Question

Question 3
Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Answer 1

BC = 48m, CD = 17m, AD = 40m and perimeter = 120 m
Perimeter of trapezium ABCD = AB + BC + CD + DA
$\Rightarrow$ 120 = AB + 48 + 17 + 40
$\Rightarrow$ 120 = AB + 105
$\Rightarrow$ AB = 120 - 105 = 15m
Area of a trapezium is $\frac{1}{2}\times (a+b)\times h$
Now Area of the field = $\frac{1}{2}\times (BC+AD)\times AB$
= $\frac{1}{2}\times (48+40)\times 15=\frac{1}{2}\times 88\times 15$
= $660m^2$
Hence are of the field ABCD is = $660m^2$