Question

A field is in a shape of a quadrilateral ABCD, in which AB= 18m, AD= 24m, BC= 40m, DC= 50m and angle A= 90 degree. Find the area of the field.

Answer 1

Since ∠ = ° A 90
 ∴ By Pythagoras theorem,
In ∆ ABD,
 BD² = AB² + AD²
BD=√ AB ²+AD²=√18²+24²
= √ 324+ 576 =√900 = 30
Area of Δ ABD=1/2 (18) (24) =(18) × (12) = 216 m²
In ∆ BCD, sides are 30, 40, 50
 ⇒ By Pythagoras theorem ∠CBD =90 °
∴ DC² = BD² + BC ² ,
∴ (50)²= (30)²+ (40)²
Thus, area of Δ BCD =1/2(40) (30) 600 m ²
Hence, area of quadrilateral ABCD = Area of ∆ ABD + area of ∆ BCD = 216 + 600 = 816 m²

Answer 2

see, draw diagonal BD which divide the quad ABCD into two triangle, ΔABD and ΔBDC
given,∠A =90° so area of ΔADB = 1/2 ×base ×height
                                                    =1/2×18×24
                                                    =216 m²
 now using pythagoras' formula, as ΔABD is right anged triangle,
BD= √((24)²+ (18)²) =√900 = 30 m
since , BD is 30 m then ΔDBC become right angled triangled, ∠B= 90°
so, area ΔBDC = 1/2×base ×height
                          =1/2×30×40= 600 m²
area of quad ABCD = area ΔABD + area of ΔBDC
                                 = 216 +600= 816 m²