# Construct a

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.

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