AREAS
11.1 – 1 D is midpoint of AC, let’s draw a perpendicular on AB from D to E, It will be parallel to BC and half of BC, so DE = ½ x 6.5 =3.25 cm Since angle AED = 90 0 , AB is base and ED is height of ΔABD base = AB = 12 cm, height = DE = 3.25 cm area(ΔABD) = ½ x 3.25 x 12 = 19.5 cm 2 11.1 – 2 In ΔPQS, angle P = 90 0 , so, base = 12 cm, height = 9 cm area(ΔPQS) = ½ x 12 x 9 = 54 cm 2 In ΔQRS, QS 2 + 8 2 = 17 2 , so, QS = 15 cm angle SQR = 90 0 , so, base = 8 cm, height = 15 cm area(ΔQRS) = ½ x 8 x 15 = 60 cm 2 area(PQRS) = 54+60 = 114 cm 2 11.1 – 3 In rectangle ADCE, length = 8 cm, breadth = 3 cm area(ADCE) = 8*3 = 24 cm 2 In ΔBEC, angle BEC = 90 0 , so, base = 8 cm, height = 3 cm area(ΔBEC) = ½ x 8 x 3 = 12 cm 2 Hence, area(ABCD) = 24+12 = 36 cm 2 11.1 – 4 AB || CD, AB = CD, AD = BC Since diagonals bisect, AO = OC and DO = OB Therefore ΔAOD and ΔBOC are congruent (SSS congruency) Hence, area(ΔAOD) ...