A rectangle ABCD is drawn such that AB > BC. Two points E and F are taken on AB and CD, respectively, such that BEDF forms a rhombus. If AB = 10cm and BC = 5cm, then find the area (in sq. cm) of the rhombus BEDF.

Let the side of the rhombus be x, and let the length AE on side AB be 10 - x. Because BEDF is a rhombus, the opposite side is parallel, meaning triangle AED is an isosceles right triangle where ED = x. Using the Pythagorean theorem in triangle AED, x^2 = (10 - x)^2 + 5^2, which expands to x^2 = 100 - 20x + x^2 + 25. Solving 20x = 125 gives x = 6.25 cm. The area of the rhombus is base times height, so 6.25 * 5 = 31.25 sq cm.

• Thanks
• Wrong explanation