# Please do for me no.79 n write steps also..

To draw a parallelogram ABCD in which BC = 5 cm and AB = 3 cm and angle abc= 60 degree, follow the following steps:

(1) Draw a line segment AB = 3 cm
(2) At B, constrcut angle ABM = 60  degree.
(3) From BM, cut-off a line segment BC = 5 cm
(4) Now through the point A, draw AN||BC
(5) Through the point C, draw CD || BA, meeting AN at D.
Thus ABCD is the required parallelogram.
(6) Through diagonal BD, divide the parallelogram ABCD into two triangles BCD and ABD.
To construct the triangle BD'C' similar to triangle BDC with scale factor 4:3, we have

(1) Below BD make an acute angle DBX.
(2) Along BX, mark four points B1, B2, B3, and B4 such that BB1=B1B2=B2B3=B3B4
(3) Join B3D
(4) From B4 , draw B4D' || B3D meeting BD produced at D'
(5) From D', draw C'D' || CD meeting BD produced at C'.
Thus, BC'D' is the required triangle whose sides are 4/3 times the corresponding side of triangle BCD.
Now, Draw thw line segment D'A' parallel to DA, where A' lies on the extended side BA.
Justification:
Since AB || CD and CD || C'D'
Therefore, AB|| C'D' $⇒$AB' || C'D'
Similarly,
BC|| DA and DA |||D'A'
So, BC || D'A' $⇒$BC' || D'A'
Thus, BA'C'D' is a parallelogram.

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