Answer :

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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