CBSE - Mathematics - Herons Formula

Solution

∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
BD is joined. 

In ΔBCD,
By applying Pythagoras theorem,
BD² = BC² + CD²  
⇒ BD² = 12² + 5² 
⇒ BD² = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m²
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD  = √s (s-a) (s-b) (s-c)
                                       = √15(15 - 13) (15 - 9) (15 - 8) m²
                                       = √15 × 2 × 6 × 7 m²
                                       = 6√35 m² = 35.5 m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m² + 35.5m² = 65.5m²