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  • Draw a triangle ABC with side BC = 7 cm,  angle B = 45°,   angle A = 105°. Then, construct a       triangle whose sides are 4/3  times the corresponding sides of D ABC.

6.   Draw a triangle ABC with side BC = 7 cm, \angle B = 45°,  \angle A = 105°. Then, construct a triangle whose sides are 4/3  times the corresponding sides of ABC.

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Steps of construction:- 

(i) Draw a line segment BC.

(ii) Now draw an angle \angle B  =  45o   and  \angle  C  =  30o  and draw rays in these directions.

(iii) Name the intersection of these lines as A.

(iv) Thus \bigtriangleup ABCis the required triangle.

(v) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(vi) Cut four equal parts of line BX namely  BB1, BB2, BB3, and BB4.

(vii) Now join B3 to C. Draw a line B4C' parallel to B3C.

(viii) And then draw a line A'C' parallel to AC.

Hence \Delta A'BC' is the required triangle.

 

Posted by

Devendra Khairwa

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