Coloured papers, sketch pens, geometry box, a pair of scissors, fevicol and eraser.
1.Draw □ ABCD with length d2 and breadth d1 units on a coloured paper.
2. Mark points E, F, G and H as midpoints of the sides AD, DC, CB and BA respectively of sides of the □ ABCD.
3. Join HF and EG. Mark their intersection as point O. Fold the rectangle ABCD along EG and HF dividing the □ ABCD into four congruent rectangles, namely OEAH, OEDF, OFCG and OGBH.
4. Divide each of the four rectangles into two congruent triangles by drawing their respective diagonals.
1. As the small rectangles are congruent, their diagonals EH, HG, GF, FE are equal. Thus EHGF is a rhombus.
2. In the rectangle AHOE, triangles AHE and EHO are congruent and hence equal in area.
3. Thus area of the right triangle EOH is half the area of the rectangle AEOH and similarly, the area of right triangles HOG, GOF, FOE are half the area of the rectangles HBGO, OGCF and FOED respectively.
4. Thus the area of rhombus EFGH = (1/2 X Area of rectangle AEOH) + (1/2 X Area of rectangle HBGO) + (1/2 X Area of rectangle OGCF) + (1/2 X Area of rectangle FOED)
= 1/2 X ( Area of rectangle AEOH + Area of rectangle HBGO + Area of rectangle OGCF + Area of rectangle FOED)
= 1/2 x Area of rectangle ABCD
= 1/2 x d1 x d2
= 1/2 X HF X EG
= 1/2 x product of diagonals of rhombus EFGH
Area of rhombus is half the product of its diagonals.