Question

2. Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and
ABC = 60, divide it into triangles BCD and ABD by the diagonal BD.Construct the triangle BD ' C ' similar to ΔBDC with scale factor 4 3. Draw theline segment D 'A ' parallel to DA where A ' lies on extended side BA. Is A 'BC 'D 'a parallelogram?

Answer 1

Solution:-

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

(1) Firstly draw a line 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.

Explanation:-

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, A'BC'D' is a parallelogram.


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@GauravSaxena01

Answer 2

Step-by-step explanation:

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

1. Draw a line segment AB = 3 cm

2. At B, construct angle ABM = 60 degree

3. From Bm, cut a line segment BC = 5 cm

4. Now, from point A, draw AN || BC

5. From the point C, draw CD || BA, meeting AN at D

So, ABCD is the required parallelogram.

Now, from diagonal BD, divide the parallelogram ABCD into two triangles BCD and ABD

To construct the triangle BD1 C1 similar to triangle BDC with scale factor 4 : 3, we do the following steps:

1. Below BD, make an acute angle DBX

2. Along BX, make four points B1 , B2 ,B3 , B4 such that BB1 , B1 B2 , B2 B3 , B3 B4

3. Join B3 D

4. From B4 , draw B3 D1 || B3 D meeting BD produced at D1

5. From D1 , draw C1 D1 || CD meeting BD produced at C1

Thus B D1 C1 is the required triangle whose sides are 4/3 times the corresponding side of triangle BDC.