Answer 3. Steps of Construction:

1. Draw a line segment AB = 5 cm

2. Draw ∠ABX = 60 degree and ∠ABY = 90 degree at B

3. With B as center and radius 3 cm, mark an arc intersecting the ray BY in Z

4. Draw a line passing through Z and parallel to AB. Let it intersect the ray BX in C

5. Through C, draw an altitude to AB intersecting in D. Here CD = BF = 3 cm

6. Join AC

Thus ΔABC is the required traingle.

Answer 4.

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.

Now, draw the line segment D₁ A₁ parallel to DA where A₁ lies on the extended side BA

Justification:

Since AB || CD and CD || C₁ D₁ 

=> AB || C₁ D₁  

=> AB₁ || C₁ D₁ 

Similarly

Since BC || DA and DA || D₁ A₁ 

=> BC || DA  

=> BC₁ || D₁ A₁

Hence, A₁ B₁ C₁ D₁ is a parallelogram

NOTE: Please DO NOT ask multiple parts in one questions.

Please ask the remaining questions as separate questions.

Thanks!