Let AD = BC = x & AB = CD = y, then Area of rectangle = xy From ΔABC AC² = AB² + BC² ⇒ AC² = y²+x² ⇒ AC = √(x²+y²)................................e...

Let AD = BC = x & AB = CD = y, then Area of rectangle = xy From ΔABC AC² = AB² + BC² ⇒ AC² = y²+x² ⇒ AC = √(x²+y²)................................eq(3) From ΔABC & AMU ∠ABC = ∠AMU = 90° ∠CAB = ∠MAU {common angles of ΔABC & AMU} ⇒ ∠ACB = ∠AUM ⇒ ΔABC & AMU are similar triangles that is, BC/MU = AB/AM = AC/AU ⇒ x/21 = AB/AM = AC/(y-21) ⇒ x/21 = AC/(y-21) ⇒ x(y-21) = 21AC ⇒ xy-21x = 21AC ⇒ 21 = xy/(x+AC)...............................eq(2) From ΔACD & NCV ∠ADC = ∠NVC = 90° ∠ACD = ∠NCV {common angles of ΔACD & NCV} ⇒ ∠CAD = ∠CNV ⇒ ΔACD & ΔMCV are similar triangles that is, AD/VN = AC/NC = CD/VCxQ ⇒ x/24 = AC/NC = y/(y-24) ⇒ x/24 = y/(y-24) ⇒ x(y-24) = 24y ⇒ xy - 24x = 24y ⇒ 24x+24y = xy ⇒ xy = 24x+24y.......................................eq(1) x = ¾y ⇒ x = 42 cm Area of rectangle = xy ⇒ xy = 42×56 ⇒ xy = 2352