Draw a rectangle of width 5 centimetres and height 3 centimetres.

i) Draw a rectangle of the same area with width 6 centimetres.

ii) Draw a square of the same area.

Steps of construction:

1. Draw a line BC = 5 cm, taking B and C as centres, draw

∠ABC = ∠BCD = 90°, such that AB = CD = 3 cm

2. Join AD to get rectangle ABCD.

(i) Steps of construction:

1. In the previous rectangle, Extend DC by 6 cm to DC' and and extend BC by 3 cm to BC''.

2. Join C''C' and BC'

3. Draw the perpendicular bisector of C''C' and BC', they intersect each other at O.

4. Taking O as centre and OC' as radius draw a circle which intersect the previous rectangle at D'.

5. Now, taking CD' as height and CA'=CC'' as length, draw a rectangle A'B'D'C, which is required here.

Concept used: If the chords of a circle intersect each other within a circle, then the areas of rectangle formed by the parts of the same chord are equal.

And in the diagram, we made

BC' and DC" are two chord,

Therefore

Area(ABCD) = Area(A'B'D'C) and

CA' = CC'' = 6 cm

Therefore, all the required conditions are verified here.

(ii) Steps of construction:

1. From the rectangle drawn, extend AD by 3 cm to D'.

2. From the mid–point of AD'(Let it be O such that OA = OD'=4 cm) Draw a circle taking OA = OD' as radius.

3. Now, draw a chord PQ ⊥ AD', passing through D.

4. Taking DP as side, draw a square DPRS, which is required.

Concept used:

The area of rectangle form of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area made by the square of half the chord.

And in the diagram, we made, AD' is diameter and PQ is perpendicular chord, therefore

Area(ABCD) = ar(DPRS), and DPRS is a square

Hence, all the required conditions are verified here.