K as centre and with the same radius as above (step a) draw an arc which cuts the previous arc at point L.

c. With K and L as centres and with radius more than half the length of KL, draw two arcs which intersect at M.

d. Join CM which makes an angle of 30° with the line BC. Extend BJ and CM to join at point A. This is the required triangle Δ ABC.

4. Now with B as centre draw a ray BX making an acute angle with BC on the opposite side of vertex A. (as the scale factor is greater than 1). Mark 4 points B₁,B₂,B₃ and B₄ on BX which are equidistant. i.e. BB₁ = B₁B₂ = B₂B₃ = B₃B₄.

5. Join B₃C. Then draw a line B₄C’ which is parallel to B₃C and meets the extended line BC at C’.

6. Now draw a line C’A’ which is parallel to CA and meets the extended line AC at A’. The Δ A’BC’ is the required triangle of scale factor .

Justification:

Consider,

---- (1)

Also, as AC ∥ A’C’,

We can say that ∠ A’C’B = ∠ ACB (corresponding angles) – (2)

Now, consider the triangles Δ ABC and Δ A’BC’

∠ B = ∠ B (Common angle)

∠ A’C’B = ∠ ACB (from (2))

From the Angle Angle Similarity of triangles, we can clearly say that

Δ ABC ≅ Δ A’BC’

By CPCT (Corresponding Parts of Congruent Triangles), we can say that,

From (1) we have

Hence justified.